源码网商城,靠谱的源码在线交易网站 我的订单 购物车 帮助

源码网商城

python中将阿拉伯数字转换成中文的实现代码

  • 时间:2020-12-31 22:40 编辑: 来源: 阅读:
  • 扫一扫,手机访问
摘要:python中将阿拉伯数字转换成中文的实现代码
[u]复制代码[/u] 代码如下:
#!/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'十', u'百', 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)
把金额小写转换成大写的Python代码 功能将小于十万亿元的小写金额转换为大写 代码
[u]复制代码[/u] 代码如下:
  def IIf( b, s1, s2):   if b:     return s1   else:     return s2 def num2chn(nin=None):     cs = ('零','壹','贰','叁','肆','伍','陆','柒','捌','玖','◇','分','角','圆','拾','佰','仟', '万','拾','佰','仟','亿','拾','佰','仟','万')     st = ''; st1=''     s = '%0.2f' % (nin)         sln =len(s)     if sln >; 15: return None     fg = (nin<1)     for i in range(0, sln-3):         ns = ord(s[sln-i-4]) - ord('0')         st=IIf((ns==0)and(fg or (i==8)or(i==4)or(i==0)), '', cs[ns])       + IIf((ns==0)and((i<>;8)and(i<>;4)and(i<>;0)or fg and(i==0)),'', cs[i+13])       + st         fg = (ns==0)     fg = False     for i in [1,2]:         ns = ord(s[sln-i]) - ord('0')         st1 = IIf((ns==0)and((i==1)or(i==2)and(fg or (nin<1))), '', cs[ns])        + IIf((ns>;0), cs[i+10], IIf((i==2) or fg, '', '整'))        + st1         fg = (ns==0)     st.replace('亿万','万')     return IIf( nin==0, '零', st + st1) if __name__ == '__main__':   num = 12340.1   print num   print num2chn(num)
  • 全部评论(0)
联系客服
客服电话:
400-000-3129
微信版

扫一扫进微信版
返回顶部