代码如下:

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

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 = (nin1)

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 (nin1))), ”, 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)

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