将用户输入与列表“A”进行比较，并根据列表“A”的索引将输入的字符转换为列表“B”

Question

一个学校项目的程序设计基于两个列表的索引对用户输入进行加密/解密。我的问题是为列表编制索引，并将用户输入与索引进行比较，以便屏幕的输出=list2

所有的代码都非常基本，在变量名中使用'str'和'input'，以免混淆我自己。你知道吗

list1 = ['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','.','1','2','3','4','5','6','7','8','9','0']

list2 = ['4','R','5','G','Z','3','2','D','A','E','X','Y','U','I','6','W','7','O','V','8','F','Q','0','L','J','.','H','9','C','B','N','S','P','M','1','T','K']

strInput = input("Type the message you would like to Encrypt  ").upper()
inputList = split(strInput)
print(inputList)
i = 0

for char in inputList:
    if inputList[i] != list1[i]:
    i = i + 1

现在，它应该接受用户输入，通过索引比较每个列表上的位置，然后使用另一个列表将文本打印到屏幕上。我就是想不出我的索引问题。 提前感谢您的帮助！你知道吗

Answer 1

您可以使用.index(char)在第一个列表中找到字符的索引

list1 = ['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','.','1','2','3','4','5','6','7','8','9','0']

list2 = ['4','R','5','G','Z','3','2','D','A','E','X','Y','U','I','6','W','7','O','V','8','F','Q','0','L','J','.','H','9','C','B','N','S','P','M','1','T','K']

str_input = 'TEST'

encrypted_chars = []

for char in str_input:
    if char == ' ':
        encrypted_chars.append(char)
    else:
        encrypted_chars.append(list2[list1.index(char)])

encrypted_message = ''.join(encrypted_chars)  # 8ZV8

您还可以使用python的列表理解docs

encrypted_chars = [(list2[list1.index(char)] if char != ' ' else ' ') for char in str_input]
encrypted_message = ''.join(encrypted_chars)

要解密你基本上会做相反的事情。你知道吗

decrypted_chars = [(list1[list2.index(char)] if char != ' ' else ' ') for char in encrypted_message]
decrypted_message = ''.join(decrypted_chars)