我想在Python中创建一个简单的恺撒密码功能转移基于用户输入的字母，并创建在年底最后，新的字符串。唯一的问题是，最后的密文只显示最后移动字符，而不是整个字符串的所有移动字符。
这里是我的代码：
plainText = raw_input("What is your plaintext? ")
shift = int(raw_input("What is your shift? "))
def caesar(plainText, shift):
for ch in plainText:
if ch.isalpha():
stayInAlphabet = ord(ch) + shift
if stayInAlphabet > ord('z'):
stayInAlphabet -= 26
finalLetter = chr(stayInAlphabet)
cipherText = ""
cipherText += finalLetter
print "Your ciphertext is: ", cipherText
return cipherText
caesar(plainText, shift)
我知道这个答案并没有真正回答你的问题，但我认为这是有帮助的呢。下面就来实现与字符串方法凯撒密码的另一种方式：
def caesar(plaintext, shift):
alphabet = string.ascii_lowercase
shifted_alphabet = alphabet[shift:] + alphabet[:shift]
table = string.maketrans(alphabet, shifted_alphabet)
return plaintext.translate(table)
事实上，因为字符串方法是用C语言实现，我们将看到这个版本的性能提高。这是我会考虑这样做的“Python化”的方式。
>>> def rotate(txt, key):
... def cipher(i, low=range(97,123), upper=range(65,91)):
... if i in low or i in upper:
... s = 65 if i in upper else 97
... i = (i - s + key) % 26 + s
... return chr(i)
... return ''.join([cipher(ord(s)) for s in txt])
# test
>>> rotate('abc', 2)
'cde'
>>> rotate('xyz', 2)
'zab'
>>> rotate('ab', 26)
'ab'
>>> rotate('Hello, World!', 7)
'Olssv, Dvysk!'
def encrypt():
plainText = input("What is your plaintext? ")
shift = int(input("What is your shift? "))
cipherText = ""
for ch in plainText:
if ch.isalpha():
stayInAlphabet = ord(ch) + shift
if stayInAlphabet > ord('z'):
stayInAlphabet -= 26
finalLetter = chr(stayInAlphabet)
cipherText += finalLetter
print ("Your ciphertext is: ", cipherText,"with a shift of",shift)
def decrypte():
encryption=input("enter in your encrypted code")
encryption_shift=int(input("enter in your encryption shift"))
cipherText1 = ""
for c in encryption:
if c.isalpha():
stayInAlphabet1 = ord(c) - encryption_shift
if stayInAlphabet1 > ord('z'):
stayInAlphabet1 += 26
finalLetter1 = chr(stayInAlphabet1)
cipherText1 += finalLetter1
print ("Your ciphertext is: ", cipherText1,"with negative shift of",encryption_shift)
from tkinter import *
menu=Tk()
menu.title("menu")
menu.geometry("300x300")
button1= Button(menu,text="encrypt",command=encrypt)
button1.pack()
button2= Button(menu,text="decrypt",command=decrypte)
button2.pack()
button3= Button(menu,text="exit",command=exit)
button3.pack()
menu.mainloop()
这是the answer of @amillerrhodes与不同的字母，不只是小写的工作代码的改进版本：
def caesar(text, step, alphabets):
def shift(alphabet):
return alphabet[step:] + alphabet[:step]
shifted_alphabets = tuple(map(shift, alphabets))
joined_aphabets = ''.join(alphabets)
joined_shifted_alphabets = ''.join(shifted_alphabets)
table = str.maketrans(joined_aphabets, joined_shifted_alphabets)
return text.translate(table)
使用示例：
>>> import string
>>> alphabets = (string.ascii_lowercase, string.ascii_uppercase, string.digits)
>>> caesar('Abc-xyZ.012:789?жñç', step=4, alphabets=alphabets)
'Efg-bcD.456:123?жñç'
参考文献：
Docs on str.maketrans。
Docs on str.translate。
Docs on the string library
from string import ascii_lowercase as alphabet
class CaesarCypher:
alpha_len = len(alphabet)
min_guess_rate = 0.2
加密和解密是同样的东西。当你想用这意味着你可以用移26进行加密移10来解密例如 - 10在这种情况下周期将重复的，如果你要移动整个字母也将是相同的。另外在这里，我继续大写和非字符
def __call__(self, text, offset, encrypt=True):
if not encrypt:
offset = self.alpha_len - offset
result = []
for letter in text:
if not letter.isalpha():
result.append(letter)
continue
letter_to_process = letter.lower()
processed_letter = self._encrypt_letter(letter_to_process, offset)
if letter.isupper():
processed_letter = processed_letter.upper()
result.append(processed_letter)
return ''.join(result)
所有加密到这里最多。
def _encrypt_letter(self, letter, offset=0):
position = (alphabet.find(letter) + offset) % self.alpha_len
return alphabet[position]
这部分是蛮力，并通过字典猜测频率。
@staticmethod
def __how_many_do_i_know(text):
clean_words = filter(lambda x: x.isalpha(), text.split())
clean_words = ['\'{}\''.format(x) for x in clean_words]
cursor = conn.cursor()
query = 'SELECT COUNT(*) FROM mydictionary WHERE word IN ({})'.format(",".join(clean_words))
cursor.execute(query)
response = cursor.fetchone()[0]
return response / len(clean_words)
def guess_encode(self, text):
options = [self(text, offset, encrypt=False) for offset in range(self.alpha_len)]
best_option = [self.__how_many_do_i_know(option) for option in options]
best_key, guess_rate = max(enumerate(best_option), key=lambda x: x[-1])
guess_text = options[best_key]
return best_key, guess_rate, guess_text
import string
wrd=raw_input("Enter word").lower()
fwrd=""
for let in wrd:
fwrd+=string.ascii_lowercase[(string.ascii_lowercase).index(let)+3]
print"Original word",wrd
print"New word",fwrd
根据我这个答案对你来说是有用的：
def casear(a,key):
str=""
if key>26:
key%=26
for i in range(0,len(a)):
if a[i].isalpha():
b=ord(a[i])
b+=key
#if b>90: #if upper case letter ppear in your string
# c=b-90 #if upper case letter ppear in your string
# str+=chr(64+c) #if upper case letter ppear in your string
if b>122:
c=b-122
str+=chr(96+c)
else:
str+=chr(b)
else:
str+=a[i]
print str
a=raw_input()
key=int(input())
casear(a,key)
此功能转移根据给定键的所有字母向右。
我有一个很难记住的字符为int的转换，以便这可以优化
def decryptCaesar(encrypted, shift):
minRange = ord('a')
decrypted = ""
for char in encrypted:
decrypted += chr(((ord(char) - minRange + shift) % 26) + minRange)
return decrypted
为什么不使用上移输入功能反转，并和参加与移plain_text，并输入它作为密文：
Plain = int(input("enter a number "))
Rev = plain[::-1]
Cipher = " ".join(for cipher_text in Rev)
message = 'The quick brown fox jumped over the lazy dog. 1234567890 !@#$%^&*()_+-'
encrypted = ''.join(chr(ord(char)+3) for char in message)
decrypted = ''.join(chr(ord(char)-3) for char in encrypted)
print(encrypted)
print(decrypted)
# Wkh#txlfn#eurzq#ir{#mxpshg#ryhu#wkh#od}|#grj1#456789:;<3#$C&'(a)-+,b.0
# The quick brown fox jumped over the lazy dog. 1234567890 !@#$%^&*()_+-
key = 3
def wub():
def choice():
choice = input("Do you wish to Encrypt of Decrypt?")
choice = choice.lower()
if choice == "e" or "encrypt":
return choice
elif choice == "d" or "decrypt":
return choice
else:
print("Invalid response, please try again.")
choice()
def message():
user = input("Enter your message: ")
return user
def waffle(choice, message, key):
translated = ""
if choice == "e" or "encrypt":
for character in message:
num = ord(character)
num += key
translated += chr(num)
derek = open('Encrypted.txt', 'w')
derek.write(translated)
derek.close()
return translated
else:
for character in message:
num = ord(character)
num -= key
translated += chr(num)
return translated
choice = choice() #Runs function for encrypt/decrypt selection. Saves choice made.
message = message() #Run function for user to enter message. Saves message.
final = waffle(choice, message, key) #Runs function to translate message, using the choice, message and key variables)
print("\n Operation complete!")
print(final)
wub()
你需要的for循环开始之前移动cipherText = ""。你通过循环每次重新设定它。
def caesar(plainText, shift):
cipherText = ""
for ch in plainText:
if ch.isalpha():
stayInAlphabet = ord(ch) + shift
if stayInAlphabet > ord('z'):
stayInAlphabet -= 26
finalLetter = chr(stayInAlphabet)
cipherText += finalLetter
print "Your ciphertext is: ", cipherText
return cipherText
使用一些ASCII一些花样：
# See http://ascii.cl/
upper = {ascii:chr(ascii) for ascii in range(65,91)}
lower = {ascii:chr(ascii) for ascii in range(97,123)}
digit = {ascii:chr(ascii) for ascii in range(48,58)}
def ceasar(s, k):
for c in s:
o = ord(c)
# Do not change symbols and digits
if (o not in upper and o not in lower) or o in digit:
yield o
else:
# If it's in the upper case and
# that the rotation is within the uppercase
if o in upper and o + k % 26 in upper:
yield o + k % 26
# If it's in the lower case and
# that the rotation is within the lowercase
elif o in lower and o + k % 26 in lower:
yield o + k % 26
# Otherwise move back 26 spaces after rotation.
else: # alphabet.
yield o + k % 26 -26
x = (''.join(map(chr, ceasar(s, k))))
print (x)
包括电池
while 1:
phrase = raw_input("Could you please give me a phrase to encrypt?\n")
if phrase == "" : break
print "Here it is your phrase, encrypted:"
print phrase.encode("rot_13")
print "Have a nice afternoon!"
https://docs.python.org/2/library/codecs.html#python-specific-encodings
Python 3的更新
[现在
rot_13]编解码器提供了一个文本转换：一个str到str映射。它不被支持str.encode()（其仅产生字节输出）。
或者，换句话说，你必须从encode模块导入codecs并用串用它来进行编码的第一个参数
from codecs import decode
...
print(encode(phrase, 'rot13'))
问题是，你在每个循环迭代中设置的密文空字符串，行
cipherText = ""
必须在循环开始前被移动。
正如指出的其他人，你是在复位的for循环迭代的密文。 for循环开始前放置密文将解决你的问题。
此外，还有解决使用Python的标准库这一问题的另一种方法。 Python标准库定义了一个函数maketrans（）和一个翻译方法，关于串操作。
功能maketrans（）创建可以与平移的方法用于将一个字符集的更有效地改变到另一个的转换表。 （由实施例从Python标准库的引用）。
import string
def caesar(plaintext, shift):
shift %= 26 # Values greater than 26 will wrap around
alphabet_lower = string.ascii_lowercase
alphabet_upper = string.ascii_uppercase
shifted_alphabet_lower = alphabet_lower[shift:] + alphabet_lower[:shift]
shifted_alphabet_upper = alphabet_upper[shift:] + alphabet_upper[:shift]
alphabet = alphabet_lower + alphabet_upper
shifted_alphabet = shifted_alphabet_lower + shifted_alphabet_upper
table = string.maketrans(alphabet, shifted_alphabet)
return plaintext.translate(table)
plainText = raw_input("What is your plaintext? ")
shift = int(raw_input("What is your shift? "))
def caesar(plainText, shift):
for ch in plainText:
if ch.isalpha():
stayInAlphabet = ord(ch) + shift
if stayInAlphabet > ord('z'):
stayInAlphabet -= 26
finalLetter = chr(stayInAlphabet)
#####HERE YOU RESET CIPHERTEXT IN EACH ITERATION#####
cipherText = ""
cipherText += finalLetter
print "Your ciphertext is: ", cipherText
return cipherText
caesar(plainText, shift)
作为一个别的，如果ch.isalpha()你可以把finalLetter=ch。
您应该删除行：cipherText = ""
干杯。
正如@ I82much说，你需要采取cipherText = ""你之外的for循环。将其放置在函数的开始。此外，你的程序，这将导致它，当你得到大写字母输入生成加密错误的bug。尝试：
if ch.isalpha():
finalLetter = chr((ord(ch.lower()) - 97 + shift) % 26 + 97)
在这里，一个功能更强大的方式：（如果你用shift我来编码，然后使用-i来解码）
def ceasar(story, shift):
return ''.join([ # concentrate list to string
(lambda c, is_upper: c.upper() if is_upper else c) # if original char is upper case than convert result to upper case too
(
("abcdefghijklmnopqrstuvwxyz"*2)[ord(char.lower()) - ord('a') + shift % 26], # rotate char, this is extra easy since Python accepts list indexs below 0
char.isupper()
)
if char.isalpha() else char # if not in alphabet then don't change it
for char in story
])