python中将阿拉伯数字转换成中文的实现代码

#!/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

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)