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 = 2def 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: encodeEnter in a word to encode: Yoruke-Stack-OverflowAqtwmg-Uvcem-Qxgthnqy用法示例decode：Would you like to encode or decode text? encode/decode: decodeEnter in the encoded word: Aqtwmg-Uvcem-QxgthnqyYoruke-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&nbsp;encoded_letter&nbsp;>&nbsp;"z":&nbsp; &nbsp;&nbsp;#&nbsp;do&nbsp;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&nbsp;=&nbsp;chr(letter_ord&nbsp;+&nbsp;2)和encoded_ord&nbsp;=&nbsp;letter_ord&nbsp;+&nbsp;2if&nbsp;encoded_ord&nbsp;>&nbsp;ord('z'):&nbsp;&nbsp;#&nbsp;after&nbsp;encoding&nbsp;if&nbsp;exceeds&nbsp;from&nbsp;'z' &nbsp;&nbsp;&nbsp;&nbsp;difference&nbsp;=&nbsp;encoded_ord&nbsp;-&nbsp;ord('z')&nbsp;&nbsp;#&nbsp;finding&nbsp;how&nbsp;much&nbsp;exceeded&nbsp;from&nbsp;'z' &nbsp;&nbsp;&nbsp;&nbsp;encoded_ord&nbsp;=&nbsp;ord('a')&nbsp;+&nbsp;difference&nbsp;+&nbsp;1&nbsp;&nbsp;#&nbsp;restart&nbsp;from&nbsp;'a'&nbsp;again.encoded_letter&nbsp;=&nbsp;chr(encoded_ord)解码替换decoded_letter&nbsp;=&nbsp;chr(letter_ord-2)和decoded_ord&nbsp;=&nbsp;letter_ord&nbsp;-&nbsp;2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if&nbsp;decoded_ord&nbsp;<&nbsp;ord('a'):&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;#&nbsp;after&nbsp;decoding&nbsp;if&nbsp;deceeds&nbsp;from&nbsp;'a' &nbsp;&nbsp;&nbsp;&nbsp;difference&nbsp;=&nbsp;ord('a')&nbsp;-&nbsp;decoded_ord&nbsp;&nbsp;&nbsp;#&nbsp;finding&nbsp;how&nbsp;much&nbsp;deceeded&nbsp;from&nbsp;'a' &nbsp;&nbsp;&nbsp;&nbsp;decoded_ord&nbsp;=&nbsp;ord('z')&nbsp;-&nbsp;difference&nbsp;+&nbsp;1&nbsp;&nbsp;&nbsp;#&nbsp;restart&nbsp;from&nbsp;'z'&nbsp;again&nbsp;but&nbsp;backwarddecoded_letter&nbsp;=&nbsp;chr(decoded_ord)应该管用