import numpy as np from scipy.linalg import solve, LinAlgWarning import warnings __all__ = ['nnls'] def nnls(A, b, maxiter=None, *, atol=None): """ Solve ``argmin_x || Ax - b ||_2`` for ``x>=0``. This problem, often called as NonNegative Least Squares, is a convex optimization problem with convex constraints. It typically arises when the ``x`` models quantities for which only nonnegative values are attainable; weight of ingredients, component costs and so on. Parameters ---------- A : (m, n) ndarray Coefficient array b : (m,) ndarray, float Right-hand side vector. maxiter: int, optional Maximum number of iterations, optional. Default value is ``3 * n``. atol: float Tolerance value used in the algorithm to assess closeness to zero in the projected residual ``(A.T @ (A x - b)`` entries. Increasing this value relaxes the solution constraints. A typical relaxation value can be selected as ``max(m, n) * np.linalg.norm(a, 1) * np.spacing(1.)``. This value is not set as default since the norm operation becomes expensive for large problems hence can be used only when necessary. Returns ------- x : ndarray Solution vector. rnorm : float The 2-norm of the residual, ``|| Ax-b ||_2``. See Also -------- lsq_linear : Linear least squares with bounds on the variables Notes ----- The code is based on [2]_ which is an improved version of the classical algorithm of [1]_. It utilizes an active set method and solves the KKT (Karush-Kuhn-Tucker) conditions for the non-negative least squares problem. References ---------- .. [1] : Lawson C., Hanson R.J., "Solving Least Squares Problems", SIAM, 1995, :doi:`10.1137/1.9781611971217` .. [2] : Bro, Rasmus and de Jong, Sijmen, "A Fast Non-Negativity- Constrained Least Squares Algorithm", Journal Of Chemometrics, 1997, :doi:`10.1002/(SICI)1099-128X(199709/10)11:5<393::AID-CEM483>3.0.CO;2-L` Examples -------- >>> import numpy as np >>> from scipy.optimize import nnls ... >>> A = np.array([[1, 0], [1, 0], [0, 1]]) >>> b = np.array([2, 1, 1]) >>> nnls(A, b) (array([1.5, 1. ]), 0.7071067811865475) >>> b = np.array([-1, -1, -1]) >>> nnls(A, b) (array([0., 0.]), 1.7320508075688772) """ A = np.asarray_chkfinite(A) b = np.asarray_chkfinite(b) if len(A.shape) != 2: raise ValueError("Expected a two-dimensional array (matrix)" + f", but the shape of A is {A.shape}") if len(b.shape) != 1: raise ValueError("Expected a one-dimensional array (vector)" + f", but the shape of b is {b.shape}") m, n = A.shape if m != b.shape[0]: raise ValueError( "Incompatible dimensions. The first dimension of " + f"A is {m}, while the shape of b is {(b.shape[0], )}") x, rnorm, mode = _nnls(A, b, maxiter, tol=atol) if mode != 1: raise RuntimeError("Maximum number of iterations reached.") return x, rnorm def _nnls(A, b, maxiter=None, tol=None): """ This is a single RHS algorithm from ref [2] above. For multiple RHS support, the algorithm is given in :doi:`10.1002/cem.889` """ m, n = A.shape AtA = A.T @ A Atb = b @ A # Result is 1D - let NumPy figure it out if not maxiter: maxiter = 3*n if tol is None: tol = 10 * max(m, n) * np.spacing(1.) # Initialize vars x = np.zeros(n, dtype=np.float64) s = np.zeros(n, dtype=np.float64) # Inactive constraint switches P = np.zeros(n, dtype=bool) # Projected residual w = Atb.copy().astype(np.float64) # x=0. Skip (-AtA @ x) term # Overall iteration counter # Outer loop is not counted, inner iter is counted across outer spins iter = 0 while (not P.all()) and (w[~P] > tol).any(): # B # Get the "most" active coeff index and move to inactive set k = np.argmax(w * (~P)) # B.2 P[k] = True # B.3 # Iteration solution s[:] = 0. # B.4 with warnings.catch_warnings(): warnings.filterwarnings('ignore', message='Ill-conditioned matrix', category=LinAlgWarning) s[P] = solve(AtA[np.ix_(P, P)], Atb[P], assume_a='sym', check_finite=False) # Inner loop while (iter < maxiter) and (s[P].min() < 0): # C.1 iter += 1 inds = P * (s < 0) alpha = (x[inds] / (x[inds] - s[inds])).min() # C.2 x *= (1 - alpha) x += alpha*s P[x <= tol] = False with warnings.catch_warnings(): warnings.filterwarnings('ignore', message='Ill-conditioned matrix', category=LinAlgWarning) s[P] = solve(AtA[np.ix_(P, P)], Atb[P], assume_a='sym', check_finite=False) s[~P] = 0 # C.6 x[:] = s[:] w[:] = Atb - AtA @ x if iter == maxiter: # Typically following line should return # return x, np.linalg.norm(A@x - b), -1 # however at the top level, -1 raises an exception wasting norm # Instead return dummy number 0. return x, 0., -1 return x, np.linalg.norm(A@x - b), 1