242 lines
8.3 KiB
Python
242 lines
8.3 KiB
Python
# -*- coding: utf-8 -*-
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# Copyright (c) 2003, Taro Ogawa. All Rights Reserved.
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# Copyright (c) 2013, Savoir-faire Linux inc. All Rights Reserved.
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# This library is free software; you can redistribute it and/or
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# modify it under the terms of the GNU Lesser General Public
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# License as published by the Free Software Foundation; either
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# version 2.1 of the License, or (at your option) any later version.
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# This library is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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# Lesser General Public License for more details.
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# You should have received a copy of the GNU Lesser General Public
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# License along with this library; if not, write to the Free Software
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# Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston,
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# MA 02110-1301 USA
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from __future__ import division, unicode_literals
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import re
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from .lang_EU import Num2Word_EU
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DOLLAR = ('dólar', 'dólares')
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CENTS = ('cêntimo', 'cêntimos')
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class Num2Word_PT(Num2Word_EU):
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CURRENCY_FORMS = {
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'AUD': (DOLLAR, CENTS),
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'CAD': (DOLLAR, CENTS),
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'EUR': (('euro', 'euros'), CENTS),
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'GBP': (('libra', 'libras'), ('péni', 'pence')),
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'USD': (DOLLAR, CENTS),
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}
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GIGA_SUFFIX = None
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MEGA_SUFFIX = "ilião"
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def setup(self):
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super(Num2Word_PT, self).setup()
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lows = ["quatr", "tr", "b", "m"]
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self.high_numwords = self.gen_high_numwords([], [], lows)
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self.negword = "menos "
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self.pointword = "vírgula"
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self.exclude_title = ["e", "vírgula", "menos"]
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self.mid_numwords = [
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(1000, "mil"), (100, "cem"), (90, "noventa"),
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(80, "oitenta"), (70, "setenta"), (60, "sessenta"),
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(50, "cinquenta"), (40, "quarenta"), (30, "trinta")
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]
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self.low_numwords = [
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"vinte", "dezanove", "dezoito", "dezassete", "dezasseis",
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"quinze", "catorze", "treze", "doze", "onze", "dez",
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"nove", "oito", "sete", "seis", "cinco", "quatro", "três", "dois",
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"um", "zero"
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]
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self.ords = [
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{
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0: "",
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1: "primeiro",
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2: "segundo",
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3: "terceiro",
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4: "quarto",
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5: "quinto",
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6: "sexto",
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7: "sétimo",
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8: "oitavo",
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9: "nono",
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},
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{
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0: "",
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1: "décimo",
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2: "vigésimo",
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3: "trigésimo",
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4: "quadragésimo",
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5: "quinquagésimo",
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6: "sexagésimo",
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7: "septuagésimo",
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8: "octogésimo",
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9: "nonagésimo",
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},
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{
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0: "",
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1: "centésimo",
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2: "ducentésimo",
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3: "tricentésimo",
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4: "quadrigentésimo",
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5: "quingentésimo",
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6: "seiscentésimo",
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7: "septigentésimo",
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8: "octigentésimo",
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9: "nongentésimo",
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},
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]
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self.thousand_separators = {
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3: "milésimo",
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6: "milionésimo",
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9: "milésimo milionésimo",
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12: "bilionésimo",
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15: "milésimo bilionésimo"
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}
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self.hundreds = {
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1: "cento",
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2: "duzentos",
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3: "trezentos",
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4: "quatrocentos",
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5: "quinhentos",
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6: "seiscentos",
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7: "setecentos",
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8: "oitocentos",
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9: "novecentos",
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}
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def merge(self, curr, next):
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ctext, cnum, ntext, nnum = curr + next
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if cnum == 1:
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if nnum < 1000000:
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return next
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ctext = "um"
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elif cnum == 100 and not nnum % 1000 == 0:
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ctext = "cento"
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if nnum < cnum:
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if cnum < 100:
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return ("%s e %s" % (ctext, ntext), cnum + nnum)
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return ("%s e %s" % (ctext, ntext), cnum + nnum)
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elif (not nnum % 1000000000) and cnum > 1:
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ntext = ntext[:-4] + "liões"
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elif (not nnum % 1000000) and cnum > 1:
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ntext = ntext[:-4] + "lhões"
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# correct "milião" to "milhão"
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if ntext == 'milião':
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ntext = 'milhão'
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if nnum == 100:
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ctext = self.hundreds[cnum]
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ntext = ""
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else:
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ntext = " " + ntext
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return (ctext + ntext, cnum * nnum)
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def to_cardinal(self, value):
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result = super(Num2Word_PT, self).to_cardinal(value)
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# Transforms "mil e cento e catorze" into "mil cento e catorze"
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# Transforms "cem milhões e duzentos mil e duzentos e dez" em "cem
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# milhões duzentos mil duzentos e dez" but "cem milhões e duzentos
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# mil e duzentos" in "cem milhões duzentos mil e duzentos" and not in
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# "cem milhões duzentos mil duzentos"
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for ext in (
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'mil', 'milhão', 'milhões', 'mil milhões',
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'bilião', 'biliões', 'mil biliões'):
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if re.match('.*{} e \\w*entos? (?=.*e)'.format(ext), result):
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result = result.replace(
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'{} e'.format(ext), '{}'.format(ext)
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)
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return result
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# for the ordinal conversion the code is similar to pt_BR code,
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# although there are other rules that are probably more correct in
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# Portugal. Concerning numbers from 2000th on, saying "dois
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# milésimos" instead of "segundo milésimo" (the first number
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# would be used in the cardinal form instead of the ordinal) is better.
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# This was not implemented.
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# source:
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# https://ciberduvidas.iscte-iul.pt/consultorio/perguntas/a-forma-por-extenso-de-2000-e-de-outros-ordinais/16428
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def to_ordinal(self, value):
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# Before changing this function remember this is used by pt-BR
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# so act accordingly
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self.verify_ordinal(value)
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result = []
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value = str(value)
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thousand_separator = ''
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for idx, char in enumerate(value[::-1]):
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if idx and idx % 3 == 0:
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thousand_separator = self.thousand_separators[idx]
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if char != '0' and thousand_separator:
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# avoiding "segundo milionésimo milésimo" for 6000000,
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# for instance
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result.append(thousand_separator)
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thousand_separator = ''
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result.append(self.ords[idx % 3][int(char)])
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result = ' '.join(result[::-1])
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result = result.strip()
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result = re.sub('\\s+', ' ', result)
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if result.startswith('primeiro') and value != '1':
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# avoiding "primeiro milésimo", "primeiro milionésimo" and so on
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result = result[9:]
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return result
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def to_ordinal_num(self, value):
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# Before changing this function remember this is used by pt-BR
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# so act accordingly
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self.verify_ordinal(value)
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return "%sº" % (value)
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def to_year(self, val, longval=True):
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# Before changing this function remember this is used by pt-BR
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# so act accordingly
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if val < 0:
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return self.to_cardinal(abs(val)) + ' antes de Cristo'
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return self.to_cardinal(val)
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def to_currency(self, val, currency='EUR', cents=True, separator=' e',
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adjective=False):
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# change negword because base.to_currency() does not need space after
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backup_negword = self.negword
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self.negword = self.negword[:-1]
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result = super(Num2Word_PT, self).to_currency(
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val, currency=currency, cents=cents, separator=separator,
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adjective=adjective)
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# undo the change on negword
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self.negword = backup_negword
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# transforms "milhões euros" em "milhões de euros"
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cr1, _ = self.CURRENCY_FORMS[currency]
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for ext in (
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'milhão', 'milhões', 'bilião',
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'biliões', 'trilião', 'triliões'):
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if re.match('.*{} (?={})'.format(ext, cr1[1]), result):
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result = result.replace(
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'{}'.format(ext), '{} de'.format(ext), 1
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)
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# do not print "e zero cêntimos"
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result = result.replace(' e zero cêntimos', '')
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return result
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