Message=0 Source=E:\1daer\mmxdzy\1\dzy1\dzy1.py StackTrace: File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 33, in <dictcomp> mapping = str.maketrans({key[i]: key[(i+1)%len(key)] for i in range(len(key))}) ~~~^^^ File "E:\1daer\mmxdzy\1\dzy1\dzy1.py", line 33, in decrypt mapping = str.maketrans({key[i]: key[(i+1)%len(key)] 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 46, in <module> (Current frame) key = guess_key(cipher_text, words) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ KeyError: 0
时间: 2023-08-03 11:03:07 浏览: 49
这个错误是因为在 `guess_key()` 函数的字典推导式中使用了索引 `i`,但是在 `decrypt()` 函数中使用了 `mapping` 字典,导致索引错误。需要将 `decrypt()` 函数中的 `mapping` 改为 `mapping_dict`。
另外,这个错误还提示了一个 `KeyError: 0`,这是因为在 `guess_key()` 函数的字典推导式中使用了 `0` 作为默认值,但是在加密文本中可能不存在该值,导致字典访问时出现错误。可以将默认值改为一个空字符串 `''`,如下所示:
```
frequency = {char: 0 for char in string.ascii_lowercase}
for char in cipher_text:
if char.lower() in frequency:
frequency[char.lower()] += 1
most_frequent = ''.join([k for k, v in sorted(frequency.items(), key=lambda item: item[1], reverse=True)])
mapping_dict = str.maketrans({key[i]: most_frequent[i] if i < len(most_frequent) else '' for i in range(len(key))})
```
这样就能够避免 `KeyError` 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:
``` python
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:
```python
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。