如何在python中将最多63位的数字转换成word？

Question

我想在python中将数字转换成单词，而不导入任何模块或库

我已经试过编写一些代码，但是当在这段时间内出现这么多的零时，它就不能正常工作了

print("upto sixty three digits only: ")
a=input("enter the number: ")
c=a
e=int(a)
b=""
l=[]
l2=[]
while len(c)%3 != 0:
    c="0"+c
d=len(c)/3
while len(c)!=0:
    l.append(c[0:3])
    c=c[3:len(c)]

on=["", "one","two","three", "four","five","six","seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]

tens=["","","twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]

h="hundred"
grt=["thousand","million","billion","trillion","quadrillion","quintillion","sextillion","septillion","octillion","nonillion","decillion","undecillion","duodecillion","tredecillion","quattuor-decillion","quindecillion","sexdecillion","septen-decillion","octodecillion","novemdecillion","vigintillion"                                        ]

for i in l:
    k=int(i)

    if i[0]=="0":
        if k<20:
            l2.append(on[k])
        else:
            l2.append(str(tens[k//10])+" "+str(on[k%10]))
    else:
        if int(i[1:3])<20:
            l2.append(str(on[k//100])+" "+h+" "+str(on[k%100]))
        else:
            l2.append(str(on[k//100])+" "+h+" "+str(tens[(k%100)//10])+" "+str(on[k%10]))


for i in range(0,len(l2)):

    b=b+l2[i]+" "+grt[int(d-2)]+" "
    d=d-1
l=b.split()
l.pop()
d=""
for i in l:
    d=d+i+" "
print(d)

当输入的数字之间没有零时，代码工作正常

（例如123456789=>；一亿二千三百四十五万六千七百八十九）

但当中间有许多零时，它就不起作用了

（例如120000000987=>；一亿两千万九百八十七）

Answer 1

正如@gelonida所建议的，如果你用函数来做“肮脏的工作”会更好。例如，我修改了您的脚本，添加了两个函数，一个用于将数字划分为长度为3的块，另一个用于对前面提到的块进行“文字化”。就像这样，我相信代码会变得更简单、更简短、更易读：

def wordify(n):
    s=''
    if len(n)==3:
        s+=on[int(n[0])]+' hundred '
    if len(n)>1:
        if n[-2]=='1':
            s+=teen[int(n[-1])]
            return s
        else:
            s+=tens[int(n[-2])]+' '
    s+=on[int(n[-1])]
    return s

a=input('Insert number: ')
gb3 = lambda x: [x[i:i+3] for i in range(0,len(x),3)]
rem = len(a)%3
tokens=[a[:rem]]+gb3(a[rem:]) if rem!=0 else gb3(a[rem:])
l = len(tokens)

on=["", "one","two","three", "four","five","six","seven", "eight", "nine"]
teen=["ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]
tens=["","","twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
grt=["","thousand","million","billion","trillion","quadrillion","quintillion","sextillion","septillion","octillion","nonillion","decillion","undecillion","duodecillion","tredecillion","quattuor-decillion","quindecillion","sexdecillion","septen-decillion","octodecillion","novemdecillion","vigintillion"]

print()
word_number = ''
for k in range(l):
    if int(tokens[k])!=0:
        word_number += wordify(tokens[k])+' '+grt[l-k-1]+' '
print(' '.join([w for w in word_number.split(' ') if len(w)>1]))

上一次打印调用中，我使用了一个快速的方法来均匀地排列单词。 告诉我你是怎么想的！你知道吗