Python3 - 索引超出范围（Vigenere Cipher）

Question

我开始编写代码来执行Vigenere密码加密。首先，我想做出关键。密钥需要一遍又一遍地重复，直到它与要加密的消息的长度相匹配，因此我创建了一个函数来为我执行此操作：

def makekey(key, message):
    finalkey = []
    print(message) # to see if the key fits under the message correctly
    key = list(key)
    while len(finalkey) < len(message): # while the key is shorter than the message
        for i in range(0, len(key)): # iterate through characters in the key
            finalkey.append(key[i]) # append the characters in the processed key list
                if len(finalkey) > len(message): # if the final key is still bigger than the message
                    difference = len(finalkey) - len(message) # finds the difference between the two
                    del key[-difference] # removes the difference
    return ''.join(finalkey) # joins the final key into a string

print(makekey("LOVE", "Python")) # calling the function

输出应如下所示：

Python
LOVELO

但该程序只是给我一个超出范围错误的索引，我不知道发生了什么！

错误信息：

Traceback (most recent call last):
  File "test.py", line 14, in <module>
    print(makekey("LOVE", "Python")) # calling the function
  File "test.py", line 8, in makekey
    finalkey.append(key[i]) # append the characters in the processed key list
IndexError: list index out of range

Answer 1

您的代码不起作用的原因：

del key[-difference]

本来应该：

del finalkey[-difference]

在删除该元素后，你会在IndexError（其中i = 3）中尝试访问key[3]时获得finalkey.append(key[i])。

而且只是为了好玩，这是一个替代实现。

def make_key(key, message):
    """ Returns a string that repeats the `key` string until it's
        the length of the `message` string
    """
    if len(key) < len(message):  # only lengthen key if it's too short
        # In python, "abc" * 3 == "abcabcabc"
        # so what would we need to multiply our `key` by to get
        # it to be as long as `message`, or longer?
        # A guaranteed answer is: floor(len(message) / len(key)) + 1
        multiplier = (len(message) // len(key)) + 1
        key = key * multiplier
    # now we have a `key` that is at least as long as `message`
    # so return a slice of the key where the end of the slice == len(message)
    return key[:len(message)]

print(makekey("LOVE", "Python"))

印刷品：LOVELO

Edit - cryptic one-liner solution

如果你想让每个阅读你代码的人都盯着你，你可以试试这个：

from itertools import islice, cycle
key = "LOVE"
message = "Python"
finalkey = ''.join(islice(cycle(key), len(message)))

cycle函数采用iterable对象 - 在我们的例子中，key字符串 - 并在无限循环中重复它。所以，如果我们创建cycle("LOVE")，它将永远生成"L", "O", "V", "E", "L", "O", "V", "E", "L" ...。

islice函数允许我们对迭代器对象进行“切片”。在Python中，“切片”是[0:3]表达式的new = old[0:3]部分的术语 - 我们“切掉”原始的子集。因为我们不希望我们的字符串无限长 - 这不会非常有用 - 我们只想获取我们创建的cycle片段：

islice(cycle(key), len(message)

这需要我们的迭代器 - cycle(key) - 并从索引0开始切片并以索引len(message)结束。这将返回另一个迭代器 - 这次，它不是无限的。迭代器的内容现在是："L", "O", "V", "E", "L", "O"。

现在，我们只需将islice缝合成一个完整的字符串：

''.join(islice...) == "LOVELO"

只是为了在工具箱中提供另一个工具！