不可变字符串，类型错误：“str”对象不支持项目分配

Question

我正在使用其他人的代码来制作密码求解器，但它给了我

typeerror: 'str' object does not support item assignment"

在

key[keyIndex] = cipherletter

有没有办法与错误保持相同的含义？ :)

def decryptWithCipherletterMapping(ciphertext, letterMapping):
# Return a string of the ciphertext decrypted with the letter mapping,
# with any ambiguous decrypted letters replaced with an _ underscore.

# First create a simple sub key from the letterMapping mapping.
    key = ['x'] * len(LETTERS)
    for cipherletter in LETTERS:
        if len(letterMapping[cipherletter]) == 1:
# If there's only one letter, add it to the key.
            keyIndex = LETTERS.find(letterMapping[cipherletter][0])
            key[keyIndex] = cipherletter
        else:
            ciphertext = ciphertext.replace(cipherletter.lower(), '_')
            ciphertext = ciphertext.replace(cipherletter.upper(), '_')
            key = ''.join(key)

# With the key we've created, decrypt the ciphertext.

return simpleSubCipher.decryptMessage(key, ciphertext)

Answer 1

简短的回答是您遇到了缩进错误。而不是：

# First create a simple sub key from the letterMapping mapping.
    key = ['x'] * len(LETTERS)
    for cipherletter in LETTERS:
        if len(letterMapping[cipherletter]) == 1:
# If there's only one letter, add it to the key.
            keyIndex = LETTERS.find(letterMapping[cipherletter][0])
            key[keyIndex] = cipherletter
        else:
            ciphertext = ciphertext.replace(cipherletter.lower(), '_')
            ciphertext = ciphertext.replace(cipherletter.upper(), '_')
            key = ''.join(key)

你应该有：

# First create a simple sub key from the letterMapping mapping.
keyList = ['x'] * len(LETTERS)

for cipherletter in LETTERS:
    if len(letterMapping[cipherletter]) == 1:
        # If there's only one letter, add it to the key.
        keyIndex = LETTERS.find(letterMapping[cipherletter][0])
        keyList[keyIndex] = cipherletter
    else:
        ciphertext = ciphertext.replace(cipherletter.lower(), '_')
        ciphertext = ciphertext.replace(cipherletter.upper(), '_')

key = ''.join(keyList)

更深层次的问题是您在原始代码中使用key， 是一个字符串 和 一个列表。显然这很令人困惑，因为您最终会感到困惑！在我上面的版本中，我将它们分成两个不同的变量，我想你会发现它们更清楚。

Answer 2

因为key 是str 的类型。

在 Python 中，字符串不支持像列表那样的项目分配（正如错误消息明确指出的那样）。

要将字符串的字符更新为给定索引，您可以执行以下操作：

string = "foobar"
charindex = 2 # so you wants to replace the second "o"
print(string[0:charindex] + 'O' + string[charindex+1:])
# gives foObar

或者把它转入函数：

def replace_at_index(string, replace, index):
    return string[0:index] + replace + string[index+1:]

print(replace_at_index('EggsandBacon', 'A', 4))
# gives EggsAndBacon

所以你会像这样使用它：

key = replace_at_index(key, cipherletter, keyIndex)