尝试包装凯撒密码（Python）

Question

我尝试使用“if”语句来跳过非字母顺序的字符，但它始终忽略代码并添加移位的字符。我试图将用户输入移动 2，但编码字不应该有任何特殊字符，因此移动将从 za 绕回。下面的代码是在我尝试“if”语句之前的。encoded_word = ("")decoded_word = ("")def Encoding(text):    global encoded_word    letter_list = list(text)    length_list = len(text)     for i in range (0,length_list):        letter_ord = ord(letter_list[i])        encoded_letter = chr(letter_ord+2)        encoded_word = (encoded_word + encoded_letter)    print ("The encoded word is now:",encoded_word)def Decoding(text):    global decoded_word    letter_list = list(text)    length_list = len(text)     for i in range (0,length_list):        letter_ord = ord(letter_list[i])        decoded_letter = chr(letter_ord-2)        decoded_word = (decoded_word + decoded_letter)    print ("The decoded word is:",decoded_word)decode_encode = str(input("Would you like to encode or decode text? encode/decode: "))if decode_encode == "encode":    user_word = str(input("Enter in a word to encode: "))    Encoding(user_word)if decode_encode == "decode":    user_word = str(input("Enter in the encoded word: "))    Decoding(user_word)

Answer 1

要检查字符是否是字母，可以使用 isalpha()。例如应该只打印 d 和 f：

    list = ["3","2","-","1","4","5","-","3","d", "f"]

    character_list = []

    for i in list:

        if i.isalpha():

            character_list.append(i)

        

    print (character_list)

Answer 2

我认为你走在正确的轨道上，你只是要处理更多的边缘情况，例如移位量是否大于 26，或者移位需要环绕等。此外，使用字符串连接效率很低，+因为副本每次连接都需要对现有字符串进行连接。因此，请考虑附加到字符列表，并仅在末尾创建输出编码/解码字符串：

SHIFT_AMOUNT = 2

def encode(text):

    res = []

    actual_shift_amount = SHIFT_AMOUNT % 26

    for ch in text:

        new_letter = ch

        if ord('a') <= ord(ch) <= ord('z'):

            if (ord(ch) + actual_shift_amount) <= ord('z'):

                new_letter = chr(ord(ch) + actual_shift_amount)

            else:

                new_letter = chr(ord('a') + actual_shift_amount - (ord('z') - ord(ch) + 1))

        elif ord('A') <= ord(ch) <= ord('Z'):

            if (ord(ch) + actual_shift_amount) <= ord('Z'):

                new_letter = chr(ord(ch) + actual_shift_amount)

            else:

                new_letter = chr(ord('A') + actual_shift_amount - (ord('Z') - ord(ch) + 1))

        res.append(new_letter)

    return ''.join(res)

def decode(text):

    res = []

    actual_shift_amount = SHIFT_AMOUNT % 26

    for ch in text:

        new_letter = ch

        if ord('a') <= ord(ch) <= ord('z'):

            if (ord(ch) - actual_shift_amount) >= ord('a'):

                new_letter = chr(ord(ch) - actual_shift_amount)

            else:

                new_letter = chr(ord('z') - actual_shift_amount + (ord(ch) - ord('a') + 1))

        elif ord('A') <= ord(ch) <= ord('Z'):

            if (ord(ch) - actual_shift_amount) >= ord('A'):

                new_letter = chr(ord(ch) - actual_shift_amount)

            else:

                new_letter = chr(ord('Z') - actual_shift_amount + (ord(ch) - ord('A') + 1))

        res.append(new_letter)

    return ''.join(res)

decode_or_encode = input("Would you like to encode or decode text? encode/decode: ").lower()

if decode_or_encode == "encode":

    user_word = input("Enter in a word to encode: ")

    print(encode(user_word))

elif decode_or_encode == "decode":

    user_word = input("Enter in the encoded word: ")

    print(decode(user_word))

用法示例encode：

Would you like to encode or decode text? encode/decode: encode

Enter in a word to encode: Yoruke-Stack-Overflow

Aqtwmg-Uvcem-Qxgthnqy

用法示例decode：

Would you like to encode or decode text? encode/decode: decode

Enter in the encoded word: Aqtwmg-Uvcem-Qxgthnqy

Yoruke-Stack-Overflow

Answer 3

问题是，当加或减 2 时，您不会检查是否达到了字母表的限制（“a”或“z”）。你应该做类似的事情

if encoded_letter > "z": 
  # do something

Answer 4

您所遵循的方法并不是最好的方法。但是，假设您仅考虑小写字符。

编码

替换

encoded_letter = chr(letter_ord + 2)

和

encoded_ord = letter_ord + 2if encoded_ord > ord('z'):  # after encoding if exceeds from 'z'
    difference = encoded_ord - ord('z')  # finding how much exceeded from 'z'
    encoded_ord = ord('a') + difference + 1  # restart from 'a' again.encoded_letter = chr(encoded_ord)

解码

替换

decoded_letter = chr(letter_ord-2)

和

decoded_ord = letter_ord - 2        if decoded_ord < ord('a'):     # after decoding if deceeds from 'a'
    difference = ord('a') - decoded_ord   # finding how much deceeded from 'a'
    decoded_ord = ord('z') - difference + 1   # restart from 'z' again but backwarddecoded_letter = chr(decoded_ord)

应该管用