def Num2MoneyFormat( change_number ):
    """
    .轉換數字為大寫貨幣格式( format_word.__len__() - 3 + 2位小數 )
    change_number 支持 float, int, long, string
    """
    format_word = ["分", "角", "元",
               "拾","百","千","萬",
               "拾","百","千","億",
               "拾","百","千","萬",
               "拾","百","千","兆"]

    format_num = ["零","壹","貳","叁","肆","伍","陸","柒","捌","玖"]
    if type( change_number ) == str:
        # - 如果是字符串,先嘗試轉換成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                       #標記連續0次數，以刪除萬字，或適時插入零字
    start = len(real_numbers) - 3
    change_words = []
    for i in range(start, -3, -1):  #使i對應實際位數，負數為角分
        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 萬/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]):
        # - 最后兩位非"角,分"則補"整"
        change_words.append("整")

    return ''.join(change_words)

　　def IIf( b, s1, s2):
　　if b:
　　　　return s1
　　else:
　　　　return s2
def num2chn(nin=None):
　　　　cs =
('零','壹','貳','叁','肆','伍','陸','柒','捌','玖','◇','分','角','圓','拾','佰','仟',
'萬','拾','佰','仟','億','拾','佰','仟','萬')
　　　　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('億萬','萬')
　　　　return IIf( nin==0, '零', st + st1)
if __name__ == '__main__':
　　num = 12340.1
　　print num
　　print num2chn(num)