python将人民币转换大写的脚本代码

def Num2MoneyFormat( change_number ):
    """
    .转换数字为大写货币格式( format_word.__len__() - 3 + 2位小数 )
    change_number 支持 float, int, long, string
    """
    format_word = ["分", "角", "元",
               "拾","百","千","万",
               "拾","百","千","亿",
               "拾","百","千","万",
               "拾","百","千","兆"]

    format_num = ["零","壹","贰","叁","肆","伍","陆","柒","捌","玖"]
    if type( change_number ) == str:
        # - 如果是字符串,先尝试转换成float或int.
        if '.' in change_number:
            try:    change_number = float( change_number )
            except: raise ValueError, '％s   can\'t change'％change_number
        else:
            try:    change_number = int( change_number )
            except: raise ValueError, '％s   can\'t change'％change_number

    if type( change_number ) == float:
        real_numbers = []
        for i in range( len( format_word ) - 3, -3, -1 ):
            if change_number >= 10 ** i or i < 1:
                real_numbers.append( int( round( change_number/( 10**i ), 2)％10 ) )

    elif isinstance( change_number, (int, long) ):
        real_numbers = [ int( i ) for i in str( change_number ) + '00' ]

    else:
        raise ValueError, '％s   can\'t change'％change_number

    zflag = 0                       #标记连续0次数，以删除万字，或适时插入零字
    start = len(real_numbers) - 3
    change_words = []
    for i in range(start, -3, -1):  #使i对应实际位数，负数为角分
        if 0 <> real_numbers[start-i] or len(change_words) == 0:
            if zflag:
                change_words.append(format_num[0])
                zflag = 0
            change_words.append( format_num[ real_numbers[ start - i ] ] )
            change_words.append(format_word[i+2])

        elif 0 == i or (0 == i％4 and zflag < 3):    #控制 万/元
            change_words.append(format_word[i+2])
            zflag = 0
        else:
            zflag += 1

    if change_words[-1] not in ( format_word[0], format_word[1]):
        # - 最后两位非"角,分"则补"整"
        change_words.append("整")

    return ''.join(change_words)

Python 把金额小写转换成大写2

功能将小于十万亿元的小写金额转换为大写

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)