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python将人民币转换大写的脚本代码
来源: 互联网 发布时间:2014-09-04
本文导语: 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","...
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 = (nin