python - Python - 为什么这个 for 循环只打印 1 个字母？

Question

def caesar(plaintext,shift):  
    alphabet=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]  

    #Create our substitution dictionary  
    dic={}  
    for i in range(0,len(alphabet)):  
        dic[alphabet[i]]=alphabet[(i+shift)%len(alphabet)]  

    #Convert each letter of plaintext to the corrsponding  
    #encrypted letter in our dictionary creating the cryptext  
    ciphertext=("")  
    for l in plaintext.lower():  
            if l in dic:  
                l=dic[l]  
                ciphertext+=l
            return ciphertext  

#Example useage  
plaintext="the cat sat on the mat"  
print "Plaintext:", plaintext  
print "Cipertext:", (caesar(plaintext,29)) 

密文只打印一个字母，而不是在凯撒移位中打印“明文”变量。我希望它打印整个句子。

谢谢

Answer 1

这是因为你return ciphertext的缩进错误。您从 for 循环的第一次迭代返回。（缩进在 Python 中很重要！）

    for l in plaintext.lower():  
            if l in dic:  
                l=dic[l]  
                ciphertext+=l
            return ciphertext  # Indented to match level of `if`.

修复它。

    for l in plaintext.lower():  
        if l in dic:  
            l=dic[l]  
            ciphertext+=l
    return ciphertext

几个指针

1. 无需列出代码中的所有字母，您只需设置alphabets = string.ascii_lowercase.
2. 您不需要存储变量dic[l]，只需执行ciphertext += dic[l].

Answer 2

用 string.translate 做得更好:)

import string
def caesar(plaintext,shift):  
    alphabet=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
    alphabet_shifted = alphabet[shift:]+alphabet[:shift]
    tab =    string.maketrans("".join(alphabet),"".join(alphabet_shifted))
    return plaintext.translate(tab)

Answer 3

您需要修复 return 语句的缩进：

for l in plaintext.lower():  
    if l in dic:  
       l=dic[l]  
       ciphertext+=l
    # <-- if you return here, then you will loop only once.      
return ciphertext