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