为凯撒密码构建字典

Question

假设输入了一条消息，并且您想通过将单词“up the alphabet”中的每个字母移动 x 个空格来对其进行加密。例如“cat”移动了一个空格变成了“dbu”：

我正在尝试构建一个字典，其中大写和小写字母作为键，重复字母（移动“移位”字母数）作为值。例如 shift=1 d={a:b, b:c...A:B, B:C}。每次“班次”改变时，这个字母都会改变。请让我知道我哪里出错了。请参阅：def build_shift_dict(self, shift) 下面。提前致谢！：

class Message(object):
### DO NOT MODIFY THIS METHOD ###
def __init__(self, text):
    '''
    Initializes a Message object
            
    text (string): the message's text

    a Message object has two attributes:
        self.message_text (string, determined by input text)
        self.valid_words (list, determined using helper function load_words
    '''
    self.message_text = text
    self.valid_words = load_words(WORDLIST_FILENAME)

### DO NOT MODIFY THIS METHOD ###
def get_message_text(self):
    '''
    Used to safely access self.message_text outside of the class
    
    Returns: self.message_text
    '''
    return self.message_text

### DO NOT MODIFY THIS METHOD ###
def get_valid_words(self):
    '''
    Used to safely access a copy of self.valid_words outside of the class
    
    Returns: a COPY of self.valid_words
    '''
    return self.valid_words[:]
    
import string    
def build_shift_dict(self, shift):
    '''
    Creates a dictionary that can be used to apply a cipher to a letter.
    The dictionary maps every uppercase and lowercase letter to a
    character shifted down the alphabet by the input shift. The dictionary
    should have 52 keys of all the uppercase letters and all the lowercase
    letters only.        
    
    shift (integer): the amount by which to shift every letter of the 
    alphabet. 0 <= shift < 26

    Returns: a dictionary mapping a letter (string) to 
             another letter (string). 
    '''
    stringLower = string.ascii_lowercase 
    stringUpper = string.ascii_uppercase
    Alphabet = list(stringLower + stringUpper)
    dictionary = {}
    string2Lower = list(string.ascii_lowercase[shift:] + string.ascii_lowercase[0:shift])
    string2Upper = list(string.ascii_uppercase[shift:] + string.ascii_uppercase[0:shift])
    combinedList = string2Lower.append(string2Upper)
    
    for key in Alphabet: 
        for value in combinedList: 
            dictionary[key] = value 
            combinedList.remove(value) 
             
    return dictionary
    

    
    

def apply_shift(self, shift):
    '''
    Applies the Caesar Cipher to self.message_text with the input shift.
    Creates a new string that is self.message_text shifted down the
    alphabet by some number of characters determined by the input shift        
    
    shift (integer): the shift with which to encrypt the message.
    0 <= shift < 26

    Returns: the message text (string) in which every character is shifted
         down the alphabet by the input shift
    '''
    result=''
    dictionary = self.build_shift_dict(shift)
    messageString = str(Message)
    for letter in messageString:
        if letter in dictionary.keys:
         result += dictionary[i]
    return result

Answer 1

您可以删除很多基于类的东西，因为除了添加样板代码和使事情复杂化之外，它并没有做太多事情。你可以在之后把它放回去，但是在这里我以一种更 Python 的方式为你写了核心功能：

import string

def build_shift_dict(shift):

    alphabet = string.ascii_lowercase + string.ascii_uppercase

    shifted_lower = string.ascii_lowercase[shift:] + string.ascii_lowercase[0:shift]
    shifted_upper = string.ascii_uppercase[shift:] + string.ascii_uppercase[0:shift]

    shifted_alphabet = shifted_lower + shifted_upper

    return {key: value for key, value in zip(alphabet, shifted_alphabet)}

获得shifted_alphabet 后，您可以将zip 与原始字母表一起使用，然后使用Dictionary Comprehension 创建字典。这是一种更 Pythonic 的方式来制作 dict，而不是 for 循环

---

然后要应用移位，我们可以使用list comprehension 来查找消息中每个字母的字典，并创建一个列表，然后将join 重新组合成一个字符串并返回。

def apply_shift(shift, message):
    dictionary = build_shift_dict(shift)

    return ''.join(dictionary[letter] for letter in message)

---

然后使用：

>>> print(apply_shift(2, "Hello"))
Jgnnq

---

在这里-> https://repl.it/@LukeStorry/63042493

---

更新：您可以通过使用 ord 和 chr 来避免创建和使用字典：

def apply_shift_v2(shift, message):
    return ''.join(chr(ord(letter) + shift) for letter in message)

这具有适用于任何字符的额外好处，而不仅限于您设置字典时使用的字母数字字符。

Answer 2

我的建议是不要创建字典，这会更简单

def apply_shift(shift, letter):
    asci = ord(letter)
    new_ascii = 97 + (asci + shift -97) % 26
    return chr(new_ascii)

print(apply_shift(1, 'z') )

输出：

a