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ai-content-maker/.venv/Lib/site-packages/scipy/stats/_morestats.py

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from __future__ import annotations
import math
import warnings
from collections import namedtuple
import numpy as np
from numpy import (isscalar, r_, log, around, unique, asarray, zeros,
arange, sort, amin, amax, sqrt, array, atleast_1d, # noqa: F401
compress, pi, exp, ravel, count_nonzero, sin, cos, # noqa: F401
arctan2, hypot)
from scipy import optimize, special, interpolate, stats
from scipy._lib._bunch import _make_tuple_bunch
from scipy._lib._util import _rename_parameter, _contains_nan, _get_nan
from ._ansari_swilk_statistics import gscale, swilk
from . import _stats_py, _wilcoxon
from ._fit import FitResult
from ._stats_py import find_repeats, _get_pvalue, SignificanceResult # noqa: F401
from .contingency import chi2_contingency
from . import distributions
from ._distn_infrastructure import rv_generic
from ._axis_nan_policy import _axis_nan_policy_factory
__all__ = ['mvsdist',
'bayes_mvs', 'kstat', 'kstatvar', 'probplot', 'ppcc_max', 'ppcc_plot',
'boxcox_llf', 'boxcox', 'boxcox_normmax', 'boxcox_normplot',
'shapiro', 'anderson', 'ansari', 'bartlett', 'levene',
'fligner', 'mood', 'wilcoxon', 'median_test',
'circmean', 'circvar', 'circstd', 'anderson_ksamp',
'yeojohnson_llf', 'yeojohnson', 'yeojohnson_normmax',
'yeojohnson_normplot', 'directional_stats',
'false_discovery_control'
]
Mean = namedtuple('Mean', ('statistic', 'minmax'))
Variance = namedtuple('Variance', ('statistic', 'minmax'))
Std_dev = namedtuple('Std_dev', ('statistic', 'minmax'))
def bayes_mvs(data, alpha=0.90):
r"""
Bayesian confidence intervals for the mean, var, and std.
Parameters
----------
data : array_like
Input data, if multi-dimensional it is flattened to 1-D by `bayes_mvs`.
Requires 2 or more data points.
alpha : float, optional
Probability that the returned confidence interval contains
the true parameter.
Returns
-------
mean_cntr, var_cntr, std_cntr : tuple
The three results are for the mean, variance and standard deviation,
respectively. Each result is a tuple of the form::
(center, (lower, upper))
with `center` the mean of the conditional pdf of the value given the
data, and `(lower, upper)` a confidence interval, centered on the
median, containing the estimate to a probability ``alpha``.
See Also
--------
mvsdist
Notes
-----
Each tuple of mean, variance, and standard deviation estimates represent
the (center, (lower, upper)) with center the mean of the conditional pdf
of the value given the data and (lower, upper) is a confidence interval
centered on the median, containing the estimate to a probability
``alpha``.
Converts data to 1-D and assumes all data has the same mean and variance.
Uses Jeffrey's prior for variance and std.
Equivalent to ``tuple((x.mean(), x.interval(alpha)) for x in mvsdist(dat))``
References
----------
T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and
standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278,
2006.
Examples
--------
First a basic example to demonstrate the outputs:
>>> from scipy import stats
>>> data = [6, 9, 12, 7, 8, 8, 13]
>>> mean, var, std = stats.bayes_mvs(data)
>>> mean
Mean(statistic=9.0, minmax=(7.103650222612533, 10.896349777387467))
>>> var
Variance(statistic=10.0, minmax=(3.176724206..., 24.45910382...))
>>> std
Std_dev(statistic=2.9724954732045084,
minmax=(1.7823367265645143, 4.945614605014631))
Now we generate some normally distributed random data, and get estimates of
mean and standard deviation with 95% confidence intervals for those
estimates:
>>> n_samples = 100000
>>> data = stats.norm.rvs(size=n_samples)
>>> res_mean, res_var, res_std = stats.bayes_mvs(data, alpha=0.95)
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.hist(data, bins=100, density=True, label='Histogram of data')
>>> ax.vlines(res_mean.statistic, 0, 0.5, colors='r', label='Estimated mean')
>>> ax.axvspan(res_mean.minmax[0],res_mean.minmax[1], facecolor='r',
... alpha=0.2, label=r'Estimated mean (95% limits)')
>>> ax.vlines(res_std.statistic, 0, 0.5, colors='g', label='Estimated scale')
>>> ax.axvspan(res_std.minmax[0],res_std.minmax[1], facecolor='g', alpha=0.2,
... label=r'Estimated scale (95% limits)')
>>> ax.legend(fontsize=10)
>>> ax.set_xlim([-4, 4])
>>> ax.set_ylim([0, 0.5])
>>> plt.show()
"""
m, v, s = mvsdist(data)
if alpha >= 1 or alpha <= 0:
raise ValueError("0 < alpha < 1 is required, but alpha=%s was given."
% alpha)
m_res = Mean(m.mean(), m.interval(alpha))
v_res = Variance(v.mean(), v.interval(alpha))
s_res = Std_dev(s.mean(), s.interval(alpha))
return m_res, v_res, s_res
def mvsdist(data):
"""
'Frozen' distributions for mean, variance, and standard deviation of data.
Parameters
----------
data : array_like
Input array. Converted to 1-D using ravel.
Requires 2 or more data-points.
Returns
-------
mdist : "frozen" distribution object
Distribution object representing the mean of the data.
vdist : "frozen" distribution object
Distribution object representing the variance of the data.
sdist : "frozen" distribution object
Distribution object representing the standard deviation of the data.
See Also
--------
bayes_mvs
Notes
-----
The return values from ``bayes_mvs(data)`` is equivalent to
``tuple((x.mean(), x.interval(0.90)) for x in mvsdist(data))``.
In other words, calling ``<dist>.mean()`` and ``<dist>.interval(0.90)``
on the three distribution objects returned from this function will give
the same results that are returned from `bayes_mvs`.
References
----------
T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and
standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278,
2006.
Examples
--------
>>> from scipy import stats
>>> data = [6, 9, 12, 7, 8, 8, 13]
>>> mean, var, std = stats.mvsdist(data)
We now have frozen distribution objects "mean", "var" and "std" that we can
examine:
>>> mean.mean()
9.0
>>> mean.interval(0.95)
(6.6120585482655692, 11.387941451734431)
>>> mean.std()
1.1952286093343936
"""
x = ravel(data)
n = len(x)
if n < 2:
raise ValueError("Need at least 2 data-points.")
xbar = x.mean()
C = x.var()
if n > 1000: # gaussian approximations for large n
mdist = distributions.norm(loc=xbar, scale=math.sqrt(C / n))
sdist = distributions.norm(loc=math.sqrt(C), scale=math.sqrt(C / (2. * n)))
vdist = distributions.norm(loc=C, scale=math.sqrt(2.0 / n) * C)
else:
nm1 = n - 1
fac = n * C / 2.
val = nm1 / 2.
mdist = distributions.t(nm1, loc=xbar, scale=math.sqrt(C / nm1))
sdist = distributions.gengamma(val, -2, scale=math.sqrt(fac))
vdist = distributions.invgamma(val, scale=fac)
return mdist, vdist, sdist
@_axis_nan_policy_factory(
lambda x: x, result_to_tuple=lambda x: (x,), n_outputs=1, default_axis=None
)
def kstat(data, n=2):
r"""
Return the nth k-statistic (1<=n<=4 so far).
The nth k-statistic k_n is the unique symmetric unbiased estimator of the
nth cumulant kappa_n.
Parameters
----------
data : array_like
Input array. Note that n-D input gets flattened.
n : int, {1, 2, 3, 4}, optional
Default is equal to 2.
Returns
-------
kstat : float
The nth k-statistic.
See Also
--------
kstatvar : Returns an unbiased estimator of the variance of the k-statistic
moment : Returns the n-th central moment about the mean for a sample.
Notes
-----
For a sample size n, the first few k-statistics are given by:
.. math::
k_{1} = \mu
k_{2} = \frac{n}{n-1} m_{2}
k_{3} = \frac{ n^{2} } {(n-1) (n-2)} m_{3}
k_{4} = \frac{ n^{2} [(n + 1)m_{4} - 3(n - 1) m^2_{2}]} {(n-1) (n-2) (n-3)}
where :math:`\mu` is the sample mean, :math:`m_2` is the sample
variance, and :math:`m_i` is the i-th sample central moment.
References
----------
http://mathworld.wolfram.com/k-Statistic.html
http://mathworld.wolfram.com/Cumulant.html
Examples
--------
>>> from scipy import stats
>>> from numpy.random import default_rng
>>> rng = default_rng()
As sample size increases, n-th moment and n-th k-statistic converge to the
same number (although they aren't identical). In the case of the normal
distribution, they converge to zero.
>>> for n in [2, 3, 4, 5, 6, 7]:
... x = rng.normal(size=10**n)
... m, k = stats.moment(x, 3), stats.kstat(x, 3)
... print("%.3g %.3g %.3g" % (m, k, m-k))
-0.631 -0.651 0.0194 # random
0.0282 0.0283 -8.49e-05
-0.0454 -0.0454 1.36e-05
7.53e-05 7.53e-05 -2.26e-09
0.00166 0.00166 -4.99e-09
-2.88e-06 -2.88e-06 8.63e-13
"""
if n > 4 or n < 1:
raise ValueError("k-statistics only supported for 1<=n<=4")
n = int(n)
S = np.zeros(n + 1, np.float64)
data = ravel(data)
N = data.size
# raise ValueError on empty input
if N == 0:
raise ValueError("Data input must not be empty")
# on nan input, return nan without warning
if np.isnan(np.sum(data)):
return np.nan
for k in range(1, n + 1):
S[k] = np.sum(data**k, axis=0)
if n == 1:
return S[1] * 1.0/N
elif n == 2:
return (N*S[2] - S[1]**2.0) / (N*(N - 1.0))
elif n == 3:
return (2*S[1]**3 - 3*N*S[1]*S[2] + N*N*S[3]) / (N*(N - 1.0)*(N - 2.0))
elif n == 4:
return ((-6*S[1]**4 + 12*N*S[1]**2 * S[2] - 3*N*(N-1.0)*S[2]**2 -
4*N*(N+1)*S[1]*S[3] + N*N*(N+1)*S[4]) /
(N*(N-1.0)*(N-2.0)*(N-3.0)))
else:
raise ValueError("Should not be here.")
@_axis_nan_policy_factory(
lambda x: x, result_to_tuple=lambda x: (x,), n_outputs=1, default_axis=None
)
def kstatvar(data, n=2):
r"""Return an unbiased estimator of the variance of the k-statistic.
See `kstat` for more details of the k-statistic.
Parameters
----------
data : array_like
Input array. Note that n-D input gets flattened.
n : int, {1, 2}, optional
Default is equal to 2.
Returns
-------
kstatvar : float
The nth k-statistic variance.
See Also
--------
kstat : Returns the n-th k-statistic.
moment : Returns the n-th central moment about the mean for a sample.
Notes
-----
The variances of the first few k-statistics are given by:
.. math::
var(k_{1}) = \frac{\kappa^2}{n}
var(k_{2}) = \frac{\kappa^4}{n} + \frac{2\kappa^2_{2}}{n - 1}
var(k_{3}) = \frac{\kappa^6}{n} + \frac{9 \kappa_2 \kappa_4}{n - 1} +
\frac{9 \kappa^2_{3}}{n - 1} +
\frac{6 n \kappa^3_{2}}{(n-1) (n-2)}
var(k_{4}) = \frac{\kappa^8}{n} + \frac{16 \kappa_2 \kappa_6}{n - 1} +
\frac{48 \kappa_{3} \kappa_5}{n - 1} +
\frac{34 \kappa^2_{4}}{n-1} +
\frac{72 n \kappa^2_{2} \kappa_4}{(n - 1) (n - 2)} +
\frac{144 n \kappa_{2} \kappa^2_{3}}{(n - 1) (n - 2)} +
\frac{24 (n + 1) n \kappa^4_{2}}{(n - 1) (n - 2) (n - 3)}
""" # noqa: E501
data = ravel(data)
N = len(data)
if n == 1:
return kstat(data, n=2) * 1.0/N
elif n == 2:
k2 = kstat(data, n=2)
k4 = kstat(data, n=4)
return (2*N*k2**2 + (N-1)*k4) / (N*(N+1))
else:
raise ValueError("Only n=1 or n=2 supported.")
def _calc_uniform_order_statistic_medians(n):
"""Approximations of uniform order statistic medians.
Parameters
----------
n : int
Sample size.
Returns
-------
v : 1d float array
Approximations of the order statistic medians.
References
----------
.. [1] James J. Filliben, "The Probability Plot Correlation Coefficient
Test for Normality", Technometrics, Vol. 17, pp. 111-117, 1975.
Examples
--------
Order statistics of the uniform distribution on the unit interval
are marginally distributed according to beta distributions.
The expectations of these order statistic are evenly spaced across
the interval, but the distributions are skewed in a way that
pushes the medians slightly towards the endpoints of the unit interval:
>>> import numpy as np
>>> n = 4
>>> k = np.arange(1, n+1)
>>> from scipy.stats import beta
>>> a = k
>>> b = n-k+1
>>> beta.mean(a, b)
array([0.2, 0.4, 0.6, 0.8])
>>> beta.median(a, b)
array([0.15910358, 0.38572757, 0.61427243, 0.84089642])
The Filliben approximation uses the exact medians of the smallest
and greatest order statistics, and the remaining medians are approximated
by points spread evenly across a sub-interval of the unit interval:
>>> from scipy.stats._morestats import _calc_uniform_order_statistic_medians
>>> _calc_uniform_order_statistic_medians(n)
array([0.15910358, 0.38545246, 0.61454754, 0.84089642])
This plot shows the skewed distributions of the order statistics
of a sample of size four from a uniform distribution on the unit interval:
>>> import matplotlib.pyplot as plt
>>> x = np.linspace(0.0, 1.0, num=50, endpoint=True)
>>> pdfs = [beta.pdf(x, a[i], b[i]) for i in range(n)]
>>> plt.figure()
>>> plt.plot(x, pdfs[0], x, pdfs[1], x, pdfs[2], x, pdfs[3])
"""
v = np.empty(n, dtype=np.float64)
v[-1] = 0.5**(1.0 / n)
v[0] = 1 - v[-1]
i = np.arange(2, n)
v[1:-1] = (i - 0.3175) / (n + 0.365)
return v
def _parse_dist_kw(dist, enforce_subclass=True):
"""Parse `dist` keyword.
Parameters
----------
dist : str or stats.distributions instance.
Several functions take `dist` as a keyword, hence this utility
function.
enforce_subclass : bool, optional
If True (default), `dist` needs to be a
`_distn_infrastructure.rv_generic` instance.
It can sometimes be useful to set this keyword to False, if a function
wants to accept objects that just look somewhat like such an instance
(for example, they have a ``ppf`` method).
"""
if isinstance(dist, rv_generic):
pass
elif isinstance(dist, str):
try:
dist = getattr(distributions, dist)
except AttributeError as e:
raise ValueError("%s is not a valid distribution name" % dist) from e
elif enforce_subclass:
msg = ("`dist` should be a stats.distributions instance or a string "
"with the name of such a distribution.")
raise ValueError(msg)
return dist
def _add_axis_labels_title(plot, xlabel, ylabel, title):
"""Helper function to add axes labels and a title to stats plots."""
try:
if hasattr(plot, 'set_title'):
# Matplotlib Axes instance or something that looks like it
plot.set_title(title)
plot.set_xlabel(xlabel)
plot.set_ylabel(ylabel)
else:
# matplotlib.pyplot module
plot.title(title)
plot.xlabel(xlabel)
plot.ylabel(ylabel)
except Exception:
# Not an MPL object or something that looks (enough) like it.
# Don't crash on adding labels or title
pass
def probplot(x, sparams=(), dist='norm', fit=True, plot=None, rvalue=False):
"""
Calculate quantiles for a probability plot, and optionally show the plot.
Generates a probability plot of sample data against the quantiles of a
specified theoretical distribution (the normal distribution by default).
`probplot` optionally calculates a best-fit line for the data and plots the
results using Matplotlib or a given plot function.
Parameters
----------
x : array_like
Sample/response data from which `probplot` creates the plot.
sparams : tuple, optional
Distribution-specific shape parameters (shape parameters plus location
and scale).
dist : str or stats.distributions instance, optional
Distribution or distribution function name. The default is 'norm' for a
normal probability plot. Objects that look enough like a
stats.distributions instance (i.e. they have a ``ppf`` method) are also
accepted.
fit : bool, optional
Fit a least-squares regression (best-fit) line to the sample data if
True (default).
plot : object, optional
If given, plots the quantiles.
If given and `fit` is True, also plots the least squares fit.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
rvalue : bool, optional
If `plot` is provided and `fit` is True, setting `rvalue` to True
includes the coefficient of determination on the plot.
Default is False.
Returns
-------
(osm, osr) : tuple of ndarrays
Tuple of theoretical quantiles (osm, or order statistic medians) and
ordered responses (osr). `osr` is simply sorted input `x`.
For details on how `osm` is calculated see the Notes section.
(slope, intercept, r) : tuple of floats, optional
Tuple containing the result of the least-squares fit, if that is
performed by `probplot`. `r` is the square root of the coefficient of
determination. If ``fit=False`` and ``plot=None``, this tuple is not
returned.
Notes
-----
Even if `plot` is given, the figure is not shown or saved by `probplot`;
``plt.show()`` or ``plt.savefig('figname.png')`` should be used after
calling `probplot`.
`probplot` generates a probability plot, which should not be confused with
a Q-Q or a P-P plot. Statsmodels has more extensive functionality of this
type, see ``statsmodels.api.ProbPlot``.
The formula used for the theoretical quantiles (horizontal axis of the
probability plot) is Filliben's estimate::
quantiles = dist.ppf(val), for
0.5**(1/n), for i = n
val = (i - 0.3175) / (n + 0.365), for i = 2, ..., n-1
1 - 0.5**(1/n), for i = 1
where ``i`` indicates the i-th ordered value and ``n`` is the total number
of values.
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> nsample = 100
>>> rng = np.random.default_rng()
A t distribution with small degrees of freedom:
>>> ax1 = plt.subplot(221)
>>> x = stats.t.rvs(3, size=nsample, random_state=rng)
>>> res = stats.probplot(x, plot=plt)
A t distribution with larger degrees of freedom:
>>> ax2 = plt.subplot(222)
>>> x = stats.t.rvs(25, size=nsample, random_state=rng)
>>> res = stats.probplot(x, plot=plt)
A mixture of two normal distributions with broadcasting:
>>> ax3 = plt.subplot(223)
>>> x = stats.norm.rvs(loc=[0,5], scale=[1,1.5],
... size=(nsample//2,2), random_state=rng).ravel()
>>> res = stats.probplot(x, plot=plt)
A standard normal distribution:
>>> ax4 = plt.subplot(224)
>>> x = stats.norm.rvs(loc=0, scale=1, size=nsample, random_state=rng)
>>> res = stats.probplot(x, plot=plt)
Produce a new figure with a loggamma distribution, using the ``dist`` and
``sparams`` keywords:
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> x = stats.loggamma.rvs(c=2.5, size=500, random_state=rng)
>>> res = stats.probplot(x, dist=stats.loggamma, sparams=(2.5,), plot=ax)
>>> ax.set_title("Probplot for loggamma dist with shape parameter 2.5")
Show the results with Matplotlib:
>>> plt.show()
"""
x = np.asarray(x)
if x.size == 0:
if fit:
return (x, x), (np.nan, np.nan, 0.0)
else:
return x, x
osm_uniform = _calc_uniform_order_statistic_medians(len(x))
dist = _parse_dist_kw(dist, enforce_subclass=False)
if sparams is None:
sparams = ()
if isscalar(sparams):
sparams = (sparams,)
if not isinstance(sparams, tuple):
sparams = tuple(sparams)
osm = dist.ppf(osm_uniform, *sparams)
osr = sort(x)
if fit:
# perform a linear least squares fit.
slope, intercept, r, prob, _ = _stats_py.linregress(osm, osr)
if plot is not None:
plot.plot(osm, osr, 'bo')
if fit:
plot.plot(osm, slope*osm + intercept, 'r-')
_add_axis_labels_title(plot, xlabel='Theoretical quantiles',
ylabel='Ordered Values',
title='Probability Plot')
# Add R^2 value to the plot as text
if fit and rvalue:
xmin = amin(osm)
xmax = amax(osm)
ymin = amin(x)
ymax = amax(x)
posx = xmin + 0.70 * (xmax - xmin)
posy = ymin + 0.01 * (ymax - ymin)
plot.text(posx, posy, "$R^2=%1.4f$" % r**2)
if fit:
return (osm, osr), (slope, intercept, r)
else:
return osm, osr
def ppcc_max(x, brack=(0.0, 1.0), dist='tukeylambda'):
"""Calculate the shape parameter that maximizes the PPCC.
The probability plot correlation coefficient (PPCC) plot can be used
to determine the optimal shape parameter for a one-parameter family
of distributions. ``ppcc_max`` returns the shape parameter that would
maximize the probability plot correlation coefficient for the given
data to a one-parameter family of distributions.
Parameters
----------
x : array_like
Input array.
brack : tuple, optional
Triple (a,b,c) where (a<b<c). If bracket consists of two numbers (a, c)
then they are assumed to be a starting interval for a downhill bracket
search (see `scipy.optimize.brent`).
dist : str or stats.distributions instance, optional
Distribution or distribution function name. Objects that look enough
like a stats.distributions instance (i.e. they have a ``ppf`` method)
are also accepted. The default is ``'tukeylambda'``.
Returns
-------
shape_value : float
The shape parameter at which the probability plot correlation
coefficient reaches its max value.
See Also
--------
ppcc_plot, probplot, boxcox
Notes
-----
The brack keyword serves as a starting point which is useful in corner
cases. One can use a plot to obtain a rough visual estimate of the location
for the maximum to start the search near it.
References
----------
.. [1] J.J. Filliben, "The Probability Plot Correlation Coefficient Test
for Normality", Technometrics, Vol. 17, pp. 111-117, 1975.
.. [2] Engineering Statistics Handbook, NIST/SEMATEC,
https://www.itl.nist.gov/div898/handbook/eda/section3/ppccplot.htm
Examples
--------
First we generate some random data from a Weibull distribution
with shape parameter 2.5:
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> rng = np.random.default_rng()
>>> c = 2.5
>>> x = stats.weibull_min.rvs(c, scale=4, size=2000, random_state=rng)
Generate the PPCC plot for this data with the Weibull distribution.
>>> fig, ax = plt.subplots(figsize=(8, 6))
>>> res = stats.ppcc_plot(x, c/2, 2*c, dist='weibull_min', plot=ax)
We calculate the value where the shape should reach its maximum and a
red line is drawn there. The line should coincide with the highest
point in the PPCC graph.
>>> cmax = stats.ppcc_max(x, brack=(c/2, 2*c), dist='weibull_min')
>>> ax.axvline(cmax, color='r')
>>> plt.show()
"""
dist = _parse_dist_kw(dist)
osm_uniform = _calc_uniform_order_statistic_medians(len(x))
osr = sort(x)
# this function computes the x-axis values of the probability plot
# and computes a linear regression (including the correlation)
# and returns 1-r so that a minimization function maximizes the
# correlation
def tempfunc(shape, mi, yvals, func):
xvals = func(mi, shape)
r, prob = _stats_py.pearsonr(xvals, yvals)
return 1 - r
return optimize.brent(tempfunc, brack=brack,
args=(osm_uniform, osr, dist.ppf))
def ppcc_plot(x, a, b, dist='tukeylambda', plot=None, N=80):
"""Calculate and optionally plot probability plot correlation coefficient.
The probability plot correlation coefficient (PPCC) plot can be used to
determine the optimal shape parameter for a one-parameter family of
distributions. It cannot be used for distributions without shape
parameters
(like the normal distribution) or with multiple shape parameters.
By default a Tukey-Lambda distribution (`stats.tukeylambda`) is used. A
Tukey-Lambda PPCC plot interpolates from long-tailed to short-tailed
distributions via an approximately normal one, and is therefore
particularly useful in practice.
Parameters
----------
x : array_like
Input array.
a, b : scalar
Lower and upper bounds of the shape parameter to use.
dist : str or stats.distributions instance, optional
Distribution or distribution function name. Objects that look enough
like a stats.distributions instance (i.e. they have a ``ppf`` method)
are also accepted. The default is ``'tukeylambda'``.
plot : object, optional
If given, plots PPCC against the shape parameter.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
N : int, optional
Number of points on the horizontal axis (equally distributed from
`a` to `b`).
Returns
-------
svals : ndarray
The shape values for which `ppcc` was calculated.
ppcc : ndarray
The calculated probability plot correlation coefficient values.
See Also
--------
ppcc_max, probplot, boxcox_normplot, tukeylambda
References
----------
J.J. Filliben, "The Probability Plot Correlation Coefficient Test for
Normality", Technometrics, Vol. 17, pp. 111-117, 1975.
Examples
--------
First we generate some random data from a Weibull distribution
with shape parameter 2.5, and plot the histogram of the data:
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> rng = np.random.default_rng()
>>> c = 2.5
>>> x = stats.weibull_min.rvs(c, scale=4, size=2000, random_state=rng)
Take a look at the histogram of the data.
>>> fig1, ax = plt.subplots(figsize=(9, 4))
>>> ax.hist(x, bins=50)
>>> ax.set_title('Histogram of x')
>>> plt.show()
Now we explore this data with a PPCC plot as well as the related
probability plot and Box-Cox normplot. A red line is drawn where we
expect the PPCC value to be maximal (at the shape parameter ``c``
used above):
>>> fig2 = plt.figure(figsize=(12, 4))
>>> ax1 = fig2.add_subplot(1, 3, 1)
>>> ax2 = fig2.add_subplot(1, 3, 2)
>>> ax3 = fig2.add_subplot(1, 3, 3)
>>> res = stats.probplot(x, plot=ax1)
>>> res = stats.boxcox_normplot(x, -4, 4, plot=ax2)
>>> res = stats.ppcc_plot(x, c/2, 2*c, dist='weibull_min', plot=ax3)
>>> ax3.axvline(c, color='r')
>>> plt.show()
"""
if b <= a:
raise ValueError("`b` has to be larger than `a`.")
svals = np.linspace(a, b, num=N)
ppcc = np.empty_like(svals)
for k, sval in enumerate(svals):
_, r2 = probplot(x, sval, dist=dist, fit=True)
ppcc[k] = r2[-1]
if plot is not None:
plot.plot(svals, ppcc, 'x')
_add_axis_labels_title(plot, xlabel='Shape Values',
ylabel='Prob Plot Corr. Coef.',
title='(%s) PPCC Plot' % dist)
return svals, ppcc
def _log_mean(logx):
# compute log of mean of x from log(x)
return special.logsumexp(logx, axis=0) - np.log(len(logx))
def _log_var(logx):
# compute log of variance of x from log(x)
logmean = _log_mean(logx)
pij = np.full_like(logx, np.pi * 1j, dtype=np.complex128)
logxmu = special.logsumexp([logx, logmean + pij], axis=0)
return np.real(special.logsumexp(2 * logxmu, axis=0)) - np.log(len(logx))
def boxcox_llf(lmb, data):
r"""The boxcox log-likelihood function.
Parameters
----------
lmb : scalar
Parameter for Box-Cox transformation. See `boxcox` for details.
data : array_like
Data to calculate Box-Cox log-likelihood for. If `data` is
multi-dimensional, the log-likelihood is calculated along the first
axis.
Returns
-------
llf : float or ndarray
Box-Cox log-likelihood of `data` given `lmb`. A float for 1-D `data`,
an array otherwise.
See Also
--------
boxcox, probplot, boxcox_normplot, boxcox_normmax
Notes
-----
The Box-Cox log-likelihood function is defined here as
.. math::
llf = (\lambda - 1) \sum_i(\log(x_i)) -
N/2 \log(\sum_i (y_i - \bar{y})^2 / N),
where ``y`` is the Box-Cox transformed input data ``x``.
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes
Generate some random variates and calculate Box-Cox log-likelihood values
for them for a range of ``lmbda`` values:
>>> rng = np.random.default_rng()
>>> x = stats.loggamma.rvs(5, loc=10, size=1000, random_state=rng)
>>> lmbdas = np.linspace(-2, 10)
>>> llf = np.zeros(lmbdas.shape, dtype=float)
>>> for ii, lmbda in enumerate(lmbdas):
... llf[ii] = stats.boxcox_llf(lmbda, x)
Also find the optimal lmbda value with `boxcox`:
>>> x_most_normal, lmbda_optimal = stats.boxcox(x)
Plot the log-likelihood as function of lmbda. Add the optimal lmbda as a
horizontal line to check that that's really the optimum:
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.plot(lmbdas, llf, 'b.-')
>>> ax.axhline(stats.boxcox_llf(lmbda_optimal, x), color='r')
>>> ax.set_xlabel('lmbda parameter')
>>> ax.set_ylabel('Box-Cox log-likelihood')
Now add some probability plots to show that where the log-likelihood is
maximized the data transformed with `boxcox` looks closest to normal:
>>> locs = [3, 10, 4] # 'lower left', 'center', 'lower right'
>>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs):
... xt = stats.boxcox(x, lmbda=lmbda)
... (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt)
... ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc)
... ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-')
... ax_inset.set_xticklabels([])
... ax_inset.set_yticklabels([])
... ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda)
>>> plt.show()
"""
data = np.asarray(data)
N = data.shape[0]
if N == 0:
return np.nan
logdata = np.log(data)
# Compute the variance of the transformed data.
if lmb == 0:
logvar = np.log(np.var(logdata, axis=0))
else:
# Transform without the constant offset 1/lmb. The offset does
# not affect the variance, and the subtraction of the offset can
# lead to loss of precision.
# Division by lmb can be factored out to enhance numerical stability.
logx = lmb * logdata
logvar = _log_var(logx) - 2 * np.log(abs(lmb))
return (lmb - 1) * np.sum(logdata, axis=0) - N/2 * logvar
def _boxcox_conf_interval(x, lmax, alpha):
# Need to find the lambda for which
# f(x,lmbda) >= f(x,lmax) - 0.5*chi^2_alpha;1
fac = 0.5 * distributions.chi2.ppf(1 - alpha, 1)
target = boxcox_llf(lmax, x) - fac
def rootfunc(lmbda, data, target):
return boxcox_llf(lmbda, data) - target
# Find positive endpoint of interval in which answer is to be found
newlm = lmax + 0.5
N = 0
while (rootfunc(newlm, x, target) > 0.0) and (N < 500):
newlm += 0.1
N += 1
if N == 500:
raise RuntimeError("Could not find endpoint.")
lmplus = optimize.brentq(rootfunc, lmax, newlm, args=(x, target))
# Now find negative interval in the same way
newlm = lmax - 0.5
N = 0
while (rootfunc(newlm, x, target) > 0.0) and (N < 500):
newlm -= 0.1
N += 1
if N == 500:
raise RuntimeError("Could not find endpoint.")
lmminus = optimize.brentq(rootfunc, newlm, lmax, args=(x, target))
return lmminus, lmplus
def boxcox(x, lmbda=None, alpha=None, optimizer=None):
r"""Return a dataset transformed by a Box-Cox power transformation.
Parameters
----------
x : ndarray
Input array to be transformed.
If `lmbda` is not None, this is an alias of
`scipy.special.boxcox`.
Returns nan if ``x < 0``; returns -inf if ``x == 0 and lmbda < 0``.
If `lmbda` is None, array must be positive, 1-dimensional, and
non-constant.
lmbda : scalar, optional
If `lmbda` is None (default), find the value of `lmbda` that maximizes
the log-likelihood function and return it as the second output
argument.
If `lmbda` is not None, do the transformation for that value.
alpha : float, optional
If `lmbda` is None and `alpha` is not None (default), return the
``100 * (1-alpha)%`` confidence interval for `lmbda` as the third
output argument. Must be between 0.0 and 1.0.
If `lmbda` is not None, `alpha` is ignored.
optimizer : callable, optional
If `lmbda` is None, `optimizer` is the scalar optimizer used to find
the value of `lmbda` that minimizes the negative log-likelihood
function. `optimizer` is a callable that accepts one argument:
fun : callable
The objective function, which evaluates the negative
log-likelihood function at a provided value of `lmbda`
and returns an object, such as an instance of
`scipy.optimize.OptimizeResult`, which holds the optimal value of
`lmbda` in an attribute `x`.
See the example in `boxcox_normmax` or the documentation of
`scipy.optimize.minimize_scalar` for more information.
If `lmbda` is not None, `optimizer` is ignored.
Returns
-------
boxcox : ndarray
Box-Cox power transformed array.
maxlog : float, optional
If the `lmbda` parameter is None, the second returned argument is
the `lmbda` that maximizes the log-likelihood function.
(min_ci, max_ci) : tuple of float, optional
If `lmbda` parameter is None and `alpha` is not None, this returned
tuple of floats represents the minimum and maximum confidence limits
given `alpha`.
See Also
--------
probplot, boxcox_normplot, boxcox_normmax, boxcox_llf
Notes
-----
The Box-Cox transform is given by::
y = (x**lmbda - 1) / lmbda, for lmbda != 0
log(x), for lmbda = 0
`boxcox` requires the input data to be positive. Sometimes a Box-Cox
transformation provides a shift parameter to achieve this; `boxcox` does
not. Such a shift parameter is equivalent to adding a positive constant to
`x` before calling `boxcox`.
The confidence limits returned when `alpha` is provided give the interval
where:
.. math::
llf(\hat{\lambda}) - llf(\lambda) < \frac{1}{2}\chi^2(1 - \alpha, 1),
with ``llf`` the log-likelihood function and :math:`\chi^2` the chi-squared
function.
References
----------
G.E.P. Box and D.R. Cox, "An Analysis of Transformations", Journal of the
Royal Statistical Society B, 26, 211-252 (1964).
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
We generate some random variates from a non-normal distribution and make a
probability plot for it, to show it is non-normal in the tails:
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(211)
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> prob = stats.probplot(x, dist=stats.norm, plot=ax1)
>>> ax1.set_xlabel('')
>>> ax1.set_title('Probplot against normal distribution')
We now use `boxcox` to transform the data so it's closest to normal:
>>> ax2 = fig.add_subplot(212)
>>> xt, _ = stats.boxcox(x)
>>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2)
>>> ax2.set_title('Probplot after Box-Cox transformation')
>>> plt.show()
"""
x = np.asarray(x)
if lmbda is not None: # single transformation
return special.boxcox(x, lmbda)
if x.ndim != 1:
raise ValueError("Data must be 1-dimensional.")
if x.size == 0:
return x
if np.all(x == x[0]):
raise ValueError("Data must not be constant.")
if np.any(x <= 0):
raise ValueError("Data must be positive.")
# If lmbda=None, find the lmbda that maximizes the log-likelihood function.
lmax = boxcox_normmax(x, method='mle', optimizer=optimizer)
y = boxcox(x, lmax)
if alpha is None:
return y, lmax
else:
# Find confidence interval
interval = _boxcox_conf_interval(x, lmax, alpha)
return y, lmax, interval
def _boxcox_inv_lmbda(x, y):
# compute lmbda given x and y for Box-Cox transformation
num = special.lambertw(-(x ** (-1 / y)) * np.log(x) / y, k=-1)
return np.real(-num / np.log(x) - 1 / y)
class _BigFloat:
def __repr__(self):
return "BIG_FLOAT"
def boxcox_normmax(
x, brack=None, method='pearsonr', optimizer=None, *, ymax=_BigFloat()
):
"""Compute optimal Box-Cox transform parameter for input data.
Parameters
----------
x : array_like
Input array. All entries must be positive, finite, real numbers.
brack : 2-tuple, optional, default (-2.0, 2.0)
The starting interval for a downhill bracket search for the default
`optimize.brent` solver. Note that this is in most cases not
critical; the final result is allowed to be outside this bracket.
If `optimizer` is passed, `brack` must be None.
method : str, optional
The method to determine the optimal transform parameter (`boxcox`
``lmbda`` parameter). Options are:
'pearsonr' (default)
Maximizes the Pearson correlation coefficient between
``y = boxcox(x)`` and the expected values for ``y`` if `x` would be
normally-distributed.
'mle'
Maximizes the log-likelihood `boxcox_llf`. This is the method used
in `boxcox`.
'all'
Use all optimization methods available, and return all results.
Useful to compare different methods.
optimizer : callable, optional
`optimizer` is a callable that accepts one argument:
fun : callable
The objective function to be minimized. `fun` accepts one argument,
the Box-Cox transform parameter `lmbda`, and returns the value of
the function (e.g., the negative log-likelihood) at the provided
argument. The job of `optimizer` is to find the value of `lmbda`
that *minimizes* `fun`.
and returns an object, such as an instance of
`scipy.optimize.OptimizeResult`, which holds the optimal value of
`lmbda` in an attribute `x`.
See the example below or the documentation of
`scipy.optimize.minimize_scalar` for more information.
ymax : float, optional
The unconstrained optimal transform parameter may cause Box-Cox
transformed data to have extreme magnitude or even overflow.
This parameter constrains MLE optimization such that the magnitude
of the transformed `x` does not exceed `ymax`. The default is
the maximum value of the input dtype. If set to infinity,
`boxcox_normmax` returns the unconstrained optimal lambda.
Ignored when ``method='pearsonr'``.
Returns
-------
maxlog : float or ndarray
The optimal transform parameter found. An array instead of a scalar
for ``method='all'``.
See Also
--------
boxcox, boxcox_llf, boxcox_normplot, scipy.optimize.minimize_scalar
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
We can generate some data and determine the optimal ``lmbda`` in various
ways:
>>> rng = np.random.default_rng()
>>> x = stats.loggamma.rvs(5, size=30, random_state=rng) + 5
>>> y, lmax_mle = stats.boxcox(x)
>>> lmax_pearsonr = stats.boxcox_normmax(x)
>>> lmax_mle
2.217563431465757
>>> lmax_pearsonr
2.238318660200961
>>> stats.boxcox_normmax(x, method='all')
array([2.23831866, 2.21756343])
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.boxcox_normplot(x, -10, 10, plot=ax)
>>> ax.axvline(lmax_mle, color='r')
>>> ax.axvline(lmax_pearsonr, color='g', ls='--')
>>> plt.show()
Alternatively, we can define our own `optimizer` function. Suppose we
are only interested in values of `lmbda` on the interval [6, 7], we
want to use `scipy.optimize.minimize_scalar` with ``method='bounded'``,
and we want to use tighter tolerances when optimizing the log-likelihood
function. To do this, we define a function that accepts positional argument
`fun` and uses `scipy.optimize.minimize_scalar` to minimize `fun` subject
to the provided bounds and tolerances:
>>> from scipy import optimize
>>> options = {'xatol': 1e-12} # absolute tolerance on `x`
>>> def optimizer(fun):
... return optimize.minimize_scalar(fun, bounds=(6, 7),
... method="bounded", options=options)
>>> stats.boxcox_normmax(x, optimizer=optimizer)
6.000...
"""
x = np.asarray(x)
end_msg = "exceed specified `ymax`."
if isinstance(ymax, _BigFloat):
dtype = x.dtype if np.issubdtype(x.dtype, np.floating) else np.float64
# 10000 is a safety factor because `special.boxcox` overflows prematurely.
ymax = np.finfo(dtype).max / 10000
end_msg = f"overflow in {dtype}."
elif ymax <= 0:
raise ValueError("`ymax` must be strictly positive")
# If optimizer is not given, define default 'brent' optimizer.
if optimizer is None:
# Set default value for `brack`.
if brack is None:
brack = (-2.0, 2.0)
def _optimizer(func, args):
return optimize.brent(func, args=args, brack=brack)
# Otherwise check optimizer.
else:
if not callable(optimizer):
raise ValueError("`optimizer` must be a callable")
if brack is not None:
raise ValueError("`brack` must be None if `optimizer` is given")
# `optimizer` is expected to return a `OptimizeResult` object, we here
# get the solution to the optimization problem.
def _optimizer(func, args):
def func_wrapped(x):
return func(x, *args)
return getattr(optimizer(func_wrapped), 'x', None)
def _pearsonr(x):
osm_uniform = _calc_uniform_order_statistic_medians(len(x))
xvals = distributions.norm.ppf(osm_uniform)
def _eval_pearsonr(lmbda, xvals, samps):
# This function computes the x-axis values of the probability plot
# and computes a linear regression (including the correlation) and
# returns ``1 - r`` so that a minimization function maximizes the
# correlation.
y = boxcox(samps, lmbda)
yvals = np.sort(y)
r, prob = _stats_py.pearsonr(xvals, yvals)
return 1 - r
return _optimizer(_eval_pearsonr, args=(xvals, x))
def _mle(x):
def _eval_mle(lmb, data):
# function to minimize
return -boxcox_llf(lmb, data)
return _optimizer(_eval_mle, args=(x,))
def _all(x):
maxlog = np.empty(2, dtype=float)
maxlog[0] = _pearsonr(x)
maxlog[1] = _mle(x)
return maxlog
methods = {'pearsonr': _pearsonr,
'mle': _mle,
'all': _all}
if method not in methods.keys():
raise ValueError("Method %s not recognized." % method)
optimfunc = methods[method]
try:
res = optimfunc(x)
except ValueError as e:
if "infs or NaNs" in str(e):
message = ("The `x` argument of `boxcox_normmax` must contain "
"only positive, finite, real numbers.")
raise ValueError(message) from e
else:
raise e
if res is None:
message = ("The `optimizer` argument of `boxcox_normmax` must return "
"an object containing the optimal `lmbda` in attribute `x`.")
raise ValueError(message)
elif not np.isinf(ymax): # adjust the final lambda
# x > 1, boxcox(x) > 0; x < 1, boxcox(x) < 0
xmax, xmin = np.max(x), np.min(x)
if xmin >= 1:
x_treme = xmax
elif xmax <= 1:
x_treme = xmin
else: # xmin < 1 < xmax
indicator = special.boxcox(xmax, res) > abs(special.boxcox(xmin, res))
if isinstance(res, np.ndarray):
indicator = indicator[1] # select corresponds with 'mle'
x_treme = xmax if indicator else xmin
mask = abs(special.boxcox(x_treme, res)) > ymax
if np.any(mask):
message = (
f"The optimal lambda is {res}, but the returned lambda is the "
f"constrained optimum to ensure that the maximum or the minimum "
f"of the transformed data does not " + end_msg
)
warnings.warn(message, stacklevel=2)
# Return the constrained lambda to ensure the transformation
# does not cause overflow or exceed specified `ymax`
constrained_res = _boxcox_inv_lmbda(x_treme, ymax * np.sign(x_treme - 1))
if isinstance(res, np.ndarray):
res[mask] = constrained_res
else:
res = constrained_res
return res
def _normplot(method, x, la, lb, plot=None, N=80):
"""Compute parameters for a Box-Cox or Yeo-Johnson normality plot,
optionally show it.
See `boxcox_normplot` or `yeojohnson_normplot` for details.
"""
if method == 'boxcox':
title = 'Box-Cox Normality Plot'
transform_func = boxcox
else:
title = 'Yeo-Johnson Normality Plot'
transform_func = yeojohnson
x = np.asarray(x)
if x.size == 0:
return x
if lb <= la:
raise ValueError("`lb` has to be larger than `la`.")
if method == 'boxcox' and np.any(x <= 0):
raise ValueError("Data must be positive.")
lmbdas = np.linspace(la, lb, num=N)
ppcc = lmbdas * 0.0
for i, val in enumerate(lmbdas):
# Determine for each lmbda the square root of correlation coefficient
# of transformed x
z = transform_func(x, lmbda=val)
_, (_, _, r) = probplot(z, dist='norm', fit=True)
ppcc[i] = r
if plot is not None:
plot.plot(lmbdas, ppcc, 'x')
_add_axis_labels_title(plot, xlabel='$\\lambda$',
ylabel='Prob Plot Corr. Coef.',
title=title)
return lmbdas, ppcc
def boxcox_normplot(x, la, lb, plot=None, N=80):
"""Compute parameters for a Box-Cox normality plot, optionally show it.
A Box-Cox normality plot shows graphically what the best transformation
parameter is to use in `boxcox` to obtain a distribution that is close
to normal.
Parameters
----------
x : array_like
Input array.
la, lb : scalar
The lower and upper bounds for the ``lmbda`` values to pass to `boxcox`
for Box-Cox transformations. These are also the limits of the
horizontal axis of the plot if that is generated.
plot : object, optional
If given, plots the quantiles and least squares fit.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
N : int, optional
Number of points on the horizontal axis (equally distributed from
`la` to `lb`).
Returns
-------
lmbdas : ndarray
The ``lmbda`` values for which a Box-Cox transform was done.
ppcc : ndarray
Probability Plot Correlelation Coefficient, as obtained from `probplot`
when fitting the Box-Cox transformed input `x` against a normal
distribution.
See Also
--------
probplot, boxcox, boxcox_normmax, boxcox_llf, ppcc_max
Notes
-----
Even if `plot` is given, the figure is not shown or saved by
`boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')``
should be used after calling `probplot`.
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
Generate some non-normally distributed data, and create a Box-Cox plot:
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.boxcox_normplot(x, -20, 20, plot=ax)
Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in
the same plot:
>>> _, maxlog = stats.boxcox(x)
>>> ax.axvline(maxlog, color='r')
>>> plt.show()
"""
return _normplot('boxcox', x, la, lb, plot, N)
def yeojohnson(x, lmbda=None):
r"""Return a dataset transformed by a Yeo-Johnson power transformation.
Parameters
----------
x : ndarray
Input array. Should be 1-dimensional.
lmbda : float, optional
If ``lmbda`` is ``None``, find the lambda that maximizes the
log-likelihood function and return it as the second output argument.
Otherwise the transformation is done for the given value.
Returns
-------
yeojohnson: ndarray
Yeo-Johnson power transformed array.
maxlog : float, optional
If the `lmbda` parameter is None, the second returned argument is
the lambda that maximizes the log-likelihood function.
See Also
--------
probplot, yeojohnson_normplot, yeojohnson_normmax, yeojohnson_llf, boxcox
Notes
-----
The Yeo-Johnson transform is given by::
y = ((x + 1)**lmbda - 1) / lmbda, for x >= 0, lmbda != 0
log(x + 1), for x >= 0, lmbda = 0
-((-x + 1)**(2 - lmbda) - 1) / (2 - lmbda), for x < 0, lmbda != 2
-log(-x + 1), for x < 0, lmbda = 2
Unlike `boxcox`, `yeojohnson` does not require the input data to be
positive.
.. versionadded:: 1.2.0
References
----------
I. Yeo and R.A. Johnson, "A New Family of Power Transformations to
Improve Normality or Symmetry", Biometrika 87.4 (2000):
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
We generate some random variates from a non-normal distribution and make a
probability plot for it, to show it is non-normal in the tails:
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(211)
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> prob = stats.probplot(x, dist=stats.norm, plot=ax1)
>>> ax1.set_xlabel('')
>>> ax1.set_title('Probplot against normal distribution')
We now use `yeojohnson` to transform the data so it's closest to normal:
>>> ax2 = fig.add_subplot(212)
>>> xt, lmbda = stats.yeojohnson(x)
>>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2)
>>> ax2.set_title('Probplot after Yeo-Johnson transformation')
>>> plt.show()
"""
x = np.asarray(x)
if x.size == 0:
return x
if np.issubdtype(x.dtype, np.complexfloating):
raise ValueError('Yeo-Johnson transformation is not defined for '
'complex numbers.')
if np.issubdtype(x.dtype, np.integer):
x = x.astype(np.float64, copy=False)
if lmbda is not None:
return _yeojohnson_transform(x, lmbda)
# if lmbda=None, find the lmbda that maximizes the log-likelihood function.
lmax = yeojohnson_normmax(x)
y = _yeojohnson_transform(x, lmax)
return y, lmax
def _yeojohnson_transform(x, lmbda):
"""Returns `x` transformed by the Yeo-Johnson power transform with given
parameter `lmbda`.
"""
dtype = x.dtype if np.issubdtype(x.dtype, np.floating) else np.float64
out = np.zeros_like(x, dtype=dtype)
pos = x >= 0 # binary mask
# when x >= 0
if abs(lmbda) < np.spacing(1.):
out[pos] = np.log1p(x[pos])
else: # lmbda != 0
# more stable version of: ((x + 1) ** lmbda - 1) / lmbda
out[pos] = np.expm1(lmbda * np.log1p(x[pos])) / lmbda
# when x < 0
if abs(lmbda - 2) > np.spacing(1.):
out[~pos] = -np.expm1((2 - lmbda) * np.log1p(-x[~pos])) / (2 - lmbda)
else: # lmbda == 2
out[~pos] = -np.log1p(-x[~pos])
return out
def yeojohnson_llf(lmb, data):
r"""The yeojohnson log-likelihood function.
Parameters
----------
lmb : scalar
Parameter for Yeo-Johnson transformation. See `yeojohnson` for
details.
data : array_like
Data to calculate Yeo-Johnson log-likelihood for. If `data` is
multi-dimensional, the log-likelihood is calculated along the first
axis.
Returns
-------
llf : float
Yeo-Johnson log-likelihood of `data` given `lmb`.
See Also
--------
yeojohnson, probplot, yeojohnson_normplot, yeojohnson_normmax
Notes
-----
The Yeo-Johnson log-likelihood function is defined here as
.. math::
llf = -N/2 \log(\hat{\sigma}^2) + (\lambda - 1)
\sum_i \text{ sign }(x_i)\log(|x_i| + 1)
where :math:`\hat{\sigma}^2` is estimated variance of the Yeo-Johnson
transformed input data ``x``.
.. versionadded:: 1.2.0
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes
Generate some random variates and calculate Yeo-Johnson log-likelihood
values for them for a range of ``lmbda`` values:
>>> x = stats.loggamma.rvs(5, loc=10, size=1000)
>>> lmbdas = np.linspace(-2, 10)
>>> llf = np.zeros(lmbdas.shape, dtype=float)
>>> for ii, lmbda in enumerate(lmbdas):
... llf[ii] = stats.yeojohnson_llf(lmbda, x)
Also find the optimal lmbda value with `yeojohnson`:
>>> x_most_normal, lmbda_optimal = stats.yeojohnson(x)
Plot the log-likelihood as function of lmbda. Add the optimal lmbda as a
horizontal line to check that that's really the optimum:
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> ax.plot(lmbdas, llf, 'b.-')
>>> ax.axhline(stats.yeojohnson_llf(lmbda_optimal, x), color='r')
>>> ax.set_xlabel('lmbda parameter')
>>> ax.set_ylabel('Yeo-Johnson log-likelihood')
Now add some probability plots to show that where the log-likelihood is
maximized the data transformed with `yeojohnson` looks closest to normal:
>>> locs = [3, 10, 4] # 'lower left', 'center', 'lower right'
>>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs):
... xt = stats.yeojohnson(x, lmbda=lmbda)
... (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt)
... ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc)
... ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-')
... ax_inset.set_xticklabels([])
... ax_inset.set_yticklabels([])
... ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda)
>>> plt.show()
"""
data = np.asarray(data)
n_samples = data.shape[0]
if n_samples == 0:
return np.nan
trans = _yeojohnson_transform(data, lmb)
trans_var = trans.var(axis=0)
loglike = np.empty_like(trans_var)
# Avoid RuntimeWarning raised by np.log when the variance is too low
tiny_variance = trans_var < np.finfo(trans_var.dtype).tiny
loglike[tiny_variance] = np.inf
loglike[~tiny_variance] = (
-n_samples / 2 * np.log(trans_var[~tiny_variance]))
loglike[~tiny_variance] += (
(lmb - 1) * (np.sign(data) * np.log1p(np.abs(data))).sum(axis=0))
return loglike
def yeojohnson_normmax(x, brack=None):
"""Compute optimal Yeo-Johnson transform parameter.
Compute optimal Yeo-Johnson transform parameter for input data, using
maximum likelihood estimation.
Parameters
----------
x : array_like
Input array.
brack : 2-tuple, optional
The starting interval for a downhill bracket search with
`optimize.brent`. Note that this is in most cases not critical; the
final result is allowed to be outside this bracket. If None,
`optimize.fminbound` is used with bounds that avoid overflow.
Returns
-------
maxlog : float
The optimal transform parameter found.
See Also
--------
yeojohnson, yeojohnson_llf, yeojohnson_normplot
Notes
-----
.. versionadded:: 1.2.0
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
Generate some data and determine optimal ``lmbda``
>>> rng = np.random.default_rng()
>>> x = stats.loggamma.rvs(5, size=30, random_state=rng) + 5
>>> lmax = stats.yeojohnson_normmax(x)
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.yeojohnson_normplot(x, -10, 10, plot=ax)
>>> ax.axvline(lmax, color='r')
>>> plt.show()
"""
def _neg_llf(lmbda, data):
llf = yeojohnson_llf(lmbda, data)
# reject likelihoods that are inf which are likely due to small
# variance in the transformed space
llf[np.isinf(llf)] = -np.inf
return -llf
with np.errstate(invalid='ignore'):
if not np.all(np.isfinite(x)):
raise ValueError('Yeo-Johnson input must be finite.')
if np.all(x == 0):
return 1.0
if brack is not None:
return optimize.brent(_neg_llf, brack=brack, args=(x,))
x = np.asarray(x)
dtype = x.dtype if np.issubdtype(x.dtype, np.floating) else np.float64
# Allow values up to 20 times the maximum observed value to be safely
# transformed without over- or underflow.
log1p_max_x = np.log1p(20 * np.max(np.abs(x)))
# Use half of floating point's exponent range to allow safe computation
# of the variance of the transformed data.
log_eps = np.log(np.finfo(dtype).eps)
log_tiny_float = (np.log(np.finfo(dtype).tiny) - log_eps) / 2
log_max_float = (np.log(np.finfo(dtype).max) + log_eps) / 2
# Compute the bounds by approximating the inverse of the Yeo-Johnson
# transform on the smallest and largest floating point exponents, given
# the largest data we expect to observe. See [1] for further details.
# [1] https://github.com/scipy/scipy/pull/18852#issuecomment-1630286174
lb = log_tiny_float / log1p_max_x
ub = log_max_float / log1p_max_x
# Convert the bounds if all or some of the data is negative.
if np.all(x < 0):
lb, ub = 2 - ub, 2 - lb
elif np.any(x < 0):
lb, ub = max(2 - ub, lb), min(2 - lb, ub)
# Match `optimize.brent`'s tolerance.
tol_brent = 1.48e-08
return optimize.fminbound(_neg_llf, lb, ub, args=(x,), xtol=tol_brent)
def yeojohnson_normplot(x, la, lb, plot=None, N=80):
"""Compute parameters for a Yeo-Johnson normality plot, optionally show it.
A Yeo-Johnson normality plot shows graphically what the best
transformation parameter is to use in `yeojohnson` to obtain a
distribution that is close to normal.
Parameters
----------
x : array_like
Input array.
la, lb : scalar
The lower and upper bounds for the ``lmbda`` values to pass to
`yeojohnson` for Yeo-Johnson transformations. These are also the
limits of the horizontal axis of the plot if that is generated.
plot : object, optional
If given, plots the quantiles and least squares fit.
`plot` is an object that has to have methods "plot" and "text".
The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
or a custom object with the same methods.
Default is None, which means that no plot is created.
N : int, optional
Number of points on the horizontal axis (equally distributed from
`la` to `lb`).
Returns
-------
lmbdas : ndarray
The ``lmbda`` values for which a Yeo-Johnson transform was done.
ppcc : ndarray
Probability Plot Correlelation Coefficient, as obtained from `probplot`
when fitting the Box-Cox transformed input `x` against a normal
distribution.
See Also
--------
probplot, yeojohnson, yeojohnson_normmax, yeojohnson_llf, ppcc_max
Notes
-----
Even if `plot` is given, the figure is not shown or saved by
`boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')``
should be used after calling `probplot`.
.. versionadded:: 1.2.0
Examples
--------
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
Generate some non-normally distributed data, and create a Yeo-Johnson plot:
>>> x = stats.loggamma.rvs(5, size=500) + 5
>>> fig = plt.figure()
>>> ax = fig.add_subplot(111)
>>> prob = stats.yeojohnson_normplot(x, -20, 20, plot=ax)
Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in
the same plot:
>>> _, maxlog = stats.yeojohnson(x)
>>> ax.axvline(maxlog, color='r')
>>> plt.show()
"""
return _normplot('yeojohnson', x, la, lb, plot, N)
ShapiroResult = namedtuple('ShapiroResult', ('statistic', 'pvalue'))
@_axis_nan_policy_factory(ShapiroResult, n_samples=1, too_small=2, default_axis=None)
def shapiro(x):
r"""Perform the Shapiro-Wilk test for normality.
The Shapiro-Wilk test tests the null hypothesis that the
data was drawn from a normal distribution.
Parameters
----------
x : array_like
Array of sample data.
Returns
-------
statistic : float
The test statistic.
p-value : float
The p-value for the hypothesis test.
See Also
--------
anderson : The Anderson-Darling test for normality
kstest : The Kolmogorov-Smirnov test for goodness of fit.
Notes
-----
The algorithm used is described in [4]_ but censoring parameters as
described are not implemented. For N > 5000 the W test statistic is
accurate, but the p-value may not be.
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm
:doi:`10.18434/M32189`
.. [2] Shapiro, S. S. & Wilk, M.B, "An analysis of variance test for
normality (complete samples)", Biometrika, 1965, Vol. 52,
pp. 591-611, :doi:`10.2307/2333709`
.. [3] Razali, N. M. & Wah, Y. B., "Power comparisons of Shapiro-Wilk,
Kolmogorov-Smirnov, Lilliefors and Anderson-Darling tests", Journal
of Statistical Modeling and Analytics, 2011, Vol. 2, pp. 21-33.
.. [4] Royston P., "Remark AS R94: A Remark on Algorithm AS 181: The
W-test for Normality", 1995, Applied Statistics, Vol. 44,
:doi:`10.2307/2986146`
.. [5] Phipson B., and Smyth, G. K., "Permutation P-values Should Never Be
Zero: Calculating Exact P-values When Permutations Are Randomly
Drawn", Statistical Applications in Genetics and Molecular Biology,
2010, Vol.9, :doi:`10.2202/1544-6115.1585`
.. [6] Panagiotakos, D. B., "The value of p-value in biomedical
research", The Open Cardiovascular Medicine Journal, 2008, Vol.2,
pp. 97-99, :doi:`10.2174/1874192400802010097`
Examples
--------
Suppose we wish to infer from measurements whether the weights of adult
human males in a medical study are not normally distributed [2]_.
The weights (lbs) are recorded in the array ``x`` below.
>>> import numpy as np
>>> x = np.array([148, 154, 158, 160, 161, 162, 166, 170, 182, 195, 236])
The normality test of [1]_ and [2]_ begins by computing a statistic based
on the relationship between the observations and the expected order
statistics of a normal distribution.
>>> from scipy import stats
>>> res = stats.shapiro(x)
>>> res.statistic
0.7888147830963135
The value of this statistic tends to be high (close to 1) for samples drawn
from a normal distribution.
The test is performed by comparing the observed value of the statistic
against the null distribution: the distribution of statistic values formed
under the null hypothesis that the weights were drawn from a normal
distribution. For this normality test, the null distribution is not easy to
calculate exactly, so it is usually approximated by Monte Carlo methods,
that is, drawing many samples of the same size as ``x`` from a normal
distribution and computing the values of the statistic for each.
>>> def statistic(x):
... # Get only the `shapiro` statistic; ignore its p-value
... return stats.shapiro(x).statistic
>>> ref = stats.monte_carlo_test(x, stats.norm.rvs, statistic,
... alternative='less')
>>> import matplotlib.pyplot as plt
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> bins = np.linspace(0.65, 1, 50)
>>> def plot(ax): # we'll reuse this
... ax.hist(ref.null_distribution, density=True, bins=bins)
... ax.set_title("Shapiro-Wilk Test Null Distribution \n"
... "(Monte Carlo Approximation, 11 Observations)")
... ax.set_xlabel("statistic")
... ax.set_ylabel("probability density")
>>> plot(ax)
>>> plt.show()
The comparison is quantified by the p-value: the proportion of values in
the null distribution less than or equal to the observed value of the
statistic.
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> plot(ax)
>>> annotation = (f'p-value={res.pvalue:.6f}\n(highlighted area)')
>>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
>>> _ = ax.annotate(annotation, (0.75, 0.1), (0.68, 0.7), arrowprops=props)
>>> i_extreme = np.where(bins <= res.statistic)[0]
>>> for i in i_extreme:
... ax.patches[i].set_color('C1')
>>> plt.xlim(0.65, 0.9)
>>> plt.ylim(0, 4)
>>> plt.show
>>> res.pvalue
0.006703833118081093
If the p-value is "small" - that is, if there is a low probability of
sampling data from a normally distributed population that produces such an
extreme value of the statistic - this may be taken as evidence against
the null hypothesis in favor of the alternative: the weights were not
drawn from a normal distribution. Note that:
- The inverse is not true; that is, the test is not used to provide
evidence *for* the null hypothesis.
- The threshold for values that will be considered "small" is a choice that
should be made before the data is analyzed [5]_ with consideration of the
risks of both false positives (incorrectly rejecting the null hypothesis)
and false negatives (failure to reject a false null hypothesis).
"""
x = np.ravel(x).astype(np.float64)
N = len(x)
if N < 3:
raise ValueError("Data must be at least length 3.")
a = zeros(N//2, dtype=np.float64)
init = 0
y = sort(x)
y -= x[N//2] # subtract the median (or a nearby value); see gh-15777
w, pw, ifault = swilk(y, a, init)
if ifault not in [0, 2]:
warnings.warn("scipy.stats.shapiro: Input data has range zero. The"
" results may not be accurate.", stacklevel=2)
if N > 5000:
warnings.warn("scipy.stats.shapiro: For N > 5000, computed p-value "
f"may not be accurate. Current N is {N}.",
stacklevel=2)
# `w` and `pw` are always Python floats, which are double precision.
# We want to ensure that they are NumPy floats, so until dtypes are
# respected, we can explicitly convert each to float64 (faster than
# `np.array([w, pw])`).
return ShapiroResult(np.float64(w), np.float64(pw))
# Values from Stephens, M A, "EDF Statistics for Goodness of Fit and
# Some Comparisons", Journal of the American Statistical
# Association, Vol. 69, Issue 347, Sept. 1974, pp 730-737
_Avals_norm = array([0.576, 0.656, 0.787, 0.918, 1.092])
_Avals_expon = array([0.922, 1.078, 1.341, 1.606, 1.957])
# From Stephens, M A, "Goodness of Fit for the Extreme Value Distribution",
# Biometrika, Vol. 64, Issue 3, Dec. 1977, pp 583-588.
_Avals_gumbel = array([0.474, 0.637, 0.757, 0.877, 1.038])
# From Stephens, M A, "Tests of Fit for the Logistic Distribution Based
# on the Empirical Distribution Function.", Biometrika,
# Vol. 66, Issue 3, Dec. 1979, pp 591-595.
_Avals_logistic = array([0.426, 0.563, 0.660, 0.769, 0.906, 1.010])
# From Richard A. Lockhart and Michael A. Stephens "Estimation and Tests of
# Fit for the Three-Parameter Weibull Distribution"
# Journal of the Royal Statistical Society.Series B(Methodological)
# Vol. 56, No. 3 (1994), pp. 491-500, table 1. Keys are c*100
_Avals_weibull = [[0.292, 0.395, 0.467, 0.522, 0.617, 0.711, 0.836, 0.931],
[0.295, 0.399, 0.471, 0.527, 0.623, 0.719, 0.845, 0.941],
[0.298, 0.403, 0.476, 0.534, 0.631, 0.728, 0.856, 0.954],
[0.301, 0.408, 0.483, 0.541, 0.640, 0.738, 0.869, 0.969],
[0.305, 0.414, 0.490, 0.549, 0.650, 0.751, 0.885, 0.986],
[0.309, 0.421, 0.498, 0.559, 0.662, 0.765, 0.902, 1.007],
[0.314, 0.429, 0.508, 0.570, 0.676, 0.782, 0.923, 1.030],
[0.320, 0.438, 0.519, 0.583, 0.692, 0.802, 0.947, 1.057],
[0.327, 0.448, 0.532, 0.598, 0.711, 0.824, 0.974, 1.089],
[0.334, 0.469, 0.547, 0.615, 0.732, 0.850, 1.006, 1.125],
[0.342, 0.472, 0.563, 0.636, 0.757, 0.879, 1.043, 1.167]]
_Avals_weibull = np.array(_Avals_weibull)
_cvals_weibull = np.linspace(0, 0.5, 11)
_get_As_weibull = interpolate.interp1d(_cvals_weibull, _Avals_weibull.T,
kind='linear', bounds_error=False,
fill_value=_Avals_weibull[-1])
def _weibull_fit_check(params, x):
# Refine the fit returned by `weibull_min.fit` to ensure that the first
# order necessary conditions are satisfied. If not, raise an error.
# Here, use `m` for the shape parameter to be consistent with [7]
# and avoid confusion with `c` as defined in [7].
n = len(x)
m, u, s = params
def dnllf_dm(m, u):
# Partial w.r.t. shape w/ optimal scale. See [7] Equation 5.
xu = x-u
return (1/m - (xu**m*np.log(xu)).sum()/(xu**m).sum()
+ np.log(xu).sum()/n)
def dnllf_du(m, u):
# Partial w.r.t. loc w/ optimal scale. See [7] Equation 6.
xu = x-u
return (m-1)/m*(xu**-1).sum() - n*(xu**(m-1)).sum()/(xu**m).sum()
def get_scale(m, u):
# Partial w.r.t. scale solved in terms of shape and location.
# See [7] Equation 7.
return ((x-u)**m/n).sum()**(1/m)
def dnllf(params):
# Partial derivatives of the NLLF w.r.t. parameters, i.e.
# first order necessary conditions for MLE fit.
return [dnllf_dm(*params), dnllf_du(*params)]
suggestion = ("Maximum likelihood estimation is known to be challenging "
"for the three-parameter Weibull distribution. Consider "
"performing a custom goodness-of-fit test using "
"`scipy.stats.monte_carlo_test`.")
if np.allclose(u, np.min(x)) or m < 1:
# The critical values provided by [7] don't seem to control the
# Type I error rate in this case. Error out.
message = ("Maximum likelihood estimation has converged to "
"a solution in which the location is equal to the minimum "
"of the data, the shape parameter is less than 2, or both. "
"The table of critical values in [7] does not "
"include this case. " + suggestion)
raise ValueError(message)
try:
# Refine the MLE / verify that first-order necessary conditions are
# satisfied. If so, the critical values provided in [7] seem reliable.
with np.errstate(over='raise', invalid='raise'):
res = optimize.root(dnllf, params[:-1])
message = ("Solution of MLE first-order conditions failed: "
f"{res.message}. `anderson` cannot continue. " + suggestion)
if not res.success:
raise ValueError(message)
except (FloatingPointError, ValueError) as e:
message = ("An error occurred while fitting the Weibull distribution "
"to the data, so `anderson` cannot continue. " + suggestion)
raise ValueError(message) from e
m, u = res.x
s = get_scale(m, u)
return m, u, s
AndersonResult = _make_tuple_bunch('AndersonResult',
['statistic', 'critical_values',
'significance_level'], ['fit_result'])
def anderson(x, dist='norm'):
"""Anderson-Darling test for data coming from a particular distribution.
The Anderson-Darling test tests the null hypothesis that a sample is
drawn from a population that follows a particular distribution.
For the Anderson-Darling test, the critical values depend on
which distribution is being tested against. This function works
for normal, exponential, logistic, weibull_min, or Gumbel (Extreme Value
Type I) distributions.
Parameters
----------
x : array_like
Array of sample data.
dist : {'norm', 'expon', 'logistic', 'gumbel', 'gumbel_l', 'gumbel_r', 'extreme1', 'weibull_min'}, optional
The type of distribution to test against. The default is 'norm'.
The names 'extreme1', 'gumbel_l' and 'gumbel' are synonyms for the
same distribution.
Returns
-------
result : AndersonResult
An object with the following attributes:
statistic : float
The Anderson-Darling test statistic.
critical_values : list
The critical values for this distribution.
significance_level : list
The significance levels for the corresponding critical values
in percents. The function returns critical values for a
differing set of significance levels depending on the
distribution that is being tested against.
fit_result : `~scipy.stats._result_classes.FitResult`
An object containing the results of fitting the distribution to
the data.
See Also
--------
kstest : The Kolmogorov-Smirnov test for goodness-of-fit.
Notes
-----
Critical values provided are for the following significance levels:
normal/exponential
15%, 10%, 5%, 2.5%, 1%
logistic
25%, 10%, 5%, 2.5%, 1%, 0.5%
gumbel_l / gumbel_r
25%, 10%, 5%, 2.5%, 1%
weibull_min
50%, 25%, 15%, 10%, 5%, 2.5%, 1%, 0.5%
If the returned statistic is larger than these critical values then
for the corresponding significance level, the null hypothesis that
the data come from the chosen distribution can be rejected.
The returned statistic is referred to as 'A2' in the references.
For `weibull_min`, maximum likelihood estimation is known to be
challenging. If the test returns successfully, then the first order
conditions for a maximum likehood estimate have been verified and
the critical values correspond relatively well to the significance levels,
provided that the sample is sufficiently large (>10 observations [7]).
However, for some data - especially data with no left tail - `anderson`
is likely to result in an error message. In this case, consider
performing a custom goodness of fit test using
`scipy.stats.monte_carlo_test`.
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm
.. [2] Stephens, M. A. (1974). EDF Statistics for Goodness of Fit and
Some Comparisons, Journal of the American Statistical Association,
Vol. 69, pp. 730-737.
.. [3] Stephens, M. A. (1976). Asymptotic Results for Goodness-of-Fit
Statistics with Unknown Parameters, Annals of Statistics, Vol. 4,
pp. 357-369.
.. [4] Stephens, M. A. (1977). Goodness of Fit for the Extreme Value
Distribution, Biometrika, Vol. 64, pp. 583-588.
.. [5] Stephens, M. A. (1977). Goodness of Fit with Special Reference
to Tests for Exponentiality , Technical Report No. 262,
Department of Statistics, Stanford University, Stanford, CA.
.. [6] Stephens, M. A. (1979). Tests of Fit for the Logistic Distribution
Based on the Empirical Distribution Function, Biometrika, Vol. 66,
pp. 591-595.
.. [7] Richard A. Lockhart and Michael A. Stephens "Estimation and Tests of
Fit for the Three-Parameter Weibull Distribution"
Journal of the Royal Statistical Society.Series B(Methodological)
Vol. 56, No. 3 (1994), pp. 491-500, Table 0.
Examples
--------
Test the null hypothesis that a random sample was drawn from a normal
distribution (with unspecified mean and standard deviation).
>>> import numpy as np
>>> from scipy.stats import anderson
>>> rng = np.random.default_rng()
>>> data = rng.random(size=35)
>>> res = anderson(data)
>>> res.statistic
0.8398018749744764
>>> res.critical_values
array([0.527, 0.6 , 0.719, 0.839, 0.998])
>>> res.significance_level
array([15. , 10. , 5. , 2.5, 1. ])
The value of the statistic (barely) exceeds the critical value associated
with a significance level of 2.5%, so the null hypothesis may be rejected
at a significance level of 2.5%, but not at a significance level of 1%.
""" # numpy/numpydoc#87 # noqa: E501
dist = dist.lower()
if dist in {'extreme1', 'gumbel'}:
dist = 'gumbel_l'
dists = {'norm', 'expon', 'gumbel_l',
'gumbel_r', 'logistic', 'weibull_min'}
if dist not in dists:
raise ValueError(f"Invalid distribution; dist must be in {dists}.")
y = sort(x)
xbar = np.mean(x, axis=0)
N = len(y)
if dist == 'norm':
s = np.std(x, ddof=1, axis=0)
w = (y - xbar) / s
fit_params = xbar, s
logcdf = distributions.norm.logcdf(w)
logsf = distributions.norm.logsf(w)
sig = array([15, 10, 5, 2.5, 1])
critical = around(_Avals_norm / (1.0 + 4.0/N - 25.0/N/N), 3)
elif dist == 'expon':
w = y / xbar
fit_params = 0, xbar
logcdf = distributions.expon.logcdf(w)
logsf = distributions.expon.logsf(w)
sig = array([15, 10, 5, 2.5, 1])
critical = around(_Avals_expon / (1.0 + 0.6/N), 3)
elif dist == 'logistic':
def rootfunc(ab, xj, N):
a, b = ab
tmp = (xj - a) / b
tmp2 = exp(tmp)
val = [np.sum(1.0/(1+tmp2), axis=0) - 0.5*N,
np.sum(tmp*(1.0-tmp2)/(1+tmp2), axis=0) + N]
return array(val)
sol0 = array([xbar, np.std(x, ddof=1, axis=0)])
sol = optimize.fsolve(rootfunc, sol0, args=(x, N), xtol=1e-5)
w = (y - sol[0]) / sol[1]
fit_params = sol
logcdf = distributions.logistic.logcdf(w)
logsf = distributions.logistic.logsf(w)
sig = array([25, 10, 5, 2.5, 1, 0.5])
critical = around(_Avals_logistic / (1.0 + 0.25/N), 3)
elif dist == 'gumbel_r':
xbar, s = distributions.gumbel_r.fit(x)
w = (y - xbar) / s
fit_params = xbar, s
logcdf = distributions.gumbel_r.logcdf(w)
logsf = distributions.gumbel_r.logsf(w)
sig = array([25, 10, 5, 2.5, 1])
critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3)
elif dist == 'gumbel_l':
xbar, s = distributions.gumbel_l.fit(x)
w = (y - xbar) / s
fit_params = xbar, s
logcdf = distributions.gumbel_l.logcdf(w)
logsf = distributions.gumbel_l.logsf(w)
sig = array([25, 10, 5, 2.5, 1])
critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3)
elif dist == 'weibull_min':
message = ("Critical values of the test statistic are given for the "
"asymptotic distribution. These may not be accurate for "
"samples with fewer than 10 observations. Consider using "
"`scipy.stats.monte_carlo_test`.")
if N < 10:
warnings.warn(message, stacklevel=2)
# [7] writes our 'c' as 'm', and they write `c = 1/m`. Use their names.
m, loc, scale = distributions.weibull_min.fit(y)
m, loc, scale = _weibull_fit_check((m, loc, scale), y)
fit_params = m, loc, scale
logcdf = stats.weibull_min(*fit_params).logcdf(y)
logsf = stats.weibull_min(*fit_params).logsf(y)
c = 1 / m # m and c are as used in [7]
sig = array([0.5, 0.75, 0.85, 0.9, 0.95, 0.975, 0.99, 0.995])
critical = _get_As_weibull(c)
# Goodness-of-fit tests should only be used to provide evidence
# _against_ the null hypothesis. Be conservative and round up.
critical = np.round(critical + 0.0005, decimals=3)
i = arange(1, N + 1)
A2 = -N - np.sum((2*i - 1.0) / N * (logcdf + logsf[::-1]), axis=0)
# FitResult initializer expects an optimize result, so let's work with it
message = '`anderson` successfully fit the distribution to the data.'
res = optimize.OptimizeResult(success=True, message=message)
res.x = np.array(fit_params)
fit_result = FitResult(getattr(distributions, dist), y,
discrete=False, res=res)
return AndersonResult(A2, critical, sig, fit_result=fit_result)
def _anderson_ksamp_midrank(samples, Z, Zstar, k, n, N):
"""Compute A2akN equation 7 of Scholz and Stephens.
Parameters
----------
samples : sequence of 1-D array_like
Array of sample arrays.
Z : array_like
Sorted array of all observations.
Zstar : array_like
Sorted array of unique observations.
k : int
Number of samples.
n : array_like
Number of observations in each sample.
N : int
Total number of observations.
Returns
-------
A2aKN : float
The A2aKN statistics of Scholz and Stephens 1987.
"""
A2akN = 0.
Z_ssorted_left = Z.searchsorted(Zstar, 'left')
if N == Zstar.size:
lj = 1.
else:
lj = Z.searchsorted(Zstar, 'right') - Z_ssorted_left
Bj = Z_ssorted_left + lj / 2.
for i in arange(0, k):
s = np.sort(samples[i])
s_ssorted_right = s.searchsorted(Zstar, side='right')
Mij = s_ssorted_right.astype(float)
fij = s_ssorted_right - s.searchsorted(Zstar, 'left')
Mij -= fij / 2.
inner = lj / float(N) * (N*Mij - Bj*n[i])**2 / (Bj*(N - Bj) - N*lj/4.)
A2akN += inner.sum() / n[i]
A2akN *= (N - 1.) / N
return A2akN
def _anderson_ksamp_right(samples, Z, Zstar, k, n, N):
"""Compute A2akN equation 6 of Scholz & Stephens.
Parameters
----------
samples : sequence of 1-D array_like
Array of sample arrays.
Z : array_like
Sorted array of all observations.
Zstar : array_like
Sorted array of unique observations.
k : int
Number of samples.
n : array_like
Number of observations in each sample.
N : int
Total number of observations.
Returns
-------
A2KN : float
The A2KN statistics of Scholz and Stephens 1987.
"""
A2kN = 0.
lj = Z.searchsorted(Zstar[:-1], 'right') - Z.searchsorted(Zstar[:-1],
'left')
Bj = lj.cumsum()
for i in arange(0, k):
s = np.sort(samples[i])
Mij = s.searchsorted(Zstar[:-1], side='right')
inner = lj / float(N) * (N * Mij - Bj * n[i])**2 / (Bj * (N - Bj))
A2kN += inner.sum() / n[i]
return A2kN
Anderson_ksampResult = _make_tuple_bunch(
'Anderson_ksampResult',
['statistic', 'critical_values', 'pvalue'], []
)
def anderson_ksamp(samples, midrank=True, *, method=None):
"""The Anderson-Darling test for k-samples.
The k-sample Anderson-Darling test is a modification of the
one-sample Anderson-Darling test. It tests the null hypothesis
that k-samples are drawn from the same population without having
to specify the distribution function of that population. The
critical values depend on the number of samples.
Parameters
----------
samples : sequence of 1-D array_like
Array of sample data in arrays.
midrank : bool, optional
Type of Anderson-Darling test which is computed. Default
(True) is the midrank test applicable to continuous and
discrete populations. If False, the right side empirical
distribution is used.
method : PermutationMethod, optional
Defines the method used to compute the p-value. If `method` is an
instance of `PermutationMethod`, the p-value is computed using
`scipy.stats.permutation_test` with the provided configuration options
and other appropriate settings. Otherwise, the p-value is interpolated
from tabulated values.
Returns
-------
res : Anderson_ksampResult
An object containing attributes:
statistic : float
Normalized k-sample Anderson-Darling test statistic.
critical_values : array
The critical values for significance levels 25%, 10%, 5%, 2.5%, 1%,
0.5%, 0.1%.
pvalue : float
The approximate p-value of the test. If `method` is not
provided, the value is floored / capped at 0.1% / 25%.
Raises
------
ValueError
If fewer than 2 samples are provided, a sample is empty, or no
distinct observations are in the samples.
See Also
--------
ks_2samp : 2 sample Kolmogorov-Smirnov test
anderson : 1 sample Anderson-Darling test
Notes
-----
[1]_ defines three versions of the k-sample Anderson-Darling test:
one for continuous distributions and two for discrete
distributions, in which ties between samples may occur. The
default of this routine is to compute the version based on the
midrank empirical distribution function. This test is applicable
to continuous and discrete data. If midrank is set to False, the
right side empirical distribution is used for a test for discrete
data. According to [1]_, the two discrete test statistics differ
only slightly if a few collisions due to round-off errors occur in
the test not adjusted for ties between samples.
The critical values corresponding to the significance levels from 0.01
to 0.25 are taken from [1]_. p-values are floored / capped
at 0.1% / 25%. Since the range of critical values might be extended in
future releases, it is recommended not to test ``p == 0.25``, but rather
``p >= 0.25`` (analogously for the lower bound).
.. versionadded:: 0.14.0
References
----------
.. [1] Scholz, F. W and Stephens, M. A. (1987), K-Sample
Anderson-Darling Tests, Journal of the American Statistical
Association, Vol. 82, pp. 918-924.
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> rng = np.random.default_rng()
>>> res = stats.anderson_ksamp([rng.normal(size=50),
... rng.normal(loc=0.5, size=30)])
>>> res.statistic, res.pvalue
(1.974403288713695, 0.04991293614572478)
>>> res.critical_values
array([0.325, 1.226, 1.961, 2.718, 3.752, 4.592, 6.546])
The null hypothesis that the two random samples come from the same
distribution can be rejected at the 5% level because the returned
test value is greater than the critical value for 5% (1.961) but
not at the 2.5% level. The interpolation gives an approximate
p-value of 4.99%.
>>> samples = [rng.normal(size=50), rng.normal(size=30),
... rng.normal(size=20)]
>>> res = stats.anderson_ksamp(samples)
>>> res.statistic, res.pvalue
(-0.29103725200789504, 0.25)
>>> res.critical_values
array([ 0.44925884, 1.3052767 , 1.9434184 , 2.57696569, 3.41634856,
4.07210043, 5.56419101])
The null hypothesis cannot be rejected for three samples from an
identical distribution. The reported p-value (25%) has been capped and
may not be very accurate (since it corresponds to the value 0.449
whereas the statistic is -0.291).
In such cases where the p-value is capped or when sample sizes are
small, a permutation test may be more accurate.
>>> method = stats.PermutationMethod(n_resamples=9999, random_state=rng)
>>> res = stats.anderson_ksamp(samples, method=method)
>>> res.pvalue
0.5254
"""
k = len(samples)
if (k < 2):
raise ValueError("anderson_ksamp needs at least two samples")
samples = list(map(np.asarray, samples))
Z = np.sort(np.hstack(samples))
N = Z.size
Zstar = np.unique(Z)
if Zstar.size < 2:
raise ValueError("anderson_ksamp needs more than one distinct "
"observation")
n = np.array([sample.size for sample in samples])
if np.any(n == 0):
raise ValueError("anderson_ksamp encountered sample without "
"observations")
if midrank:
A2kN_fun = _anderson_ksamp_midrank
else:
A2kN_fun = _anderson_ksamp_right
A2kN = A2kN_fun(samples, Z, Zstar, k, n, N)
def statistic(*samples):
return A2kN_fun(samples, Z, Zstar, k, n, N)
if method is not None:
res = stats.permutation_test(samples, statistic, **method._asdict(),
alternative='greater')
H = (1. / n).sum()
hs_cs = (1. / arange(N - 1, 1, -1)).cumsum()
h = hs_cs[-1] + 1
g = (hs_cs / arange(2, N)).sum()
a = (4*g - 6) * (k - 1) + (10 - 6*g)*H
b = (2*g - 4)*k**2 + 8*h*k + (2*g - 14*h - 4)*H - 8*h + 4*g - 6
c = (6*h + 2*g - 2)*k**2 + (4*h - 4*g + 6)*k + (2*h - 6)*H + 4*h
d = (2*h + 6)*k**2 - 4*h*k
sigmasq = (a*N**3 + b*N**2 + c*N + d) / ((N - 1.) * (N - 2.) * (N - 3.))
m = k - 1
A2 = (A2kN - m) / math.sqrt(sigmasq)
# The b_i values are the interpolation coefficients from Table 2
# of Scholz and Stephens 1987
b0 = np.array([0.675, 1.281, 1.645, 1.96, 2.326, 2.573, 3.085])
b1 = np.array([-0.245, 0.25, 0.678, 1.149, 1.822, 2.364, 3.615])
b2 = np.array([-0.105, -0.305, -0.362, -0.391, -0.396, -0.345, -0.154])
critical = b0 + b1 / math.sqrt(m) + b2 / m
sig = np.array([0.25, 0.1, 0.05, 0.025, 0.01, 0.005, 0.001])
if A2 < critical.min() and method is None:
p = sig.max()
msg = (f"p-value capped: true value larger than {p}. Consider "
"specifying `method` "
"(e.g. `method=stats.PermutationMethod()`.)")
warnings.warn(msg, stacklevel=2)
elif A2 > critical.max() and method is None:
p = sig.min()
msg = (f"p-value floored: true value smaller than {p}. Consider "
"specifying `method` "
"(e.g. `method=stats.PermutationMethod()`.)")
warnings.warn(msg, stacklevel=2)
elif method is None:
# interpolation of probit of significance level
pf = np.polyfit(critical, log(sig), 2)
p = math.exp(np.polyval(pf, A2))
else:
p = res.pvalue if method is not None else p
# create result object with alias for backward compatibility
res = Anderson_ksampResult(A2, critical, p)
res.significance_level = p
return res
AnsariResult = namedtuple('AnsariResult', ('statistic', 'pvalue'))
class _ABW:
"""Distribution of Ansari-Bradley W-statistic under the null hypothesis."""
# TODO: calculate exact distribution considering ties
# We could avoid summing over more than half the frequencies,
# but initially it doesn't seem worth the extra complexity
def __init__(self):
"""Minimal initializer."""
self.m = None
self.n = None
self.astart = None
self.total = None
self.freqs = None
def _recalc(self, n, m):
"""When necessary, recalculate exact distribution."""
if n != self.n or m != self.m:
self.n, self.m = n, m
# distribution is NOT symmetric when m + n is odd
# n is len(x), m is len(y), and ratio of scales is defined x/y
astart, a1, _ = gscale(n, m)
self.astart = astart # minimum value of statistic
# Exact distribution of test statistic under null hypothesis
# expressed as frequencies/counts/integers to maintain precision.
# Stored as floats to avoid overflow of sums.
self.freqs = a1.astype(np.float64)
self.total = self.freqs.sum() # could calculate from m and n
# probability mass is self.freqs / self.total;
def pmf(self, k, n, m):
"""Probability mass function."""
self._recalc(n, m)
# The convention here is that PMF at k = 12.5 is the same as at k = 12,
# -> use `floor` in case of ties.
ind = np.floor(k - self.astart).astype(int)
return self.freqs[ind] / self.total
def cdf(self, k, n, m):
"""Cumulative distribution function."""
self._recalc(n, m)
# Null distribution derived without considering ties is
# approximate. Round down to avoid Type I error.
ind = np.ceil(k - self.astart).astype(int)
return self.freqs[:ind+1].sum() / self.total
def sf(self, k, n, m):
"""Survival function."""
self._recalc(n, m)
# Null distribution derived without considering ties is
# approximate. Round down to avoid Type I error.
ind = np.floor(k - self.astart).astype(int)
return self.freqs[ind:].sum() / self.total
# Maintain state for faster repeat calls to ansari w/ method='exact'
_abw_state = _ABW()
@_axis_nan_policy_factory(AnsariResult, n_samples=2)
def ansari(x, y, alternative='two-sided'):
"""Perform the Ansari-Bradley test for equal scale parameters.
The Ansari-Bradley test ([1]_, [2]_) is a non-parametric test
for the equality of the scale parameter of the distributions
from which two samples were drawn. The null hypothesis states that
the ratio of the scale of the distribution underlying `x` to the scale
of the distribution underlying `y` is 1.
Parameters
----------
x, y : array_like
Arrays of sample data.
alternative : {'two-sided', 'less', 'greater'}, optional
Defines the alternative hypothesis. Default is 'two-sided'.
The following options are available:
* 'two-sided': the ratio of scales is not equal to 1.
* 'less': the ratio of scales is less than 1.
* 'greater': the ratio of scales is greater than 1.
.. versionadded:: 1.7.0
Returns
-------
statistic : float
The Ansari-Bradley test statistic.
pvalue : float
The p-value of the hypothesis test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
mood : A non-parametric test for the equality of two scale parameters
Notes
-----
The p-value given is exact when the sample sizes are both less than
55 and there are no ties, otherwise a normal approximation for the
p-value is used.
References
----------
.. [1] Ansari, A. R. and Bradley, R. A. (1960) Rank-sum tests for
dispersions, Annals of Mathematical Statistics, 31, 1174-1189.
.. [2] Sprent, Peter and N.C. Smeeton. Applied nonparametric
statistical methods. 3rd ed. Chapman and Hall/CRC. 2001.
Section 5.8.2.
.. [3] Nathaniel E. Helwig "Nonparametric Dispersion and Equality
Tests" at http://users.stat.umn.edu/~helwig/notes/npde-Notes.pdf
Examples
--------
>>> import numpy as np
>>> from scipy.stats import ansari
>>> rng = np.random.default_rng()
For these examples, we'll create three random data sets. The first
two, with sizes 35 and 25, are drawn from a normal distribution with
mean 0 and standard deviation 2. The third data set has size 25 and
is drawn from a normal distribution with standard deviation 1.25.
>>> x1 = rng.normal(loc=0, scale=2, size=35)
>>> x2 = rng.normal(loc=0, scale=2, size=25)
>>> x3 = rng.normal(loc=0, scale=1.25, size=25)
First we apply `ansari` to `x1` and `x2`. These samples are drawn
from the same distribution, so we expect the Ansari-Bradley test
should not lead us to conclude that the scales of the distributions
are different.
>>> ansari(x1, x2)
AnsariResult(statistic=541.0, pvalue=0.9762532927399098)
With a p-value close to 1, we cannot conclude that there is a
significant difference in the scales (as expected).
Now apply the test to `x1` and `x3`:
>>> ansari(x1, x3)
AnsariResult(statistic=425.0, pvalue=0.0003087020407974518)
The probability of observing such an extreme value of the statistic
under the null hypothesis of equal scales is only 0.03087%. We take this
as evidence against the null hypothesis in favor of the alternative:
the scales of the distributions from which the samples were drawn
are not equal.
We can use the `alternative` parameter to perform a one-tailed test.
In the above example, the scale of `x1` is greater than `x3` and so
the ratio of scales of `x1` and `x3` is greater than 1. This means
that the p-value when ``alternative='greater'`` should be near 0 and
hence we should be able to reject the null hypothesis:
>>> ansari(x1, x3, alternative='greater')
AnsariResult(statistic=425.0, pvalue=0.0001543510203987259)
As we can see, the p-value is indeed quite low. Use of
``alternative='less'`` should thus yield a large p-value:
>>> ansari(x1, x3, alternative='less')
AnsariResult(statistic=425.0, pvalue=0.9998643258449039)
"""
if alternative not in {'two-sided', 'greater', 'less'}:
raise ValueError("'alternative' must be 'two-sided',"
" 'greater', or 'less'.")
x, y = asarray(x), asarray(y)
n = len(x)
m = len(y)
if m < 1:
raise ValueError("Not enough other observations.")
if n < 1:
raise ValueError("Not enough test observations.")
N = m + n
xy = r_[x, y] # combine
rank = _stats_py.rankdata(xy)
symrank = amin(array((rank, N - rank + 1)), 0)
AB = np.sum(symrank[:n], axis=0)
uxy = unique(xy)
repeats = (len(uxy) != len(xy))
exact = ((m < 55) and (n < 55) and not repeats)
if repeats and (m < 55 or n < 55):
warnings.warn("Ties preclude use of exact statistic.", stacklevel=2)
if exact:
if alternative == 'two-sided':
pval = 2.0 * np.minimum(_abw_state.cdf(AB, n, m),
_abw_state.sf(AB, n, m))
elif alternative == 'greater':
# AB statistic is _smaller_ when ratio of scales is larger,
# so this is the opposite of the usual calculation
pval = _abw_state.cdf(AB, n, m)
else:
pval = _abw_state.sf(AB, n, m)
return AnsariResult(AB, min(1.0, pval))
# otherwise compute normal approximation
if N % 2: # N odd
mnAB = n * (N+1.0)**2 / 4.0 / N
varAB = n * m * (N+1.0) * (3+N**2) / (48.0 * N**2)
else:
mnAB = n * (N+2.0) / 4.0
varAB = m * n * (N+2) * (N-2.0) / 48 / (N-1.0)
if repeats: # adjust variance estimates
# compute np.sum(tj * rj**2,axis=0)
fac = np.sum(symrank**2, axis=0)
if N % 2: # N odd
varAB = m * n * (16*N*fac - (N+1)**4) / (16.0 * N**2 * (N-1))
else: # N even
varAB = m * n * (16*fac - N*(N+2)**2) / (16.0 * N * (N-1))
# Small values of AB indicate larger dispersion for the x sample.
# Large values of AB indicate larger dispersion for the y sample.
# This is opposite to the way we define the ratio of scales. see [1]_.
z = (mnAB - AB) / sqrt(varAB)
pvalue = _get_pvalue(z, distributions.norm, alternative)
return AnsariResult(AB[()], pvalue[()])
BartlettResult = namedtuple('BartlettResult', ('statistic', 'pvalue'))
@_axis_nan_policy_factory(BartlettResult, n_samples=None)
def bartlett(*samples):
r"""Perform Bartlett's test for equal variances.
Bartlett's test tests the null hypothesis that all input samples
are from populations with equal variances. For samples
from significantly non-normal populations, Levene's test
`levene` is more robust.
Parameters
----------
sample1, sample2, ... : array_like
arrays of sample data. Only 1d arrays are accepted, they may have
different lengths.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value of the test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
levene : A robust parametric test for equality of k variances
Notes
-----
Conover et al. (1981) examine many of the existing parametric and
nonparametric tests by extensive simulations and they conclude that the
tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be
superior in terms of robustness of departures from normality and power
([3]_).
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda357.htm
.. [2] Snedecor, George W. and Cochran, William G. (1989), Statistical
Methods, Eighth Edition, Iowa State University Press.
.. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
Hypothesis Testing based on Quadratic Inference Function. Technical
Report #99-03, Center for Likelihood Studies, Pennsylvania State
University.
.. [4] Bartlett, M. S. (1937). Properties of Sufficiency and Statistical
Tests. Proceedings of the Royal Society of London. Series A,
Mathematical and Physical Sciences, Vol. 160, No.901, pp. 268-282.
.. [5] C.I. BLISS (1952), The Statistics of Bioassay: With Special
Reference to the Vitamins, pp 499-503,
:doi:`10.1016/C2013-0-12584-6`.
.. [6] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be
Zero: Calculating Exact P-values When Permutations Are Randomly
Drawn." Statistical Applications in Genetics and Molecular Biology
9.1 (2010).
.. [7] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are
superior to t and F tests in biomedical research. The American
Statistician, 52(2), 127-132.
Examples
--------
In [5]_, the influence of vitamin C on the tooth growth of guinea pigs
was investigated. In a control study, 60 subjects were divided into
small dose, medium dose, and large dose groups that received
daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively.
After 42 days, the tooth growth was measured.
The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record
tooth growth measurements of the three groups in microns.
>>> import numpy as np
>>> small_dose = np.array([
... 4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7,
... 15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7
... ])
>>> medium_dose = np.array([
... 16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5,
... 19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3
... ])
>>> large_dose = np.array([
... 23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5,
... 25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23
... ])
The `bartlett` statistic is sensitive to differences in variances
between the samples.
>>> from scipy import stats
>>> res = stats.bartlett(small_dose, medium_dose, large_dose)
>>> res.statistic
0.6654670663030519
The value of the statistic tends to be high when there is a large
difference in variances.
We can test for inequality of variance among the groups by comparing the
observed value of the statistic against the null distribution: the
distribution of statistic values derived under the null hypothesis that
the population variances of the three groups are equal.
For this test, the null distribution follows the chi-square distribution
as shown below.
>>> import matplotlib.pyplot as plt
>>> k = 3 # number of samples
>>> dist = stats.chi2(df=k-1)
>>> val = np.linspace(0, 5, 100)
>>> pdf = dist.pdf(val)
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> def plot(ax): # we'll reuse this
... ax.plot(val, pdf, color='C0')
... ax.set_title("Bartlett Test Null Distribution")
... ax.set_xlabel("statistic")
... ax.set_ylabel("probability density")
... ax.set_xlim(0, 5)
... ax.set_ylim(0, 1)
>>> plot(ax)
>>> plt.show()
The comparison is quantified by the p-value: the proportion of values in
the null distribution greater than or equal to the observed value of the
statistic.
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> plot(ax)
>>> pvalue = dist.sf(res.statistic)
>>> annotation = (f'p-value={pvalue:.3f}\n(shaded area)')
>>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
>>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props)
>>> i = val >= res.statistic
>>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0')
>>> plt.show()
>>> res.pvalue
0.71696121509966
If the p-value is "small" - that is, if there is a low probability of
sampling data from distributions with identical variances that produces
such an extreme value of the statistic - this may be taken as evidence
against the null hypothesis in favor of the alternative: the variances of
the groups are not equal. Note that:
- The inverse is not true; that is, the test is not used to provide
evidence for the null hypothesis.
- The threshold for values that will be considered "small" is a choice that
should be made before the data is analyzed [6]_ with consideration of the
risks of both false positives (incorrectly rejecting the null hypothesis)
and false negatives (failure to reject a false null hypothesis).
- Small p-values are not evidence for a *large* effect; rather, they can
only provide evidence for a "significant" effect, meaning that they are
unlikely to have occurred under the null hypothesis.
Note that the chi-square distribution provides the null distribution
when the observations are normally distributed. For small samples
drawn from non-normal populations, it may be more appropriate to
perform a
permutation test: Under the null hypothesis that all three samples were
drawn from the same population, each of the measurements is equally likely
to have been observed in any of the three samples. Therefore, we can form
a randomized null distribution by calculating the statistic under many
randomly-generated partitionings of the observations into the three
samples.
>>> def statistic(*samples):
... return stats.bartlett(*samples).statistic
>>> ref = stats.permutation_test(
... (small_dose, medium_dose, large_dose), statistic,
... permutation_type='independent', alternative='greater'
... )
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> plot(ax)
>>> bins = np.linspace(0, 5, 25)
>>> ax.hist(
... ref.null_distribution, bins=bins, density=True, facecolor="C1"
... )
>>> ax.legend(['aymptotic approximation\n(many observations)',
... 'randomized null distribution'])
>>> plot(ax)
>>> plt.show()
>>> ref.pvalue # randomized test p-value
0.5387 # may vary
Note that there is significant disagreement between the p-value calculated
here and the asymptotic approximation returned by `bartlett` above.
The statistical inferences that can be drawn rigorously from a permutation
test are limited; nonetheless, they may be the preferred approach in many
circumstances [7]_.
Following is another generic example where the null hypothesis would be
rejected.
Test whether the lists `a`, `b` and `c` come from populations
with equal variances.
>>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
>>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
>>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
>>> stat, p = stats.bartlett(a, b, c)
>>> p
1.1254782518834628e-05
The very small p-value suggests that the populations do not have equal
variances.
This is not surprising, given that the sample variance of `b` is much
larger than that of `a` and `c`:
>>> [np.var(x, ddof=1) for x in [a, b, c]]
[0.007054444444444413, 0.13073888888888888, 0.008890000000000002]
"""
k = len(samples)
if k < 2:
raise ValueError("Must enter at least two input sample vectors.")
# Handle empty input and input that is not 1d
for sample in samples:
if np.asanyarray(sample).size == 0:
NaN = _get_nan(*samples) # get NaN of result_dtype of all samples
return BartlettResult(NaN, NaN)
Ni = np.empty(k)
ssq = np.empty(k, 'd')
for j in range(k):
Ni[j] = len(samples[j])
ssq[j] = np.var(samples[j], ddof=1)
Ntot = np.sum(Ni, axis=0)
spsq = np.sum((Ni - 1)*ssq, axis=0) / (1.0*(Ntot - k))
numer = (Ntot*1.0 - k) * log(spsq) - np.sum((Ni - 1.0)*log(ssq), axis=0)
denom = 1.0 + 1.0/(3*(k - 1)) * ((np.sum(1.0/(Ni - 1.0), axis=0)) -
1.0/(Ntot - k))
T = numer / denom
pval = distributions.chi2.sf(T, k - 1) # 1 - cdf
return BartlettResult(T, pval)
LeveneResult = namedtuple('LeveneResult', ('statistic', 'pvalue'))
@_axis_nan_policy_factory(LeveneResult, n_samples=None)
def levene(*samples, center='median', proportiontocut=0.05):
r"""Perform Levene test for equal variances.
The Levene test tests the null hypothesis that all input samples
are from populations with equal variances. Levene's test is an
alternative to Bartlett's test `bartlett` in the case where
there are significant deviations from normality.
Parameters
----------
sample1, sample2, ... : array_like
The sample data, possibly with different lengths. Only one-dimensional
samples are accepted.
center : {'mean', 'median', 'trimmed'}, optional
Which function of the data to use in the test. The default
is 'median'.
proportiontocut : float, optional
When `center` is 'trimmed', this gives the proportion of data points
to cut from each end. (See `scipy.stats.trim_mean`.)
Default is 0.05.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value for the test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
bartlett : A parametric test for equality of k variances in normal samples
Notes
-----
Three variations of Levene's test are possible. The possibilities
and their recommended usages are:
* 'median' : Recommended for skewed (non-normal) distributions>
* 'mean' : Recommended for symmetric, moderate-tailed distributions.
* 'trimmed' : Recommended for heavy-tailed distributions.
The test version using the mean was proposed in the original article
of Levene ([2]_) while the median and trimmed mean have been studied by
Brown and Forsythe ([3]_), sometimes also referred to as Brown-Forsythe
test.
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm
.. [2] Levene, H. (1960). In Contributions to Probability and Statistics:
Essays in Honor of Harold Hotelling, I. Olkin et al. eds.,
Stanford University Press, pp. 278-292.
.. [3] Brown, M. B. and Forsythe, A. B. (1974), Journal of the American
Statistical Association, 69, 364-367
.. [4] C.I. BLISS (1952), The Statistics of Bioassay: With Special
Reference to the Vitamins, pp 499-503,
:doi:`10.1016/C2013-0-12584-6`.
.. [5] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be
Zero: Calculating Exact P-values When Permutations Are Randomly
Drawn." Statistical Applications in Genetics and Molecular Biology
9.1 (2010).
.. [6] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are
superior to t and F tests in biomedical research. The American
Statistician, 52(2), 127-132.
Examples
--------
In [4]_, the influence of vitamin C on the tooth growth of guinea pigs
was investigated. In a control study, 60 subjects were divided into
small dose, medium dose, and large dose groups that received
daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively.
After 42 days, the tooth growth was measured.
The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record
tooth growth measurements of the three groups in microns.
>>> import numpy as np
>>> small_dose = np.array([
... 4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7,
... 15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7
... ])
>>> medium_dose = np.array([
... 16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5,
... 19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3
... ])
>>> large_dose = np.array([
... 23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5,
... 25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23
... ])
The `levene` statistic is sensitive to differences in variances
between the samples.
>>> from scipy import stats
>>> res = stats.levene(small_dose, medium_dose, large_dose)
>>> res.statistic
0.6457341109631506
The value of the statistic tends to be high when there is a large
difference in variances.
We can test for inequality of variance among the groups by comparing the
observed value of the statistic against the null distribution: the
distribution of statistic values derived under the null hypothesis that
the population variances of the three groups are equal.
For this test, the null distribution follows the F distribution as shown
below.
>>> import matplotlib.pyplot as plt
>>> k, n = 3, 60 # number of samples, total number of observations
>>> dist = stats.f(dfn=k-1, dfd=n-k)
>>> val = np.linspace(0, 5, 100)
>>> pdf = dist.pdf(val)
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> def plot(ax): # we'll reuse this
... ax.plot(val, pdf, color='C0')
... ax.set_title("Levene Test Null Distribution")
... ax.set_xlabel("statistic")
... ax.set_ylabel("probability density")
... ax.set_xlim(0, 5)
... ax.set_ylim(0, 1)
>>> plot(ax)
>>> plt.show()
The comparison is quantified by the p-value: the proportion of values in
the null distribution greater than or equal to the observed value of the
statistic.
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> plot(ax)
>>> pvalue = dist.sf(res.statistic)
>>> annotation = (f'p-value={pvalue:.3f}\n(shaded area)')
>>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
>>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props)
>>> i = val >= res.statistic
>>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0')
>>> plt.show()
>>> res.pvalue
0.5280694573759905
If the p-value is "small" - that is, if there is a low probability of
sampling data from distributions with identical variances that produces
such an extreme value of the statistic - this may be taken as evidence
against the null hypothesis in favor of the alternative: the variances of
the groups are not equal. Note that:
- The inverse is not true; that is, the test is not used to provide
evidence for the null hypothesis.
- The threshold for values that will be considered "small" is a choice that
should be made before the data is analyzed [5]_ with consideration of the
risks of both false positives (incorrectly rejecting the null hypothesis)
and false negatives (failure to reject a false null hypothesis).
- Small p-values are not evidence for a *large* effect; rather, they can
only provide evidence for a "significant" effect, meaning that they are
unlikely to have occurred under the null hypothesis.
Note that the F distribution provides an asymptotic approximation of the
null distribution.
For small samples, it may be more appropriate to perform a permutation
test: Under the null hypothesis that all three samples were drawn from
the same population, each of the measurements is equally likely to have
been observed in any of the three samples. Therefore, we can form a
randomized null distribution by calculating the statistic under many
randomly-generated partitionings of the observations into the three
samples.
>>> def statistic(*samples):
... return stats.levene(*samples).statistic
>>> ref = stats.permutation_test(
... (small_dose, medium_dose, large_dose), statistic,
... permutation_type='independent', alternative='greater'
... )
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> plot(ax)
>>> bins = np.linspace(0, 5, 25)
>>> ax.hist(
... ref.null_distribution, bins=bins, density=True, facecolor="C1"
... )
>>> ax.legend(['aymptotic approximation\n(many observations)',
... 'randomized null distribution'])
>>> plot(ax)
>>> plt.show()
>>> ref.pvalue # randomized test p-value
0.4559 # may vary
Note that there is significant disagreement between the p-value calculated
here and the asymptotic approximation returned by `levene` above.
The statistical inferences that can be drawn rigorously from a permutation
test are limited; nonetheless, they may be the preferred approach in many
circumstances [6]_.
Following is another generic example where the null hypothesis would be
rejected.
Test whether the lists `a`, `b` and `c` come from populations
with equal variances.
>>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
>>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
>>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
>>> stat, p = stats.levene(a, b, c)
>>> p
0.002431505967249681
The small p-value suggests that the populations do not have equal
variances.
This is not surprising, given that the sample variance of `b` is much
larger than that of `a` and `c`:
>>> [np.var(x, ddof=1) for x in [a, b, c]]
[0.007054444444444413, 0.13073888888888888, 0.008890000000000002]
"""
if center not in ['mean', 'median', 'trimmed']:
raise ValueError("center must be 'mean', 'median' or 'trimmed'.")
k = len(samples)
if k < 2:
raise ValueError("Must enter at least two input sample vectors.")
Ni = np.empty(k)
Yci = np.empty(k, 'd')
if center == 'median':
def func(x):
return np.median(x, axis=0)
elif center == 'mean':
def func(x):
return np.mean(x, axis=0)
else: # center == 'trimmed'
samples = tuple(_stats_py.trimboth(np.sort(sample), proportiontocut)
for sample in samples)
def func(x):
return np.mean(x, axis=0)
for j in range(k):
Ni[j] = len(samples[j])
Yci[j] = func(samples[j])
Ntot = np.sum(Ni, axis=0)
# compute Zij's
Zij = [None] * k
for i in range(k):
Zij[i] = abs(asarray(samples[i]) - Yci[i])
# compute Zbari
Zbari = np.empty(k, 'd')
Zbar = 0.0
for i in range(k):
Zbari[i] = np.mean(Zij[i], axis=0)
Zbar += Zbari[i] * Ni[i]
Zbar /= Ntot
numer = (Ntot - k) * np.sum(Ni * (Zbari - Zbar)**2, axis=0)
# compute denom_variance
dvar = 0.0
for i in range(k):
dvar += np.sum((Zij[i] - Zbari[i])**2, axis=0)
denom = (k - 1.0) * dvar
W = numer / denom
pval = distributions.f.sf(W, k-1, Ntot-k) # 1 - cdf
return LeveneResult(W, pval)
def _apply_func(x, g, func):
# g is list of indices into x
# separating x into different groups
# func should be applied over the groups
g = unique(r_[0, g, len(x)])
output = [func(x[g[k]:g[k+1]]) for k in range(len(g) - 1)]
return asarray(output)
FlignerResult = namedtuple('FlignerResult', ('statistic', 'pvalue'))
@_axis_nan_policy_factory(FlignerResult, n_samples=None)
def fligner(*samples, center='median', proportiontocut=0.05):
r"""Perform Fligner-Killeen test for equality of variance.
Fligner's test tests the null hypothesis that all input samples
are from populations with equal variances. Fligner-Killeen's test is
distribution free when populations are identical [2]_.
Parameters
----------
sample1, sample2, ... : array_like
Arrays of sample data. Need not be the same length.
center : {'mean', 'median', 'trimmed'}, optional
Keyword argument controlling which function of the data is used in
computing the test statistic. The default is 'median'.
proportiontocut : float, optional
When `center` is 'trimmed', this gives the proportion of data points
to cut from each end. (See `scipy.stats.trim_mean`.)
Default is 0.05.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value for the hypothesis test.
See Also
--------
bartlett : A parametric test for equality of k variances in normal samples
levene : A robust parametric test for equality of k variances
Notes
-----
As with Levene's test there are three variants of Fligner's test that
differ by the measure of central tendency used in the test. See `levene`
for more information.
Conover et al. (1981) examine many of the existing parametric and
nonparametric tests by extensive simulations and they conclude that the
tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be
superior in terms of robustness of departures from normality and power
[3]_.
References
----------
.. [1] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
Hypothesis Testing based on Quadratic Inference Function. Technical
Report #99-03, Center for Likelihood Studies, Pennsylvania State
University.
https://cecas.clemson.edu/~cspark/cv/paper/qif/draftqif2.pdf
.. [2] Fligner, M.A. and Killeen, T.J. (1976). Distribution-free two-sample
tests for scale. 'Journal of the American Statistical Association.'
71(353), 210-213.
.. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
Hypothesis Testing based on Quadratic Inference Function. Technical
Report #99-03, Center for Likelihood Studies, Pennsylvania State
University.
.. [4] Conover, W. J., Johnson, M. E. and Johnson M. M. (1981). A
comparative study of tests for homogeneity of variances, with
applications to the outer continental shelf bidding data.
Technometrics, 23(4), 351-361.
.. [5] C.I. BLISS (1952), The Statistics of Bioassay: With Special
Reference to the Vitamins, pp 499-503,
:doi:`10.1016/C2013-0-12584-6`.
.. [6] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be
Zero: Calculating Exact P-values When Permutations Are Randomly
Drawn." Statistical Applications in Genetics and Molecular Biology
9.1 (2010).
.. [7] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are
superior to t and F tests in biomedical research. The American
Statistician, 52(2), 127-132.
Examples
--------
In [5]_, the influence of vitamin C on the tooth growth of guinea pigs
was investigated. In a control study, 60 subjects were divided into
small dose, medium dose, and large dose groups that received
daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively.
After 42 days, the tooth growth was measured.
The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record
tooth growth measurements of the three groups in microns.
>>> import numpy as np
>>> small_dose = np.array([
... 4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7,
... 15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7
... ])
>>> medium_dose = np.array([
... 16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5,
... 19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3
... ])
>>> large_dose = np.array([
... 23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5,
... 25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23
... ])
The `fligner` statistic is sensitive to differences in variances
between the samples.
>>> from scipy import stats
>>> res = stats.fligner(small_dose, medium_dose, large_dose)
>>> res.statistic
1.3878943408857916
The value of the statistic tends to be high when there is a large
difference in variances.
We can test for inequality of variance among the groups by comparing the
observed value of the statistic against the null distribution: the
distribution of statistic values derived under the null hypothesis that
the population variances of the three groups are equal.
For this test, the null distribution follows the chi-square distribution
as shown below.
>>> import matplotlib.pyplot as plt
>>> k = 3 # number of samples
>>> dist = stats.chi2(df=k-1)
>>> val = np.linspace(0, 8, 100)
>>> pdf = dist.pdf(val)
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> def plot(ax): # we'll reuse this
... ax.plot(val, pdf, color='C0')
... ax.set_title("Fligner Test Null Distribution")
... ax.set_xlabel("statistic")
... ax.set_ylabel("probability density")
... ax.set_xlim(0, 8)
... ax.set_ylim(0, 0.5)
>>> plot(ax)
>>> plt.show()
The comparison is quantified by the p-value: the proportion of values in
the null distribution greater than or equal to the observed value of the
statistic.
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> plot(ax)
>>> pvalue = dist.sf(res.statistic)
>>> annotation = (f'p-value={pvalue:.4f}\n(shaded area)')
>>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
>>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props)
>>> i = val >= res.statistic
>>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0')
>>> plt.show()
>>> res.pvalue
0.49960016501182125
If the p-value is "small" - that is, if there is a low probability of
sampling data from distributions with identical variances that produces
such an extreme value of the statistic - this may be taken as evidence
against the null hypothesis in favor of the alternative: the variances of
the groups are not equal. Note that:
- The inverse is not true; that is, the test is not used to provide
evidence for the null hypothesis.
- The threshold for values that will be considered "small" is a choice that
should be made before the data is analyzed [6]_ with consideration of the
risks of both false positives (incorrectly rejecting the null hypothesis)
and false negatives (failure to reject a false null hypothesis).
- Small p-values are not evidence for a *large* effect; rather, they can
only provide evidence for a "significant" effect, meaning that they are
unlikely to have occurred under the null hypothesis.
Note that the chi-square distribution provides an asymptotic approximation
of the null distribution.
For small samples, it may be more appropriate to perform a
permutation test: Under the null hypothesis that all three samples were
drawn from the same population, each of the measurements is equally likely
to have been observed in any of the three samples. Therefore, we can form
a randomized null distribution by calculating the statistic under many
randomly-generated partitionings of the observations into the three
samples.
>>> def statistic(*samples):
... return stats.fligner(*samples).statistic
>>> ref = stats.permutation_test(
... (small_dose, medium_dose, large_dose), statistic,
... permutation_type='independent', alternative='greater'
... )
>>> fig, ax = plt.subplots(figsize=(8, 5))
>>> plot(ax)
>>> bins = np.linspace(0, 8, 25)
>>> ax.hist(
... ref.null_distribution, bins=bins, density=True, facecolor="C1"
... )
>>> ax.legend(['aymptotic approximation\n(many observations)',
... 'randomized null distribution'])
>>> plot(ax)
>>> plt.show()
>>> ref.pvalue # randomized test p-value
0.4332 # may vary
Note that there is significant disagreement between the p-value calculated
here and the asymptotic approximation returned by `fligner` above.
The statistical inferences that can be drawn rigorously from a permutation
test are limited; nonetheless, they may be the preferred approach in many
circumstances [7]_.
Following is another generic example where the null hypothesis would be
rejected.
Test whether the lists `a`, `b` and `c` come from populations
with equal variances.
>>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
>>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
>>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
>>> stat, p = stats.fligner(a, b, c)
>>> p
0.00450826080004775
The small p-value suggests that the populations do not have equal
variances.
This is not surprising, given that the sample variance of `b` is much
larger than that of `a` and `c`:
>>> [np.var(x, ddof=1) for x in [a, b, c]]
[0.007054444444444413, 0.13073888888888888, 0.008890000000000002]
"""
if center not in ['mean', 'median', 'trimmed']:
raise ValueError("center must be 'mean', 'median' or 'trimmed'.")
k = len(samples)
if k < 2:
raise ValueError("Must enter at least two input sample vectors.")
# Handle empty input
for sample in samples:
if sample.size == 0:
NaN = _get_nan(*samples)
return FlignerResult(NaN, NaN)
if center == 'median':
def func(x):
return np.median(x, axis=0)
elif center == 'mean':
def func(x):
return np.mean(x, axis=0)
else: # center == 'trimmed'
samples = tuple(_stats_py.trimboth(sample, proportiontocut)
for sample in samples)
def func(x):
return np.mean(x, axis=0)
Ni = asarray([len(samples[j]) for j in range(k)])
Yci = asarray([func(samples[j]) for j in range(k)])
Ntot = np.sum(Ni, axis=0)
# compute Zij's
Zij = [abs(asarray(samples[i]) - Yci[i]) for i in range(k)]
allZij = []
g = [0]
for i in range(k):
allZij.extend(list(Zij[i]))
g.append(len(allZij))
ranks = _stats_py.rankdata(allZij)
sample = distributions.norm.ppf(ranks / (2*(Ntot + 1.0)) + 0.5)
# compute Aibar
Aibar = _apply_func(sample, g, np.sum) / Ni
anbar = np.mean(sample, axis=0)
varsq = np.var(sample, axis=0, ddof=1)
Xsq = np.sum(Ni * (asarray(Aibar) - anbar)**2.0, axis=0) / varsq
pval = distributions.chi2.sf(Xsq, k - 1) # 1 - cdf
return FlignerResult(Xsq, pval)
@_axis_nan_policy_factory(lambda x1: (x1,), n_samples=4, n_outputs=1)
def _mood_inner_lc(xy, x, diffs, sorted_xy, n, m, N) -> float:
# Obtain the unique values and their frequencies from the pooled samples.
# "a_j, + b_j, = t_j, for j = 1, ... k" where `k` is the number of unique
# classes, and "[t]he number of values associated with the x's and y's in
# the jth class will be denoted by a_j, and b_j respectively."
# (Mielke, 312)
# Reuse previously computed sorted array and `diff` arrays to obtain the
# unique values and counts. Prepend `diffs` with a non-zero to indicate
# that the first element should be marked as not matching what preceded it.
diffs_prep = np.concatenate(([1], diffs))
# Unique elements are where the was a difference between elements in the
# sorted array
uniques = sorted_xy[diffs_prep != 0]
# The count of each element is the bin size for each set of consecutive
# differences where the difference is zero. Replace nonzero differences
# with 1 and then use the cumulative sum to count the indices.
t = np.bincount(np.cumsum(np.asarray(diffs_prep != 0, dtype=int)))[1:]
k = len(uniques)
js = np.arange(1, k + 1, dtype=int)
# the `b` array mentioned in the paper is not used, outside of the
# calculation of `t`, so we do not need to calculate it separately. Here
# we calculate `a`. In plain language, `a[j]` is the number of values in
# `x` that equal `uniques[j]`.
sorted_xyx = np.sort(np.concatenate((xy, x)))
diffs = np.diff(sorted_xyx)
diffs_prep = np.concatenate(([1], diffs))
diff_is_zero = np.asarray(diffs_prep != 0, dtype=int)
xyx_counts = np.bincount(np.cumsum(diff_is_zero))[1:]
a = xyx_counts - t
# "Define .. a_0 = b_0 = t_0 = S_0 = 0" (Mielke 312) so we shift `a`
# and `t` arrays over 1 to allow a first element of 0 to accommodate this
# indexing.
t = np.concatenate(([0], t))
a = np.concatenate(([0], a))
# S is built from `t`, so it does not need a preceding zero added on.
S = np.cumsum(t)
# define a copy of `S` with a prepending zero for later use to avoid
# the need for indexing.
S_i_m1 = np.concatenate(([0], S[:-1]))
# Psi, as defined by the 6th unnumbered equation on page 313 (Mielke).
# Note that in the paper there is an error where the denominator `2` is
# squared when it should be the entire equation.
def psi(indicator):
return (indicator - (N + 1)/2)**2
# define summation range for use in calculation of phi, as seen in sum
# in the unnumbered equation on the bottom of page 312 (Mielke).
s_lower = S[js - 1] + 1
s_upper = S[js] + 1
phi_J = [np.arange(s_lower[idx], s_upper[idx]) for idx in range(k)]
# for every range in the above array, determine the sum of psi(I) for
# every element in the range. Divide all the sums by `t`. Following the
# last unnumbered equation on page 312.
phis = [np.sum(psi(I_j)) for I_j in phi_J] / t[js]
# `T` is equal to a[j] * phi[j], per the first unnumbered equation on
# page 312. `phis` is already in the order based on `js`, so we index
# into `a` with `js` as well.
T = sum(phis * a[js])
# The approximate statistic
E_0_T = n * (N * N - 1) / 12
varM = (m * n * (N + 1.0) * (N ** 2 - 4) / 180 -
m * n / (180 * N * (N - 1)) * np.sum(
t * (t**2 - 1) * (t**2 - 4 + (15 * (N - S - S_i_m1) ** 2))
))
return ((T - E_0_T) / np.sqrt(varM),)
def _mood_too_small(samples, kwargs, axis=-1):
x, y = samples
n = x.shape[axis]
m = y.shape[axis]
N = m + n
return N < 3
@_axis_nan_policy_factory(SignificanceResult, n_samples=2, too_small=_mood_too_small)
def mood(x, y, axis=0, alternative="two-sided"):
"""Perform Mood's test for equal scale parameters.
Mood's two-sample test for scale parameters is a non-parametric
test for the null hypothesis that two samples are drawn from the
same distribution with the same scale parameter.
Parameters
----------
x, y : array_like
Arrays of sample data.
axis : int, optional
The axis along which the samples are tested. `x` and `y` can be of
different length along `axis`.
If `axis` is None, `x` and `y` are flattened and the test is done on
all values in the flattened arrays.
alternative : {'two-sided', 'less', 'greater'}, optional
Defines the alternative hypothesis. Default is 'two-sided'.
The following options are available:
* 'two-sided': the scales of the distributions underlying `x` and `y`
are different.
* 'less': the scale of the distribution underlying `x` is less than
the scale of the distribution underlying `y`.
* 'greater': the scale of the distribution underlying `x` is greater
than the scale of the distribution underlying `y`.
.. versionadded:: 1.7.0
Returns
-------
res : SignificanceResult
An object containing attributes:
statistic : scalar or ndarray
The z-score for the hypothesis test. For 1-D inputs a scalar is
returned.
pvalue : scalar ndarray
The p-value for the hypothesis test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
ansari : A non-parametric test for the equality of 2 variances
bartlett : A parametric test for equality of k variances in normal samples
levene : A parametric test for equality of k variances
Notes
-----
The data are assumed to be drawn from probability distributions ``f(x)``
and ``f(x/s) / s`` respectively, for some probability density function f.
The null hypothesis is that ``s == 1``.
For multi-dimensional arrays, if the inputs are of shapes
``(n0, n1, n2, n3)`` and ``(n0, m1, n2, n3)``, then if ``axis=1``, the
resulting z and p values will have shape ``(n0, n2, n3)``. Note that
``n1`` and ``m1`` don't have to be equal, but the other dimensions do.
References
----------
[1] Mielke, Paul W. "Note on Some Squared Rank Tests with Existing Ties."
Technometrics, vol. 9, no. 2, 1967, pp. 312-14. JSTOR,
https://doi.org/10.2307/1266427. Accessed 18 May 2022.
Examples
--------
>>> import numpy as np
>>> from scipy import stats
>>> rng = np.random.default_rng()
>>> x2 = rng.standard_normal((2, 45, 6, 7))
>>> x1 = rng.standard_normal((2, 30, 6, 7))
>>> res = stats.mood(x1, x2, axis=1)
>>> res.pvalue.shape
(2, 6, 7)
Find the number of points where the difference in scale is not significant:
>>> (res.pvalue > 0.1).sum()
78
Perform the test with different scales:
>>> x1 = rng.standard_normal((2, 30))
>>> x2 = rng.standard_normal((2, 35)) * 10.0
>>> stats.mood(x1, x2, axis=1)
SignificanceResult(statistic=array([-5.76174136, -6.12650783]),
pvalue=array([8.32505043e-09, 8.98287869e-10]))
"""
x = np.asarray(x, dtype=float)
y = np.asarray(y, dtype=float)
if axis < 0:
axis = x.ndim + axis
# Determine shape of the result arrays
res_shape = tuple([x.shape[ax] for ax in range(len(x.shape)) if ax != axis])
if not (res_shape == tuple([y.shape[ax] for ax in range(len(y.shape)) if
ax != axis])):
raise ValueError("Dimensions of x and y on all axes except `axis` "
"should match")
n = x.shape[axis]
m = y.shape[axis]
N = m + n
if N < 3:
raise ValueError("Not enough observations.")
xy = np.concatenate((x, y), axis=axis)
# determine if any of the samples contain ties
sorted_xy = np.sort(xy, axis=axis)
diffs = np.diff(sorted_xy, axis=axis)
if 0 in diffs:
z = np.asarray(_mood_inner_lc(xy, x, diffs, sorted_xy, n, m, N,
axis=axis))
else:
if axis != 0:
xy = np.moveaxis(xy, axis, 0)
xy = xy.reshape(xy.shape[0], -1)
# Generalized to the n-dimensional case by adding the axis argument,
# and using for loops, since rankdata is not vectorized. For improving
# performance consider vectorizing rankdata function.
all_ranks = np.empty_like(xy)
for j in range(xy.shape[1]):
all_ranks[:, j] = _stats_py.rankdata(xy[:, j])
Ri = all_ranks[:n]
M = np.sum((Ri - (N + 1.0) / 2) ** 2, axis=0)
# Approx stat.
mnM = n * (N * N - 1.0) / 12
varM = m * n * (N + 1.0) * (N + 2) * (N - 2) / 180
z = (M - mnM) / sqrt(varM)
pval = _get_pvalue(z, distributions.norm, alternative)
if res_shape == ():
# Return scalars, not 0-D arrays
z = z[0]
pval = pval[0]
else:
z.shape = res_shape
pval.shape = res_shape
return SignificanceResult(z[()], pval[()])
WilcoxonResult = _make_tuple_bunch('WilcoxonResult', ['statistic', 'pvalue'])
def wilcoxon_result_unpacker(res):
if hasattr(res, 'zstatistic'):
return res.statistic, res.pvalue, res.zstatistic
else:
return res.statistic, res.pvalue
def wilcoxon_result_object(statistic, pvalue, zstatistic=None):
res = WilcoxonResult(statistic, pvalue)
if zstatistic is not None:
res.zstatistic = zstatistic
return res
def wilcoxon_outputs(kwds):
method = kwds.get('method', 'auto')
if method == 'approx':
return 3
return 2
@_rename_parameter("mode", "method")
@_axis_nan_policy_factory(
wilcoxon_result_object, paired=True,
n_samples=lambda kwds: 2 if kwds.get('y', None) is not None else 1,
result_to_tuple=wilcoxon_result_unpacker, n_outputs=wilcoxon_outputs,
)
def wilcoxon(x, y=None, zero_method="wilcox", correction=False,
alternative="two-sided", method='auto', *, axis=0):
"""Calculate the Wilcoxon signed-rank test.
The Wilcoxon signed-rank test tests the null hypothesis that two
related paired samples come from the same distribution. In particular,
it tests whether the distribution of the differences ``x - y`` is symmetric
about zero. It is a non-parametric version of the paired T-test.
Parameters
----------
x : array_like
Either the first set of measurements (in which case ``y`` is the second
set of measurements), or the differences between two sets of
measurements (in which case ``y`` is not to be specified.) Must be
one-dimensional.
y : array_like, optional
Either the second set of measurements (if ``x`` is the first set of
measurements), or not specified (if ``x`` is the differences between
two sets of measurements.) Must be one-dimensional.
.. warning::
When `y` is provided, `wilcoxon` calculates the test statistic
based on the ranks of the absolute values of ``d = x - y``.
Roundoff error in the subtraction can result in elements of ``d``
being assigned different ranks even when they would be tied with
exact arithmetic. Rather than passing `x` and `y` separately,
consider computing the difference ``x - y``, rounding as needed to
ensure that only truly unique elements are numerically distinct,
and passing the result as `x`, leaving `y` at the default (None).
zero_method : {"wilcox", "pratt", "zsplit"}, optional
There are different conventions for handling pairs of observations
with equal values ("zero-differences", or "zeros").
* "wilcox": Discards all zero-differences (default); see [4]_.
* "pratt": Includes zero-differences in the ranking process,
but drops the ranks of the zeros (more conservative); see [3]_.
In this case, the normal approximation is adjusted as in [5]_.
* "zsplit": Includes zero-differences in the ranking process and
splits the zero rank between positive and negative ones.
correction : bool, optional
If True, apply continuity correction by adjusting the Wilcoxon rank
statistic by 0.5 towards the mean value when computing the
z-statistic if a normal approximation is used. Default is False.
alternative : {"two-sided", "greater", "less"}, optional
Defines the alternative hypothesis. Default is 'two-sided'.
In the following, let ``d`` represent the difference between the paired
samples: ``d = x - y`` if both ``x`` and ``y`` are provided, or
``d = x`` otherwise.
* 'two-sided': the distribution underlying ``d`` is not symmetric
about zero.
* 'less': the distribution underlying ``d`` is stochastically less
than a distribution symmetric about zero.
* 'greater': the distribution underlying ``d`` is stochastically
greater than a distribution symmetric about zero.
method : {"auto", "exact", "approx"} or `PermutationMethod` instance, optional
Method to calculate the p-value, see Notes. Default is "auto".
axis : int or None, default: 0
If an int, the axis of the input along which to compute the statistic.
The statistic of each axis-slice (e.g. row) of the input will appear
in a corresponding element of the output. If ``None``, the input will
be raveled before computing the statistic.
Returns
-------
An object with the following attributes.
statistic : array_like
If `alternative` is "two-sided", the sum of the ranks of the
differences above or below zero, whichever is smaller.
Otherwise the sum of the ranks of the differences above zero.
pvalue : array_like
The p-value for the test depending on `alternative` and `method`.
zstatistic : array_like
When ``method = 'approx'``, this is the normalized z-statistic::
z = (T - mn - d) / se
where ``T`` is `statistic` as defined above, ``mn`` is the mean of the
distribution under the null hypothesis, ``d`` is a continuity
correction, and ``se`` is the standard error.
When ``method != 'approx'``, this attribute is not available.
See Also
--------
kruskal, mannwhitneyu
Notes
-----
In the following, let ``d`` represent the difference between the paired
samples: ``d = x - y`` if both ``x`` and ``y`` are provided, or ``d = x``
otherwise. Assume that all elements of ``d`` are independent and
identically distributed observations, and all are distinct and nonzero.
- When ``len(d)`` is sufficiently large, the null distribution of the
normalized test statistic (`zstatistic` above) is approximately normal,
and ``method = 'approx'`` can be used to compute the p-value.
- When ``len(d)`` is small, the normal approximation may not be accurate,
and ``method='exact'`` is preferred (at the cost of additional
execution time).
- The default, ``method='auto'``, selects between the two: when
``len(d) <= 50`` and there are no zeros, the exact method is used;
otherwise, the approximate method is used.
The presence of "ties" (i.e. not all elements of ``d`` are unique) or
"zeros" (i.e. elements of ``d`` are zero) changes the null distribution
of the test statistic, and ``method='exact'`` no longer calculates
the exact p-value. If ``method='approx'``, the z-statistic is adjusted
for more accurate comparison against the standard normal, but still,
for finite sample sizes, the standard normal is only an approximation of
the true null distribution of the z-statistic. For such situations, the
`method` parameter also accepts instances `PermutationMethod`. In this
case, the p-value is computed using `permutation_test` with the provided
configuration options and other appropriate settings.
References
----------
.. [1] https://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test
.. [2] Conover, W.J., Practical Nonparametric Statistics, 1971.
.. [3] Pratt, J.W., Remarks on Zeros and Ties in the Wilcoxon Signed
Rank Procedures, Journal of the American Statistical Association,
Vol. 54, 1959, pp. 655-667. :doi:`10.1080/01621459.1959.10501526`
.. [4] Wilcoxon, F., Individual Comparisons by Ranking Methods,
Biometrics Bulletin, Vol. 1, 1945, pp. 80-83. :doi:`10.2307/3001968`
.. [5] Cureton, E.E., The Normal Approximation to the Signed-Rank
Sampling Distribution When Zero Differences are Present,
Journal of the American Statistical Association, Vol. 62, 1967,
pp. 1068-1069. :doi:`10.1080/01621459.1967.10500917`
Examples
--------
In [4]_, the differences in height between cross- and self-fertilized
corn plants is given as follows:
>>> d = [6, 8, 14, 16, 23, 24, 28, 29, 41, -48, 49, 56, 60, -67, 75]
Cross-fertilized plants appear to be higher. To test the null
hypothesis that there is no height difference, we can apply the
two-sided test:
>>> from scipy.stats import wilcoxon
>>> res = wilcoxon(d)
>>> res.statistic, res.pvalue
(24.0, 0.041259765625)
Hence, we would reject the null hypothesis at a confidence level of 5%,
concluding that there is a difference in height between the groups.
To confirm that the median of the differences can be assumed to be
positive, we use:
>>> res = wilcoxon(d, alternative='greater')
>>> res.statistic, res.pvalue
(96.0, 0.0206298828125)
This shows that the null hypothesis that the median is negative can be
rejected at a confidence level of 5% in favor of the alternative that
the median is greater than zero. The p-values above are exact. Using the
normal approximation gives very similar values:
>>> res = wilcoxon(d, method='approx')
>>> res.statistic, res.pvalue
(24.0, 0.04088813291185591)
Note that the statistic changed to 96 in the one-sided case (the sum
of ranks of positive differences) whereas it is 24 in the two-sided
case (the minimum of sum of ranks above and below zero).
In the example above, the differences in height between paired plants are
provided to `wilcoxon` directly. Alternatively, `wilcoxon` accepts two
samples of equal length, calculates the differences between paired
elements, then performs the test. Consider the samples ``x`` and ``y``:
>>> import numpy as np
>>> x = np.array([0.5, 0.825, 0.375, 0.5])
>>> y = np.array([0.525, 0.775, 0.325, 0.55])
>>> res = wilcoxon(x, y, alternative='greater')
>>> res
WilcoxonResult(statistic=5.0, pvalue=0.5625)
Note that had we calculated the differences by hand, the test would have
produced different results:
>>> d = [-0.025, 0.05, 0.05, -0.05]
>>> ref = wilcoxon(d, alternative='greater')
>>> ref
WilcoxonResult(statistic=6.0, pvalue=0.4375)
The substantial difference is due to roundoff error in the results of
``x-y``:
>>> d - (x-y)
array([2.08166817e-17, 6.93889390e-17, 1.38777878e-17, 4.16333634e-17])
Even though we expected all the elements of ``(x-y)[1:]`` to have the same
magnitude ``0.05``, they have slightly different magnitudes in practice,
and therefore are assigned different ranks in the test. Before performing
the test, consider calculating ``d`` and adjusting it as necessary to
ensure that theoretically identically values are not numerically distinct.
For example:
>>> d2 = np.around(x - y, decimals=3)
>>> wilcoxon(d2, alternative='greater')
WilcoxonResult(statistic=6.0, pvalue=0.4375)
"""
return _wilcoxon._wilcoxon_nd(x, y, zero_method, correction, alternative,
method, axis)
MedianTestResult = _make_tuple_bunch(
'MedianTestResult',
['statistic', 'pvalue', 'median', 'table'], []
)
def median_test(*samples, ties='below', correction=True, lambda_=1,
nan_policy='propagate'):
"""Perform a Mood's median test.
Test that two or more samples come from populations with the same median.
Let ``n = len(samples)`` be the number of samples. The "grand median" of
all the data is computed, and a contingency table is formed by
classifying the values in each sample as being above or below the grand
median. The contingency table, along with `correction` and `lambda_`,
are passed to `scipy.stats.chi2_contingency` to compute the test statistic
and p-value.
Parameters
----------
sample1, sample2, ... : array_like
The set of samples. There must be at least two samples.
Each sample must be a one-dimensional sequence containing at least
one value. The samples are not required to have the same length.
ties : str, optional
Determines how values equal to the grand median are classified in
the contingency table. The string must be one of::
"below":
Values equal to the grand median are counted as "below".
"above":
Values equal to the grand median are counted as "above".
"ignore":
Values equal to the grand median are not counted.
The default is "below".
correction : bool, optional
If True, *and* there are just two samples, apply Yates' correction
for continuity when computing the test statistic associated with
the contingency table. Default is True.
lambda_ : float or str, optional
By default, the statistic computed in this test is Pearson's
chi-squared statistic. `lambda_` allows a statistic from the
Cressie-Read power divergence family to be used instead. See
`power_divergence` for details.
Default is 1 (Pearson's chi-squared statistic).
nan_policy : {'propagate', 'raise', 'omit'}, optional
Defines how to handle when input contains nan. 'propagate' returns nan,
'raise' throws an error, 'omit' performs the calculations ignoring nan
values. Default is 'propagate'.
Returns
-------
res : MedianTestResult
An object containing attributes:
statistic : float
The test statistic. The statistic that is returned is determined
by `lambda_`. The default is Pearson's chi-squared statistic.
pvalue : float
The p-value of the test.
median : float
The grand median.
table : ndarray
The contingency table. The shape of the table is (2, n), where
n is the number of samples. The first row holds the counts of the
values above the grand median, and the second row holds the counts
of the values below the grand median. The table allows further
analysis with, for example, `scipy.stats.chi2_contingency`, or with
`scipy.stats.fisher_exact` if there are two samples, without having
to recompute the table. If ``nan_policy`` is "propagate" and there
are nans in the input, the return value for ``table`` is ``None``.
See Also
--------
kruskal : Compute the Kruskal-Wallis H-test for independent samples.
mannwhitneyu : Computes the Mann-Whitney rank test on samples x and y.
Notes
-----
.. versionadded:: 0.15.0
References
----------
.. [1] Mood, A. M., Introduction to the Theory of Statistics. McGraw-Hill
(1950), pp. 394-399.
.. [2] Zar, J. H., Biostatistical Analysis, 5th ed. Prentice Hall (2010).
See Sections 8.12 and 10.15.
Examples
--------
A biologist runs an experiment in which there are three groups of plants.
Group 1 has 16 plants, group 2 has 15 plants, and group 3 has 17 plants.
Each plant produces a number of seeds. The seed counts for each group
are::
Group 1: 10 14 14 18 20 22 24 25 31 31 32 39 43 43 48 49
Group 2: 28 30 31 33 34 35 36 40 44 55 57 61 91 92 99
Group 3: 0 3 9 22 23 25 25 33 34 34 40 45 46 48 62 67 84
The following code applies Mood's median test to these samples.
>>> g1 = [10, 14, 14, 18, 20, 22, 24, 25, 31, 31, 32, 39, 43, 43, 48, 49]
>>> g2 = [28, 30, 31, 33, 34, 35, 36, 40, 44, 55, 57, 61, 91, 92, 99]
>>> g3 = [0, 3, 9, 22, 23, 25, 25, 33, 34, 34, 40, 45, 46, 48, 62, 67, 84]
>>> from scipy.stats import median_test
>>> res = median_test(g1, g2, g3)
The median is
>>> res.median
34.0
and the contingency table is
>>> res.table
array([[ 5, 10, 7],
[11, 5, 10]])
`p` is too large to conclude that the medians are not the same:
>>> res.pvalue
0.12609082774093244
The "G-test" can be performed by passing ``lambda_="log-likelihood"`` to
`median_test`.
>>> res = median_test(g1, g2, g3, lambda_="log-likelihood")
>>> res.pvalue
0.12224779737117837
The median occurs several times in the data, so we'll get a different
result if, for example, ``ties="above"`` is used:
>>> res = median_test(g1, g2, g3, ties="above")
>>> res.pvalue
0.063873276069553273
>>> res.table
array([[ 5, 11, 9],
[11, 4, 8]])
This example demonstrates that if the data set is not large and there
are values equal to the median, the p-value can be sensitive to the
choice of `ties`.
"""
if len(samples) < 2:
raise ValueError('median_test requires two or more samples.')
ties_options = ['below', 'above', 'ignore']
if ties not in ties_options:
raise ValueError(f"invalid 'ties' option '{ties}'; 'ties' must be one "
f"of: {str(ties_options)[1:-1]}")
data = [np.asarray(sample) for sample in samples]
# Validate the sizes and shapes of the arguments.
for k, d in enumerate(data):
if d.size == 0:
raise ValueError("Sample %d is empty. All samples must "
"contain at least one value." % (k + 1))
if d.ndim != 1:
raise ValueError("Sample %d has %d dimensions. All "
"samples must be one-dimensional sequences." %
(k + 1, d.ndim))
cdata = np.concatenate(data)
contains_nan, nan_policy = _contains_nan(cdata, nan_policy)
if contains_nan and nan_policy == 'propagate':
return MedianTestResult(np.nan, np.nan, np.nan, None)
if contains_nan:
grand_median = np.median(cdata[~np.isnan(cdata)])
else:
grand_median = np.median(cdata)
# When the minimum version of numpy supported by scipy is 1.9.0,
# the above if/else statement can be replaced by the single line:
# grand_median = np.nanmedian(cdata)
# Create the contingency table.
table = np.zeros((2, len(data)), dtype=np.int64)
for k, sample in enumerate(data):
sample = sample[~np.isnan(sample)]
nabove = count_nonzero(sample > grand_median)
nbelow = count_nonzero(sample < grand_median)
nequal = sample.size - (nabove + nbelow)
table[0, k] += nabove
table[1, k] += nbelow
if ties == "below":
table[1, k] += nequal
elif ties == "above":
table[0, k] += nequal
# Check that no row or column of the table is all zero.
# Such a table can not be given to chi2_contingency, because it would have
# a zero in the table of expected frequencies.
rowsums = table.sum(axis=1)
if rowsums[0] == 0:
raise ValueError("All values are below the grand median (%r)." %
grand_median)
if rowsums[1] == 0:
raise ValueError("All values are above the grand median (%r)." %
grand_median)
if ties == "ignore":
# We already checked that each sample has at least one value, but it
# is possible that all those values equal the grand median. If `ties`
# is "ignore", that would result in a column of zeros in `table`. We
# check for that case here.
zero_cols = np.nonzero((table == 0).all(axis=0))[0]
if len(zero_cols) > 0:
msg = ("All values in sample %d are equal to the grand "
"median (%r), so they are ignored, resulting in an "
"empty sample." % (zero_cols[0] + 1, grand_median))
raise ValueError(msg)
stat, p, dof, expected = chi2_contingency(table, lambda_=lambda_,
correction=correction)
return MedianTestResult(stat, p, grand_median, table)
def _circfuncs_common(samples, high, low):
# Ensure samples are array-like and size is not zero
if samples.size == 0:
NaN = _get_nan(samples)
return NaN, NaN, NaN
# Recast samples as radians that range between 0 and 2 pi and calculate
# the sine and cosine
sin_samp = sin((samples - low)*2.*pi / (high - low))
cos_samp = cos((samples - low)*2.*pi / (high - low))
return samples, sin_samp, cos_samp
@_axis_nan_policy_factory(
lambda x: x, n_outputs=1, default_axis=None,
result_to_tuple=lambda x: (x,)
)
def circmean(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'):
"""Compute the circular mean for samples in a range.
Parameters
----------
samples : array_like
Input array.
high : float or int, optional
High boundary for the sample range. Default is ``2*pi``.
low : float or int, optional
Low boundary for the sample range. Default is 0.
Returns
-------
circmean : float
Circular mean.
See Also
--------
circstd : Circular standard deviation.
circvar : Circular variance.
Examples
--------
For simplicity, all angles are printed out in degrees.
>>> import numpy as np
>>> from scipy.stats import circmean
>>> import matplotlib.pyplot as plt
>>> angles = np.deg2rad(np.array([20, 30, 330]))
>>> circmean = circmean(angles)
>>> np.rad2deg(circmean)
7.294976657784009
>>> mean = angles.mean()
>>> np.rad2deg(mean)
126.66666666666666
Plot and compare the circular mean against the arithmetic mean.
>>> plt.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
... np.sin(np.linspace(0, 2*np.pi, 500)),
... c='k')
>>> plt.scatter(np.cos(angles), np.sin(angles), c='k')
>>> plt.scatter(np.cos(circmean), np.sin(circmean), c='b',
... label='circmean')
>>> plt.scatter(np.cos(mean), np.sin(mean), c='r', label='mean')
>>> plt.legend()
>>> plt.axis('equal')
>>> plt.show()
"""
samples, sin_samp, cos_samp = _circfuncs_common(samples, high, low)
sin_sum = sin_samp.sum(axis)
cos_sum = cos_samp.sum(axis)
res = arctan2(sin_sum, cos_sum)
res = np.asarray(res)
res[res < 0] += 2*pi
res = res[()]
return res*(high - low)/2.0/pi + low
@_axis_nan_policy_factory(
lambda x: x, n_outputs=1, default_axis=None,
result_to_tuple=lambda x: (x,)
)
def circvar(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'):
"""Compute the circular variance for samples assumed to be in a range.
Parameters
----------
samples : array_like
Input array.
high : float or int, optional
High boundary for the sample range. Default is ``2*pi``.
low : float or int, optional
Low boundary for the sample range. Default is 0.
Returns
-------
circvar : float
Circular variance.
See Also
--------
circmean : Circular mean.
circstd : Circular standard deviation.
Notes
-----
This uses the following definition of circular variance: ``1-R``, where
``R`` is the mean resultant vector. The
returned value is in the range [0, 1], 0 standing for no variance, and 1
for a large variance. In the limit of small angles, this value is similar
to half the 'linear' variance.
References
----------
.. [1] Fisher, N.I. *Statistical analysis of circular data*. Cambridge
University Press, 1993.
Examples
--------
>>> import numpy as np
>>> from scipy.stats import circvar
>>> import matplotlib.pyplot as plt
>>> samples_1 = np.array([0.072, -0.158, 0.077, 0.108, 0.286,
... 0.133, -0.473, -0.001, -0.348, 0.131])
>>> samples_2 = np.array([0.111, -0.879, 0.078, 0.733, 0.421,
... 0.104, -0.136, -0.867, 0.012, 0.105])
>>> circvar_1 = circvar(samples_1)
>>> circvar_2 = circvar(samples_2)
Plot the samples.
>>> fig, (left, right) = plt.subplots(ncols=2)
>>> for image in (left, right):
... image.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
... np.sin(np.linspace(0, 2*np.pi, 500)),
... c='k')
... image.axis('equal')
... image.axis('off')
>>> left.scatter(np.cos(samples_1), np.sin(samples_1), c='k', s=15)
>>> left.set_title(f"circular variance: {np.round(circvar_1, 2)!r}")
>>> right.scatter(np.cos(samples_2), np.sin(samples_2), c='k', s=15)
>>> right.set_title(f"circular variance: {np.round(circvar_2, 2)!r}")
>>> plt.show()
"""
samples, sin_samp, cos_samp = _circfuncs_common(samples, high, low)
sin_mean = sin_samp.mean(axis)
cos_mean = cos_samp.mean(axis)
# hypot can go slightly above 1 due to rounding errors
with np.errstate(invalid='ignore'):
R = np.minimum(1, hypot(sin_mean, cos_mean))
res = 1. - R
return res
@_axis_nan_policy_factory(
lambda x: x, n_outputs=1, default_axis=None,
result_to_tuple=lambda x: (x,)
)
def circstd(samples, high=2*pi, low=0, axis=None, nan_policy='propagate', *,
normalize=False):
"""
Compute the circular standard deviation for samples assumed to be in the
range [low to high].
Parameters
----------
samples : array_like
Input array.
high : float or int, optional
High boundary for the sample range. Default is ``2*pi``.
low : float or int, optional
Low boundary for the sample range. Default is 0.
normalize : boolean, optional
If True, the returned value is equal to ``sqrt(-2*log(R))`` and does
not depend on the variable units. If False (default), the returned
value is scaled by ``((high-low)/(2*pi))``.
Returns
-------
circstd : float
Circular standard deviation.
See Also
--------
circmean : Circular mean.
circvar : Circular variance.
Notes
-----
This uses a definition of circular standard deviation from [1]_.
Essentially, the calculation is as follows.
.. code-block:: python
import numpy as np
C = np.cos(samples).mean()
S = np.sin(samples).mean()
R = np.sqrt(C**2 + S**2)
l = 2*np.pi / (high-low)
circstd = np.sqrt(-2*np.log(R)) / l
In the limit of small angles, it returns a number close to the 'linear'
standard deviation.
References
----------
.. [1] Mardia, K. V. (1972). 2. In *Statistics of Directional Data*
(pp. 18-24). Academic Press. :doi:`10.1016/C2013-0-07425-7`.
Examples
--------
>>> import numpy as np
>>> from scipy.stats import circstd
>>> import matplotlib.pyplot as plt
>>> samples_1 = np.array([0.072, -0.158, 0.077, 0.108, 0.286,
... 0.133, -0.473, -0.001, -0.348, 0.131])
>>> samples_2 = np.array([0.111, -0.879, 0.078, 0.733, 0.421,
... 0.104, -0.136, -0.867, 0.012, 0.105])
>>> circstd_1 = circstd(samples_1)
>>> circstd_2 = circstd(samples_2)
Plot the samples.
>>> fig, (left, right) = plt.subplots(ncols=2)
>>> for image in (left, right):
... image.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
... np.sin(np.linspace(0, 2*np.pi, 500)),
... c='k')
... image.axis('equal')
... image.axis('off')
>>> left.scatter(np.cos(samples_1), np.sin(samples_1), c='k', s=15)
>>> left.set_title(f"circular std: {np.round(circstd_1, 2)!r}")
>>> right.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
... np.sin(np.linspace(0, 2*np.pi, 500)),
... c='k')
>>> right.scatter(np.cos(samples_2), np.sin(samples_2), c='k', s=15)
>>> right.set_title(f"circular std: {np.round(circstd_2, 2)!r}")
>>> plt.show()
"""
samples, sin_samp, cos_samp = _circfuncs_common(samples, high, low)
sin_mean = sin_samp.mean(axis) # [1] (2.2.3)
cos_mean = cos_samp.mean(axis) # [1] (2.2.3)
# hypot can go slightly above 1 due to rounding errors
with np.errstate(invalid='ignore'):
R = np.minimum(1, hypot(sin_mean, cos_mean)) # [1] (2.2.4)
res = sqrt(-2*log(R))
if not normalize:
res *= (high-low)/(2.*pi) # [1] (2.3.14) w/ (2.3.7)
return res
class DirectionalStats:
def __init__(self, mean_direction, mean_resultant_length):
self.mean_direction = mean_direction
self.mean_resultant_length = mean_resultant_length
def __repr__(self):
return (f"DirectionalStats(mean_direction={self.mean_direction},"
f" mean_resultant_length={self.mean_resultant_length})")
def directional_stats(samples, *, axis=0, normalize=True):
"""
Computes sample statistics for directional data.
Computes the directional mean (also called the mean direction vector) and
mean resultant length of a sample of vectors.
The directional mean is a measure of "preferred direction" of vector data.
It is analogous to the sample mean, but it is for use when the length of
the data is irrelevant (e.g. unit vectors).
The mean resultant length is a value between 0 and 1 used to quantify the
dispersion of directional data: the smaller the mean resultant length, the
greater the dispersion. Several definitions of directional variance
involving the mean resultant length are given in [1]_ and [2]_.
Parameters
----------
samples : array_like
Input array. Must be at least two-dimensional, and the last axis of the
input must correspond with the dimensionality of the vector space.
When the input is exactly two dimensional, this means that each row
of the data is a vector observation.
axis : int, default: 0
Axis along which the directional mean is computed.
normalize: boolean, default: True
If True, normalize the input to ensure that each observation is a
unit vector. It the observations are already unit vectors, consider
setting this to False to avoid unnecessary computation.
Returns
-------
res : DirectionalStats
An object containing attributes:
mean_direction : ndarray
Directional mean.
mean_resultant_length : ndarray
The mean resultant length [1]_.
See Also
--------
circmean: circular mean; i.e. directional mean for 2D *angles*
circvar: circular variance; i.e. directional variance for 2D *angles*
Notes
-----
This uses a definition of directional mean from [1]_.
Assuming the observations are unit vectors, the calculation is as follows.
.. code-block:: python
mean = samples.mean(axis=0)
mean_resultant_length = np.linalg.norm(mean)
mean_direction = mean / mean_resultant_length
This definition is appropriate for *directional* data (i.e. vector data
for which the magnitude of each observation is irrelevant) but not
for *axial* data (i.e. vector data for which the magnitude and *sign* of
each observation is irrelevant).
Several definitions of directional variance involving the mean resultant
length ``R`` have been proposed, including ``1 - R`` [1]_, ``1 - R**2``
[2]_, and ``2 * (1 - R)`` [2]_. Rather than choosing one, this function
returns ``R`` as attribute `mean_resultant_length` so the user can compute
their preferred measure of dispersion.
References
----------
.. [1] Mardia, Jupp. (2000). *Directional Statistics*
(p. 163). Wiley.
.. [2] https://en.wikipedia.org/wiki/Directional_statistics
Examples
--------
>>> import numpy as np
>>> from scipy.stats import directional_stats
>>> data = np.array([[3, 4], # first observation, 2D vector space
... [6, -8]]) # second observation
>>> dirstats = directional_stats(data)
>>> dirstats.mean_direction
array([1., 0.])
In contrast, the regular sample mean of the vectors would be influenced
by the magnitude of each observation. Furthermore, the result would not be
a unit vector.
>>> data.mean(axis=0)
array([4.5, -2.])
An exemplary use case for `directional_stats` is to find a *meaningful*
center for a set of observations on a sphere, e.g. geographical locations.
>>> data = np.array([[0.8660254, 0.5, 0.],
... [0.8660254, -0.5, 0.]])
>>> dirstats = directional_stats(data)
>>> dirstats.mean_direction
array([1., 0., 0.])
The regular sample mean on the other hand yields a result which does not
lie on the surface of the sphere.
>>> data.mean(axis=0)
array([0.8660254, 0., 0.])
The function also returns the mean resultant length, which
can be used to calculate a directional variance. For example, using the
definition ``Var(z) = 1 - R`` from [2]_ where ``R`` is the
mean resultant length, we can calculate the directional variance of the
vectors in the above example as:
>>> 1 - dirstats.mean_resultant_length
0.13397459716167093
"""
samples = np.asarray(samples)
if samples.ndim < 2:
raise ValueError("samples must at least be two-dimensional. "
f"Instead samples has shape: {samples.shape!r}")
samples = np.moveaxis(samples, axis, 0)
if normalize:
vectornorms = np.linalg.norm(samples, axis=-1, keepdims=True)
samples = samples/vectornorms
mean = np.mean(samples, axis=0)
mean_resultant_length = np.linalg.norm(mean, axis=-1, keepdims=True)
mean_direction = mean / mean_resultant_length
return DirectionalStats(mean_direction,
mean_resultant_length.squeeze(-1)[()])
def false_discovery_control(ps, *, axis=0, method='bh'):
"""Adjust p-values to control the false discovery rate.
The false discovery rate (FDR) is the expected proportion of rejected null
hypotheses that are actually true.
If the null hypothesis is rejected when the *adjusted* p-value falls below
a specified level, the false discovery rate is controlled at that level.
Parameters
----------
ps : 1D array_like
The p-values to adjust. Elements must be real numbers between 0 and 1.
axis : int
The axis along which to perform the adjustment. The adjustment is
performed independently along each axis-slice. If `axis` is None, `ps`
is raveled before performing the adjustment.
method : {'bh', 'by'}
The false discovery rate control procedure to apply: ``'bh'`` is for
Benjamini-Hochberg [1]_ (Eq. 1), ``'by'`` is for Benjaminini-Yekutieli
[2]_ (Theorem 1.3). The latter is more conservative, but it is
guaranteed to control the FDR even when the p-values are not from
independent tests.
Returns
-------
ps_adusted : array_like
The adjusted p-values. If the null hypothesis is rejected where these
fall below a specified level, the false discovery rate is controlled
at that level.
See Also
--------
combine_pvalues
statsmodels.stats.multitest.multipletests
Notes
-----
In multiple hypothesis testing, false discovery control procedures tend to
offer higher power than familywise error rate control procedures (e.g.
Bonferroni correction [1]_).
If the p-values correspond with independent tests (or tests with
"positive regression dependencies" [2]_), rejecting null hypotheses
corresponding with Benjamini-Hochberg-adjusted p-values below :math:`q`
controls the false discovery rate at a level less than or equal to
:math:`q m_0 / m`, where :math:`m_0` is the number of true null hypotheses
and :math:`m` is the total number of null hypotheses tested. The same is
true even for dependent tests when the p-values are adjusted accorded to
the more conservative Benjaminini-Yekutieli procedure.
The adjusted p-values produced by this function are comparable to those
produced by the R function ``p.adjust`` and the statsmodels function
`statsmodels.stats.multitest.multipletests`. Please consider the latter
for more advanced methods of multiple comparison correction.
References
----------
.. [1] Benjamini, Yoav, and Yosef Hochberg. "Controlling the false
discovery rate: a practical and powerful approach to multiple
testing." Journal of the Royal statistical society: series B
(Methodological) 57.1 (1995): 289-300.
.. [2] Benjamini, Yoav, and Daniel Yekutieli. "The control of the false
discovery rate in multiple testing under dependency." Annals of
statistics (2001): 1165-1188.
.. [3] TileStats. FDR - Benjamini-Hochberg explained - Youtube.
https://www.youtube.com/watch?v=rZKa4tW2NKs.
.. [4] Neuhaus, Karl-Ludwig, et al. "Improved thrombolysis in acute
myocardial infarction with front-loaded administration of alteplase:
results of the rt-PA-APSAC patency study (TAPS)." Journal of the
American College of Cardiology 19.5 (1992): 885-891.
Examples
--------
We follow the example from [1]_.
Thrombolysis with recombinant tissue-type plasminogen activator (rt-PA)
and anisoylated plasminogen streptokinase activator (APSAC) in
myocardial infarction has been proved to reduce mortality. [4]_
investigated the effects of a new front-loaded administration of rt-PA
versus those obtained with a standard regimen of APSAC, in a randomized
multicentre trial in 421 patients with acute myocardial infarction.
There were four families of hypotheses tested in the study, the last of
which was "cardiac and other events after the start of thrombolitic
treatment". FDR control may be desired in this family of hypotheses
because it would not be appropriate to conclude that the front-loaded
treatment is better if it is merely equivalent to the previous treatment.
The p-values corresponding with the 15 hypotheses in this family were
>>> ps = [0.0001, 0.0004, 0.0019, 0.0095, 0.0201, 0.0278, 0.0298, 0.0344,
... 0.0459, 0.3240, 0.4262, 0.5719, 0.6528, 0.7590, 1.000]
If the chosen significance level is 0.05, we may be tempted to reject the
null hypotheses for the tests corresponding with the first nine p-values,
as the first nine p-values fall below the chosen significance level.
However, this would ignore the problem of "multiplicity": if we fail to
correct for the fact that multiple comparisons are being performed, we
are more likely to incorrectly reject true null hypotheses.
One approach to the multiplicity problem is to control the family-wise
error rate (FWER), that is, the rate at which the null hypothesis is
rejected when it is actually true. A common procedure of this kind is the
Bonferroni correction [1]_. We begin by multiplying the p-values by the
number of hypotheses tested.
>>> import numpy as np
>>> np.array(ps) * len(ps)
array([1.5000e-03, 6.0000e-03, 2.8500e-02, 1.4250e-01, 3.0150e-01,
4.1700e-01, 4.4700e-01, 5.1600e-01, 6.8850e-01, 4.8600e+00,
6.3930e+00, 8.5785e+00, 9.7920e+00, 1.1385e+01, 1.5000e+01])
To control the FWER at 5%, we reject only the hypotheses corresponding
with adjusted p-values less than 0.05. In this case, only the hypotheses
corresponding with the first three p-values can be rejected. According to
[1]_, these three hypotheses concerned "allergic reaction" and "two
different aspects of bleeding."
An alternative approach is to control the false discovery rate: the
expected fraction of rejected null hypotheses that are actually true. The
advantage of this approach is that it typically affords greater power: an
increased rate of rejecting the null hypothesis when it is indeed false. To
control the false discovery rate at 5%, we apply the Benjamini-Hochberg
p-value adjustment.
>>> from scipy import stats
>>> stats.false_discovery_control(ps)
array([0.0015 , 0.003 , 0.0095 , 0.035625 , 0.0603 ,
0.06385714, 0.06385714, 0.0645 , 0.0765 , 0.486 ,
0.58118182, 0.714875 , 0.75323077, 0.81321429, 1. ])
Now, the first *four* adjusted p-values fall below 0.05, so we would reject
the null hypotheses corresponding with these *four* p-values. Rejection
of the fourth null hypothesis was particularly important to the original
study as it led to the conclusion that the new treatment had a
"substantially lower in-hospital mortality rate."
"""
# Input Validation and Special Cases
ps = np.asarray(ps)
ps_in_range = (np.issubdtype(ps.dtype, np.number)
and np.all(ps == np.clip(ps, 0, 1)))
if not ps_in_range:
raise ValueError("`ps` must include only numbers between 0 and 1.")
methods = {'bh', 'by'}
if method.lower() not in methods:
raise ValueError(f"Unrecognized `method` '{method}'."
f"Method must be one of {methods}.")
method = method.lower()
if axis is None:
axis = 0
ps = ps.ravel()
axis = np.asarray(axis)[()]
if not np.issubdtype(axis.dtype, np.integer) or axis.size != 1:
raise ValueError("`axis` must be an integer or `None`")
if ps.size <= 1 or ps.shape[axis] <= 1:
return ps[()]
ps = np.moveaxis(ps, axis, -1)
m = ps.shape[-1]
# Main Algorithm
# Equivalent to the ideas of [1] and [2], except that this adjusts the
# p-values as described in [3]. The results are similar to those produced
# by R's p.adjust.
# "Let [ps] be the ordered observed p-values..."
order = np.argsort(ps, axis=-1)
ps = np.take_along_axis(ps, order, axis=-1) # this copies ps
# Equation 1 of [1] rearranged to reject when p is less than specified q
i = np.arange(1, m+1)
ps *= m / i
# Theorem 1.3 of [2]
if method == 'by':
ps *= np.sum(1 / i)
# accounts for rejecting all null hypotheses i for i < k, where k is
# defined in Eq. 1 of either [1] or [2]. See [3]. Starting with the index j
# of the second to last element, we replace element j with element j+1 if
# the latter is smaller.
np.minimum.accumulate(ps[..., ::-1], out=ps[..., ::-1], axis=-1)
# Restore original order of axes and data
np.put_along_axis(ps, order, values=ps.copy(), axis=-1)
ps = np.moveaxis(ps, -1, axis)
return np.clip(ps, 0, 1)
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