python将人民币转换大写的脚本代码

　　复制代码 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
复制代码 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 = (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)