如何让我的 Vigenère Cipher 忽略原始消息中的空格

Question

我正在尝试制作 Vigenère Cipher，但我似乎无法找到一种方法来实现在输入消息然后打印最终结果时忽略输入空格的功能，例如：我输入起始消息：“ python 计算”然后我输入密钥：“堆栈”如果程序忽略原始消息中的空格，我希望得到：“isukzg wppannjqr”，但我得到：“isukzgwppannjqr”。任何人都知道我该如何解决这个问题。我考虑过使用 ords 但我还没有找到实现它的方法。代码如下：

def translateMessage(key, message, mode):
    translated = ""

    keyIndex = 0
    key = key.upper()

    for symbol in message:
        xyz = alphabet.find(symbol.upper())
        if xyz != -1:
            if mode == 'encrypt' or 'e':
                xyz += alphabet.find(key[keyIndex]) + 1
            elif mode == 'decrypt' or 'd':
                xyz -= alphabet.find(key[keyIndex]) + 1

            xyz %= len(alphabet)

            if symbol.isupper():
                translated += alphabet[xyz]
            elif symbol.islower():
                translated += alphabet[xyz].lower()

            keyIndex += 1
            if keyIndex == len(key):
                keyIndex = 0

    return translated

if __name__ == '__main__':
    fetch_user_inputs()

Answer 1

您只需要在translateMessage() 中添加一条else 语句，就可以像这样在输出中添加空格

def translateMessage(key, message, mode):
    translated = ""

    keyIndex = 0
    key = key.upper()

    for symbol in message:
        xyz = alphabet.find(symbol.upper())
        if xyz != -1:
            if mode == 'encrypt' or 'e':
                xyz += alphabet.find(key[keyIndex]) + 1
            elif mode == 'decrypt' or 'd':
                xyz -= alphabet.find(key[keyIndex]) + 1

            xyz %= len(alphabet)

            if symbol.isupper():
                translated += alphabet[xyz]
            elif symbol.islower():
                translated += alphabet[xyz].lower()

            keyIndex += 1
            if keyIndex == len(key):
                keyIndex = 0
        else : translated += symbol #this will add space as it is

    return translated

Answer 2

您可以做得更好，此代码甚至可以返回格式以及空格，例如。大写字母，小写字母！

def shift_dict(Caesar, Shift):
  dic_len = len(Caesar)
  Shift = Shift % dic_len
  list_dic = [(k,v) for k, v in iter(Caesar.items())]
  Shifted = {
    list_dic[x][0]: list_dic[(x - Shift) % dic_len][1]
    for x in range(dic_len)
  }
  return Shifted

Viginere = {
  "A":0,
  "B":1,
  "C":2,
  "D":3,
  "E":4,
  "F":5,
  "G":6,
  "H":7,
  "I":8,
  "J":9,
  "K":10,
  "L":11,
  "M":12,
  "N":13,
  "O":14,
  "P":15,
  "Q":16,
  "R":17,
  "S":18,
  "T":19,
  "U":20,
  "V":21,
  "W":22,
  "X":23,
  "Y":24,
  "Z":25
}
VFU = {
  0:"A",
  1:"B",
  2:"C",
  3:"D",
  4:"E",
  5:"F",
  6:"G",
  7:"H",
  8:"I",
  9:"J",
  10:"K",
  11:"L",
  12:"M",
  13:"N",
  14:"O",
  15:"P",
  16:"Q",
  17:"R",
  18:"S",
  19:"T",
  20:"U",
  21:"V",
  22:"W",
  23:"X",
  24:"Y",
  25:"Z"
}
VFL = {
  0:"a",
  1:"b",
  2:"c",
  3:"d",
  4:"e",
  5:"f",
  6:"g",
  7:"h",
  8:"i",
  9:"j",
  10:"k",
  11:"l",
  12:"m",
  13:"n",
  14:"o",
  15:"p",
  16:"q",
  17:"r",
  18:"s",
  19:"t",
  20:"u",
  21:"v",
  22:"w",
  23:"x",
  24:"y",
  25:"z"
}

Asker = int(input("Do you want to... 1. Encode, or 2. Decode? "))
X = 0
Lister = []
Text = list(str(input("")))
Key = list(str(input("")).upper())
Z = 0

if Asker == 1:
  for i in range(len(Text)):
    Shift = Viginere[Key[(Z % len(Key))]]
    if Text[X].isalpha():
      LetterNum = Viginere[Text[X].upper()]
      Helper = (LetterNum + Shift) % 26
      if Text[X].isupper():
        Lister.append(VFU[Helper])
      else:
        Lister.append(VFL[Helper])
    else:
      Lister.append(Text[X])
      Z -= 1
    X += 1
    Z += 1
  print(*Lister, sep = "")
elif Asker == 2:
  for i in range(len(Text)):
    Shift = Viginere[Key[(Z % len(Key))]]
    if Text[X].isalpha():
      LetterNum = Viginere[Text[X].upper()]
      Helper = (LetterNum - Shift) % 26
      if Text[X].isupper():
        Lister.append(VFU[Helper])
      else:
        Lister.append(VFL[Helper])
    else:
      Lister.append(Text[X])
      Z -= 1
    X += 1
    Z += 1
  print(*Lister, sep = "")

这将轻松解决您的问题，它甚至会要求您输入字符串和密钥！这就是python魔法的定义。