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 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 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 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 + 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)
