如何让我的整数到单词字典功能正常工作

Question

我能够使这个函数在某些情况下将整数转换为单词，但当输入大于120时仍然有问题；对于121122、、、和任何大于..20的值，即221233等。。它返回100 None（对于120以上的情况）。如果输入超过1020，它将不起作用。你知道吗

def int_to_word(n):
    d={0:'zero',1:'one',2:'two',3:'three',4:'four',5:'five',
      6:'six',7:'seven',8:'eight',9:'nine',10:'ten',
      11:'eleven',12:'twelve',13:'thirteen',14:'fourteen',
      15:'fifteen',16:'sixteen',17:'seventeen',18:'eighteen',
      19:'nineteen',20:'twenty',30:'thirty',40:'forty',
      50:'fifty',60:'sixty',70:'seventy',80:'eighty',
      90:'ninety'}
    k=1000
    m=k*1000
    if (n <20):
        print(d.get(n))
    elif (n >=20 and n <100):
        if n%10==0:
            print(d.get(n))
        else:
            print(d.get(n//10*10),d.get(n%10),end="")
    elif (n>=100) and (n<k):
        if n%100==0:
            print(d[n//100],end=" ")
            print("hundred")
        else :
            print(d.get(n//100),"hundred",d.get(n%100),end="")
    elif (n>=k)and (n<m):
        if n%k==0:
            print(d.get(n//k),end=" ")
            print("thousand")
        else:
            print(d.get(n//k),end=" ")
            print("thousand",end=" ")
            print(d.get(n%k))

int_to_word(n)

Answer 1

对于您的代码，我认为问题在于您没有考虑如何正确处理数百和数千数字。您的代码只处理小于1000000的数字。以下是根据您的想法修改的代码：

def int_to_word(n):
d={0:'zero',1:'one',2:'two',3:'three',4:'four',5:'five',
  6:'six',7:'seven',8:'eight',9:'nine',10:'ten',
  11:'eleven',12:'twelve',13:'thirteen',14:'fourteen',
  15:'fifteen',16:'sixteen',17:'seventeen',18:'eighteen',
  19:'nineteen',20:'twenty',30:'thirty',40:'forty',
  50:'fifty',60:'sixty',70:'seventy',80:'eighty',
  90:'ninety'}
k=1000
m=k*1000
if (n <20):
    print(d.get(n))
elif (n >=20 and n <100):
    if n%10==0:
        print(d.get(n))
    else:
        print(d.get(n//10*10),d.get(n%10),end="")
elif (n>=100) and (n<k):
    if n%100==0:
        print(d[n//100],end=" ")
        print("hundred")
    else :
        print(d.get(n//100),"hundred",d.get(n%100//10*10),d.get(n%100%10),end=" ")
elif (n>=k)and (n<m):
    if n%k==0:
        print(d.get(n//k//10*10),d.get(n//k%10),end=" ")
        print("thousand")
    else:
        if n//k > 100:
            print(d.get(n//k//100),end=" ")
            print("hundred",end=" ")
            print(d.get(n//k%100//10*10),d.get(n//k%100%10),end=" ")
            print("thousand",end=" ")
        elif n//k >10:
            print(d.get(n//k//10*10),d.get(n//k%10),end=" ")
            print("thousand",end=" ")
        else:
            print(d.get(n//k%10),end=" ")
            print("thousand",end=" ")
        print(d.get(n%k//100),"hundred",d.get(n%k%100//10*10),d.get(n%k%100%10),end=" ")

在[2]中：int\u to \u word（999999）

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