python数字转对应中文的方法总结

时间:2022-11-24 09:17:48

本文操作环境:

windows7系统,DELL G3电脑,python3.5版

python实现将阿拉伯数字转换成中文

第一种转换方式:

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    1  -->  一
    12   -->  一二
def num_to_char(num):
    """数字转中文"""
    num=str(num)
    new_str=""
    num_dict={"0":u"零","1":u"一","2":u"二","3":u"三","4":u"四","5":u"五","6":u"六","7":u"七","8":u"八","9":u"九"}
    listnum=list(num)
    # print(listnum)
    shu=[]
    for i in listnum:
        # print(num_dict[i])
        shu.append(num_dict[i])
    new_str="".join(shu)
    # print(new_str)
    return new_str

第二种转换方式:

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     1   -->   一
    12  -->   十二
    23  -->  二十三
_MAPPING = (u'零', u'一', u'二', u'三', u'四', u'五', u'六', u'七', u'八', u'九', u'十', u'十一', u'十二', u'十三', u'十四', u'十五', u'十六', u'十七',u'十八', u'十九')
_P0 = (u'', u'', u'', u'千',)
_S4 = 10 ** 4
def _to_chinese4(num):
    assert (0 <= num and num < _S4)
    if num < 20:
        return _MAPPING[num]
    else:
        lst = []
        while num >= 10:
            lst.append(num % 10)
            num = num / 10
        lst.append(num)
        c = len(lst)  # 位数
        result = u''
        for idx, val in enumerate(lst):
            val = int(val)
            if val != 0:
                result += _P0[idx] + _MAPPING[val]
                if idx < c - 1 and lst[idx + 1] == 0:
                    result += u'零'
        return result[::-1]

实例扩展:

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#!/usr/bin/python
#-*- encoding: utf-8 -*-
 
import types
 
class NotIntegerError(Exception):
  pass
 
class OutOfRangeError(Exception):
  pass
 
_MAPPING = (u'零', u'一', u'二', u'三', u'四', u'五', u'六', u'七', u'八', u'九', )
_P0 = (u'', u'', u'', u'千', )
_S4, _S8, _S16 = 10 ** 4 , 10 ** 8, 10 ** 16
_MIN, _MAX = 0, 9999999999999999
 
def _to_chinese4(num):
  '''转换[0, 10000)之间的阿拉伯数字
  '''
  assert(0 <= num and num < _S4)
  if num < 10:
    return _MAPPING[num]
  else:
    lst = [ ]
    while num >= 10:
      lst.append(num % 10)
      num = num / 10
    lst.append(num)
    c = len(lst)  # 位数
    result = u''
    
    for idx, val in enumerate(lst):
      if val != 0:
        result += _P0[idx] + _MAPPING[val]
        if idx < c - 1 and lst[idx + 1] == 0:
          result += u'零'
    
    return result[::-1].replace(u'一十', u'十')
    
def _to_chinese8(num):
  assert(num < _S8)
  to4 = _to_chinese4
  if num < _S4:
    return to4(num)
  else:
    mod = _S4
    high, low = num / mod, num % mod
    if low == 0:
      return to4(high) + u'万'
    else:
      if low < _S4 / 10:
        return to4(high) + u'万零' + to4(low)
      else:
        return to4(high) + u'万' + to4(low)
      
def _to_chinese16(num):
  assert(num < _S16)
  to8 = _to_chinese8
  mod = _S8
  high, low = num / mod, num % mod
  if low == 0:
    return to8(high) + u'亿'
  else:
    if low < _S8 / 10:
      return to8(high) + u'亿零' + to8(low)
    else:
      return to8(high) + u'亿' + to8(low)
    
def to_chinese(num):
  if type(num) != types.IntType and type(num) != types.LongType:
    raise NotIntegerError(u'%s is not a integer.' % num)
  if num < _MIN or num > _MAX:
    raise OutOfRangeError(u'%d out of range[%d, %d)' % (num, _MIN, _MAX))
  
  if num < _S4:
    return _to_chinese4(num)
  elif num < _S8:
    return _to_chinese8(num)
  else:
    return _to_chinese16(num)
  
if __name__ == '__main__':
  print to_chinese(9000)
 

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