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gumblex/num2chinese.py

Created Feb 8, 2015
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Numbers to Chinese representations converter in Python. 中文数字转换
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num2chinese.py
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# Licensed under WTFPL or the Unlicense or CC0.
# This uses Python 3, but it's easy to port to Python 2 by changing
# strings to u'xx'.
import itertools
def num2chinese(num, big=False, simp=True, o=False, twoalt=False):
"""
Converts numbers to Chinese representations.
`big` : use financial characters.
`simp` : use simplified characters instead of traditional characters.
`o` : use 〇 for zero.
`twoalt`: use 两/兩 for two when appropriate.
Note that `o` and `twoalt` is ignored when `big` is used,
and `twoalt` is ignored when `o` is used for formal representations.
"""
# check num first
nd = str(num)
if abs(float(nd)) >= 1e48:
raise ValueError('number out of range')
elif 'e' in nd:
raise ValueError('scientific notation is not supported')
c_symbol = '正负点' if simp else '正負點'
if o: # formal
twoalt = False
if big:
c_basic = '零壹贰叁肆伍陆柒捌玖' if simp else '零壹貳參肆伍陸柒捌玖'
c_unit1 = '拾佰仟'
c_twoalt = '贰' if simp else '貳'
else:
c_basic = '〇一二三四五六七八九' if o else '零一二三四五六七八九'
c_unit1 = '十百千'
if twoalt:
c_twoalt = '两' if simp else '兩'
else:
c_twoalt = '二'
c_unit2 = '万亿兆京垓秭穰沟涧正载' if simp else '萬億兆京垓秭穰溝澗正載'
revuniq = lambda l: ''.join(k for k, g in itertools.groupby(reversed(l)))
nd = str(num)
result = []
if nd[0] == '+':
result.append(c_symbol[0])
elif nd[0] == '-':
result.append(c_symbol[1])
if '.' in nd:
integer, remainder = nd.lstrip('+-').split('.')
else:
integer, remainder = nd.lstrip('+-'), None
if int(integer):
splitted = [integer[max(i - 4, 0):i]
for i in range(len(integer), 0, -4)]
intresult = []
for nu, unit in enumerate(splitted):
# special cases
if int(unit) == 0: # 0000
intresult.append(c_basic[0])
continue
elif nu > 0 and int(unit) == 2: # 0002
intresult.append(c_twoalt + c_unit2[nu - 1])
continue
ulist = []
unit = unit.zfill(4)
for nc, ch in enumerate(reversed(unit)):
if ch == '0':
if ulist: # ???0
ulist.append(c_basic[0])
elif nc == 0:
ulist.append(c_basic[int(ch)])
elif nc == 1 and ch == '1' and unit[1] == '0':
# special case for tens
# edit the 'elif' if you don't like
# 十四, 三千零十四, 三千三百一十四
ulist.append(c_unit1[0])
elif nc > 1 and ch == '2':
ulist.append(c_twoalt + c_unit1[nc - 1])
else:
ulist.append(c_basic[int(ch)] + c_unit1[nc - 1])
ustr = revuniq(ulist)
if nu == 0:
intresult.append(ustr)
else:
intresult.append(ustr + c_unit2[nu - 1])
result.append(revuniq(intresult).strip(c_basic[0]))
else:
result.append(c_basic[0])
if remainder:
result.append(c_symbol[2])
result.append(''.join(c_basic[int(ch)] for ch in remainder))
return ''.join(result)
@aceimnorstuvwxz
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aceimnorstuvwxz commented Jul 13, 2016

thkx

@sudo-shubham
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sudo-shubham commented Jul 25, 2018

Works. Thanks.

@lkfo415579
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lkfo415579 commented Aug 13, 2018

works. thanks

@GuanJianChun
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GuanJianChun commented Aug 21, 2018 •
edited

Thanks, it works. I also write a simple reverse function。

from unicodedata import numeric
def chinese2num(s):
    amount = 0
    for ch in s:
        number = numeric(ch)
        if number < 10:
            digit = number
        else:
            amount = (amount + digit) * number if number > amount else amount + digit * number         
            digit = 0
    if len(s) > 1 and numeric(s[-2]) != 0:
        return amount + digit * numeric(s[-2]) / 10
    return amount + digit

@Tristan-J
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Tristan-J commented Nov 6, 2018

Thanks, it works. I also write a simple reverse function。

from unicodedata import numeric
def chinese2num(s):
    amount = 0
    for ch in s:
        number = numeric(ch)
        if number < 10:
            digit = number
        else:
            amount = (amount + digit) * number if number > amount else amount + digit * number         
            digit = 0
    if len(s) > 1 and numeric(s[-2]) != 0:
        return amount + digit * numeric(s[-2]) / 10
    return amount + digit

Hi, maybe you need to consider "十二" and I can't say we need this: "len(s) > 1 and numeric(s[-2]) != 0", so my suugestion is:

def chinese2num(s): amount = 0 if s[0] == "十": s = "一"+s for ch in s: number = numeric(ch) if number < 10: digit = number else: amount = (amount + digit) * number if number > amount else amount + digit * number digit = 0 return int(amount + digit)

@Aurora11111
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Aurora11111 commented Oct 25, 2019

Thanks, it works. I also write a simple reverse function。

from unicodedata import numeric
def chinese2num(s):
    amount = 0
    for ch in s:
        number = numeric(ch)
        if number < 10:
            digit = number
        else:
            amount = (amount + digit) * number if number > amount else amount + digit * number         
            digit = 0
    if len(s) > 1 and numeric(s[-2]) != 0:
        return amount + digit * numeric(s[-2]) / 10
    return amount + digit

Hi, maybe you need to consider "十二" and I can't say we need this: "len(s) > 1 and numeric(s[-2]) != 0", so my suugestion is:

def chinese2num(s): amount = 0 if s[0] == "十": s = "一"+s for ch in s: number = numeric(ch) if number < 10: digit = number else: amount = (amount + digit) * number if number > amount else amount + digit * number digit = 0 return int(amount + digit)

very nice!

@thelostpeace
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thelostpeace commented Dec 5, 2019

good one

@gitfourteen
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gitfourteen commented Aug 29, 2020

try
-1297000000000101.1
it outputs:
负一千二百九十七兆零零一百零一点一

@yjyao
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yjyao commented Feb 13, 2022 •
edited

i wrote an arabic-to-chinese number converter just now for fun. upon finishing I was curious to see how others are doing this. and that led me to this gist. thought i'd share my attempt here as well:

import bisect
import re
import unittest


zhdigits = '零一二三四五六七八九'
zhplaces = {
  0: ''  ,
  1: '十',
  2: '百',
  3: '千',
  4: '万',
  8: '亿',
}
zhplace_keys = sorted(zhplaces.keys())


def numdigits(n):
  return len(str(abs(n)))

def _zhnum(n):
  if n < 10:
    return zhdigits[n]
  # Largest place length (number of trailing digits) with a name
  # that fits `n`.
  # Examples:
  #     n       | LPL | name
  #     ---     | --- | ---
  #     100     | 2   | 百
  #     10_0000 | 4   | 万
  named_place_len = zhplace_keys[bisect.bisect_right(zhplace_keys,
                                                     numdigits(n)-1) - 1]
  # Break `n` on the `named_place_len`. The final answer (without handling
  # grammar exceptions like '十' instead of '一十零') will generally be
  #     zhnum(left_part) + unit name at `named_place_len` + zhnum(right_part)
  # additionally, if the `right_part` has leading zeros, a complimenting '零'
  # should lead it.
  # Examples:
  #
  #                208|0300
  #             /           \
  #           /              \
  #        2|08      万  零 3|00
  #      /      \          /   \
  #    二 百 零 八       三 百 empty
  left_part, right_part = n // 10**named_place_len, n % 10**named_place_len
  return (_zhnum(left_part) +
          zhplaces[named_place_len] +
          ((zhdigits[0] if numdigits(right_part) != named_place_len else '') +
           _zhnum(right_part)
           if right_part else ''))

def zhnum(n):
  answer = ('负' if n < 0 else '') + _zhnum(abs(n))
  answer = re.sub(r'^一十', '十', answer)
  answer = re.sub(r'(?<![零十])二(?=[千万亿])', r'两', answer)
  return answer


class TestZhnum(unittest.TestCase):

  '''Test Chinese number formatter.'''

  def test(self):
    self.assertEqual(zhnum(-1),                    '负一')
    self.assertEqual(zhnum(0),                     '零')
    self.assertEqual(zhnum(-0),                    '零')
    self.assertEqual(zhnum(6),                     '六')
    self.assertEqual(zhnum(10),                    '十')
    self.assertEqual(zhnum(14),                    '十四')
    self.assertEqual(zhnum(28),                    '二十八')
    self.assertEqual(zhnum(59),                    '五十九')
    self.assertEqual(zhnum(100),                   '一百')
    self.assertEqual(zhnum(101),                   '一百零一')
    self.assertEqual(zhnum(110),                   '一百一十')
    self.assertEqual(zhnum(132),                   '一百三十二')
    self.assertEqual(zhnum(1000),                  '一千')
    self.assertEqual(zhnum(2001),                  '两千零一')
    self.assertEqual(zhnum(3010),                  '三千零一十')
    self.assertEqual(zhnum(4012),                  '四千零一十二')
    self.assertEqual(zhnum(5230),                  '五千二百三十')
    self.assertEqual(zhnum(6234),                  '六千二百三十四')
    self.assertEqual(zhnum(9999),                  '九千九百九十九')
    self.assertEqual(zhnum(1_0000),                '一万')
    self.assertEqual(zhnum(20_0000),               '二十万')
    self.assertEqual(zhnum(123_4567),              '一百二十三万四千五百六十七')
    self.assertEqual(zhnum(500_1024),              '五百万一千零二十四')
    self.assertEqual(zhnum(360_5000),              '三百六十万五千')
    self.assertEqual(zhnum(3_0000_0000),           '三亿')
    self.assertEqual(zhnum(2_7600_2010),           '两亿七千六百万两千零一十')
    self.assertEqual(zhnum(2_0000_0000_0000_0000), '两亿亿')
    self.assertEqual(zhnum(2_2002_2222),           '两亿两千零二万两千二百二十二')
    self.assertEqual(zhnum(22_2222_2222),          '二十二亿两千二百二十二万两千二百二十二')


if __name__ == '__main__':
  unittest.main()

core implementation (_zhnum and zhnum sans comments) ~20 lines. doesn't handle decimal numbers though.

EDIT: fixed regex for incorrect 两's in for example 二十两万. alternative to using a regex to handle 二 and 两, we can do this in _zhnum when concatenating left_part and the place unit. specifically:

unit = zhplaces[named_place_len]
f'两{unit}' if left_part == 2 and unit in set('千万亿') else (_zhnum(left_part + unit)

the idea is to use 两 only when the whole (four-digit) chunk resolves to 2, instead of just some random number ending in 2.

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