Copy Code code as follows:
def num2moneyformat (Change_number):
"""
. Convert number to uppercase currency format (format_word.__len__ ()-3 + 2 decimal)
Change_number supports float, int, long, string
"""
Format_word = ["min", "Jiao", "Yuan",
"Pick up", "hundred", "thousand", "Million",
"Pick up", "hundred", "thousand", "billion",
"Pick up", "hundred", "thousand", "Million",
"Pick up", "hundred", "thousand", "trillion"]
Format_num = [0], "one", "II", "three", "Restaurant", "WU", "Lu", "Qi", "ba", "Nine"]
If Type (change_number) = = str:
#-If it is a string, try converting to float or int first.
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 >= 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 To delete million words, or insert 0 words in a timely manner
start = Len (real_numbers)-3
Change_words = []
For I in range (start, -3,-1): #使i对应实际位数, negative numbers are corner points
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 million/yuan
Change_words.append (format_word[i+2])
Zflag = 0
Else
Zflag + 1
If CHANGE_WORDS[-1] not in (format_word[0), format_word[1]):
#-The last two digits are not "corner," then fill the "whole"
Change_words.append ("whole")
Return '. Join (Change_words)
Python converts the amount lowercase to uppercase 2
Feature converts lowercase amounts less than 10 trillion yuan to uppercase
Copy Code code as follows:
def IIf (b, S1, S2):
If B:
return S1
Else
Return s2
def num2chn (Nin=none):
CS =
(' 0 ', ' one ', ' II ', ' three ', ' The Shops ', ' Wu ', ' Lu ', ' qi ', ' ba ', ' JIU ', ' ◇ ', ' min ', ' jiao ', ' round ', ' pick ', ' bai ', ' Qian ',
' Million ', ' pick ', ' bai ', ' thousand ', ' billion ', ' pick ', ' bai ', ' thousand ', ' million '
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, ', ' whole ')
+ St1
FG = (ns==0)
St.replace (' billions ', ' million ')
Return IIf (nin==0, ' 0 ', St + st1)
if __name__ = = ' __main__ ':
num = 12340.1
Print num
Print Num2chn (num)
