python將人民幣轉(zhuǎn)換大寫的腳本代碼
更新時(shí)間:2013年02月10日 18:20:36 作者:
python將人民幣轉(zhuǎn)換大寫的代碼,有需要的朋友可以參考下
復(fù)制代碼 代碼如下:
def Num2MoneyFormat( change_number ):
"""
.轉(zhuǎn)換數(shù)字為大寫貨幣格式( format_word.__len__() - 3 + 2位小數(shù) )
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(zhuǎn)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(biāo)記連續(xù)0次數(shù),以刪除萬字,或適時(shí)插入零字
start = len(real_numbers) - 3
change_words = []
for i in range(start, -3, -1): #使i對(duì)應(yīng)實(shí)際位數(shù),負(fù)數(shù)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(bǔ)"整"
change_words.append("整")
return ''.join(change_words)
Python 把金額小寫轉(zhuǎn)換成大寫2
功能將小于十萬億元的小寫金額轉(zhuǎn)換為大寫
復(fù)制代碼 代碼如下:
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)
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