Message=0 Source=E:\1daer\mmxdzy\1\dzy1\dzy1.py StackTrace: File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 38, in <dictcomp> mapping_dict = str.maketrans({key[i]: most_frequent[i] if i < len(most_frequent) else '' for i in range(len(key))}) ~~~^^^ File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 38, in decrypt mapping_dict = str.maketrans({key[i]: most_frequent[i] if i < len(most_frequent) else '' for i in range(len(key))}) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 28, in guess_key decrypted_text = decrypt(cipher_text, key) ^^^^^^^^^^^^^^^^^^^^^^^^^ File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 51, in <module> (Current frame) key = guess_key(cipher_text, words) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ KeyError: 0

这个错误是因为在 `decrypt()` 函数中使用了 `most_frequent[i]` 来映射密文中的字符，但可能会出现 `most_frequent` 字符串的长度小于 `key` 的情况，导致索引错误。需要在字典推导式中增加一个判断，当 `i` 大于等于 `most_frequent` 的长度时，将映射值设为 `None`，如下所示：

mapping_dict = str.maketrans({key[i]: most_frequent[i] if i < len(most_frequent) else None for i in range(len(key))})

这样就能够避免出现 `KeyError: 0` 错误了。

相关问题

Message='itertools.permutations' object is not subscriptable Source=E:\1daer\mmxdzy\1\dzy1\dzy1.py StackTrace: File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 55, in guess_key1 print(next(mp[i][j]), end=" ") ~~~~~^^^ File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 79, in <module> (Current frame) print(guess_key1(cipher_text, words)) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ TypeError: 'itertools.permutations' object is not subscriptable

这个错误是因为您在使用 `next(mp[i][j])` 时，将 `mp[i][j]` 当做了一个可索引的对象，但实际上 `mp[i][j]` 是一个 `itertools.permutations` 对象，不支持索引操作。

如果您想获取 `itertools.permutations` 中的某个元素，可以将其转换为列表或元组，然后再进行索引操作。例如，您可以将 `next(mp[i][j])` 改为 `list(mp[i][j])[0]`，这样就可以获取排列中的第一个元素。

以下是修正后的示例代码：

import itertools

def guess_key1(cipher_text, words):
    # 生成字符组合的排列
    perms = itertools.permutations('ABCDEFGHIJKLMNOPQRSTUVWXYZ', 5)
    for perm in perms:
        # 将排列转换为字典
        key = {chr(i+65): perm[i] for i in range(26)}
        # 解密密文
        plain_text = decrypt(cipher_text, key)
        # 如果解密后的明文包含给定单词，则返回密钥
        if contains_words(plain_text, words):
            return key

def contains_words(text, words):
    # 判断文本中是否包含给定单词
    for word in words:
        if word in text:
            return True
    return False

def decrypt(cipher_text, key):
    # 解密密文
    plain_text = ''
    for ch in cipher_text:
        if ch.isalpha():
            plain_text += key[ch]
        else:
            plain_text += ch
    return plain_text

cipher_text = 'LXYXKXOXRKXYXKXXKLYXKXXYXKXOXRKXYXKXXKLRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKXXYXKXXYXKXOXRKXYXKXXKLRKXXYXKXXYXKXOXRKXYXKXXKLOXKX'
words = ['CRYPTO', 'GOLD', 'SILVER']

# 获取排列器
perms = itertools.permutations('ABCDEFGHIJKLMNOPQRSTUVWXYZ', 5)

# 将排列器转换为二维列表
mp = [[0] * 5 for i in range(5)]
for i in range(5):
    for j in range(5):
        mp[i][j] = perms

# 遍历排列器中的每一个排列
for i in range(5):
    for j in range(5):
        # 获取排列器中的下一个排列
        perm = next(mp[i][j])
        # 输出排列中的第一个元素
        print(list(perm)[0], end="  ")
    print()

在上面的代码中，我们使用 `list(perm)[0]` 来获取排列中的第一个元素，然后输出到控制台上。

Message='tuple' object is not an iterator Source=E:\1daer\mmxdzy\1\dzy1\dzy1.py StackTrace: File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 45, in guess_key1 index = next(mp[i][j]) ^^^^^^^^^^^^^^ File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 73, in <module> (Current frame) print(guess_key1(cipher_text, words)) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ TypeError: 'tuple' object is not an iterator

这个错误是因为 `mp[i][j]` 不是一个迭代器，而是一个元组。在使用 `next` 函数时，需要传入一个迭代器作为参数，而不能传入一个元组。

解决这个问题的方法是将 `mp[i][j]` 转换成一个迭代器，可以使用 `iter` 函数将其转换成迭代器。修改代码如下：

def guess_key1(cipher_text, words):
    letter_frequency = get_letter_frequency(cipher_text.lower())
    sorted_letters = sorted(letter_frequency.keys(), key=lambda x: letter_frequency[x], reverse=True)
    most_common_letters_m = [sorted_letters[:8], sorted_letters[8:16], sorted_letters[-10:]]
    f1 = ['e']
    f2 = ['a', 'i', 'r', 't', 'o', 'n']
    f3 = ['s', 'l', 'c']
    f4 = ['u', 'p', 'm', 'd', 'h']
    f5 = ['g', 'b', 'y', 'f', 'v']
    f6 = ['w', 'k', 'x', 'z', 'q', 'j']
    mf = [f1, f2, f3, f4, f5, f6]
    mp = [[] for _ in range(len(mf))]
    flag = True
    while(flag):
        key = {}
        for i in range(6):
            mp[i] = iter(itertools.permutations(range(len(mf[i]))))
        for i in range(6):
            for j in range(len(mf[i])):
                index = next(mp[i][j])
                key.update({most_common_letters_m[i][index]: mf[i][j]})
        decrypted_text = decrypt(cipher_text, key)
        if is_plaintext(decrypted_text, words):
            flag = False
            return key

在这个修改中，将 `mp[i][j]` 转换成了迭代器，修改为 `iter(mp[i][j])`，然后再使用 `next` 函数来获取其下一个元素，就可以避免之前的错误。