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
#---------www.iplaypy.com--------------------
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)