python - 使 Vigenére 加密/解密跳过空格

Question

我使用 Vigenére 密码制作了一个基于菜单的加密工具。到目前为止，该程序会加密空格，我怎样才能让程序跳过空格。

#creating variables to be used
text_in_use = ''
encrypt_key = ''
decrypt_key = ''

#function to encypypt input text
def encrypt(plaintext, key):
    keyLength = len(key)
    keyAsIntegers = [ord(i) for i in key] #create list with the ASCII value for each charachter in key
    plaintextAsIntegers = [ord(i) for i in plaintext] #create list with the ASCII value for each charachter in text
    encyptedtext = '' 
    for i in range(len(plaintextAsIntegers)): #
        encryptvalue = (plaintextAsIntegers[i] + keyAsIntegers[i % keyLength]) % 26 #execute encryption or characters according to vigenere definition
        encyptedtext += chr(encryptvalue + 65)
    return encyptedtext #return the encyptes tex

#function to decrypt the encrypted text
def decrypt(encyptedtext, key):
    keyLength = len(key)
    keyAsIntegers = [ord(i) for i in key] #create list with the ASCII value for each charachter in key
    encryptedTextAsIntegers = [ord(i) for i in encyptedtext] #create list with the ASCII value for each charachter in text
    plaintext = ''
    for i in range(len(encryptedTextAsIntegers)):
        value = (encryptedTextAsIntegers[i] - keyAsIntegers[i % keyLength]) % 26 #decryption of encrypted characters
        plaintext += chr(value + 65)
    return plaintext #return decrypted text

#check if user input is valid
def check_value(userEntry):
    while True:
        try: #check if userinput is an integer
            userInput = int(input(userEntry))
            if userInput not in range(1,6): #check if userinput is in valid range
                print("Invalid choice, valid choices are 1-5! Try again! \n")
        except ValueError:
            print("Invalid choice! Input can't be empty or a string! \n")
            print("""1: Input text to work with
2: Print the current text
3: Encrypt the current text
4: Decrypt the current text
5: Exit""")
        else:
            return userInput #return valid userinput


def menu():
    while True:
        print("""1: Input text to work with
2: Print the current text
3: Encrypt the current text
4: Decrypt the current text
5: Exit""")

        choice = check_value("Enter Choice: ")

        if choice == 1: #allows user to input text for use
            text_in_use = str(input("Enter Text: ")).upper()
            print("Text is set to:",text_in_use,"\n")
        elif choice == 2: #prints set text
            print("Your text:",text_in_use,"\n") 
        elif choice == 3: #ask user to set encryptionkey
            encrypt_key = str(input("Enter an encryptionkey: ")).upper()
            text_in_use = encrypt(text_in_use, encrypt_key)
            print("Your text:", text_in_use)
        elif choice == 4: #ask user for decryptionkey
            decrypt_key = str(input("Enter a the decryptionkey: ")).upper()
            text_in_use = decrypt(text_in_use, decrypt_key)
            print("Your text:", text_in_use)
        elif choice == 5:
            exit()

menu()

我希望程序像它已经运行的那样运行，但它应该跳过加密中的空格。

如:

"HELLO MY MAN" --> encryption(key = asd) --> "HWOLG MQ MSQ"

换句话说，加密文本中的空格应该仍然存在。

Answer 1

当明文是“HELLO MY MAN”并且 key 是“asd”时，不确定你是如何得到“HWOLG MQ MSQ”的。我得到了别的东西。

无论如何，也许是这样的:

def encrypt(plaintext, key):
    from itertools import cycle
    from string import ascii_uppercase as alphabet

    offset = ord("A")

    key_char = cycle(key)

    encrypted_plaintext = ""
    for char in plaintext:
        # If the current character is any kind of whitespace...
        if char.isspace():
            # Append it to the final string with no changes.
            encrypted_plaintext += char
        else:
            # The character is not whitespace, so you have to encrypt it.
            char_sum = ord(char) + ord(next(key_char))
            char_sum_wrapped = char_sum % len(alphabet)
            encrypted_plaintext += chr(char_sum_wrapped + offset)
    return encrypted_plaintext

如果当前字符是空格，则将其附加到最终字符串而不作任何更改。 str.isspace 如果字符串中的每个字符(当前字符)都是某种空白(空格、制表符、换行符、回车符等)，则返回 true。

我尽量避免使用索引和硬编码数字，所以我改变了一些东西。例如，我没有像您那样在执行任何其他操作之前将明文和 key 中的所有字符都转换为整数，而是只转换循环中的字符。顺便说一句，循环也不同——我迭代明文中的字符，而不是执行基于范围的 for 循环，然后将 i 视为当前字符的索引。其余的基本相同(除了 key_char 和 itertools.cycle 东西，请阅读下面我的笔记)。

另一件需要注意的事情是，在这个实现中，key_char 迭代器只会在明文中的当前字符不是空格时前进 - 然而，您可能希望它前进而不管。只是要记住一些事情。

没关系，这似乎是该密码所需的行为。

另外，请注意，您的程序从以下几行开始:

#creating variables to be used
text_in_use = ''
encrypt_key = ''
decrypt_key = ''

它们根本没有贡献，您可以安全地删除它们。

编辑 - 更多信息:

itertools.cycle 是一个函数，给定一个可迭代对象(如字符串或列表)，返回一个迭代器，该迭代器产生该可迭代对象中的元素。例如:

>>> from itertools import cycle
>>> char_iterator = cycle("ABC")
>>> next(char_iterator)
'A'
>>> next(char_iterator)
'B'
>>> next(char_iterator)
'C'
>>> next(char_iterator)
'A'
>>> next(char_iterator)
'B'
>>> next(char_iterator)
'C'
>>> next(char_iterator)
'A'
>>> 

如您所见，循环无限期地重复。出于这个原因，我选择使用 itertools.cycle 来替换原始代码中的 keyAsIntegers[i % keyLength]。

string.ascii_uppercase 只是一个由 A-Z 之间的所有大写字母组成的字符串。在我的代码中，我导入了 ascii_uppercase 并在同一行中将其重命名为 alphabet，但它们是相同的。

>>> from string import ascii_uppercase as alphabet
>>> alphabet
'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
>>>