Numbers to Chinese representations converter in Python. 中文数字转换

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)

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!

good one

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

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.