是您的实际字母，因此请更改为

Question

我想写一个简单的程序来交叉加密一个单词。（即改变＆＃34; A＆＃34;到＆＃34; Z＆＃34;，＆＃34; B＆＃34;到＆＃34; Y＆＃34;等等）我是python的新手并且无法获得循环工作。这就是我所拥有的。我意识到可能有更好的方法来做到这一点。但我试图用最基本的语言基础知识来做到这一点。

w = str.upper(input("Enter a word\n"))
l = list(w)
for i in l:
    if l[i] is "A":
        l[i] = "Z"
    elif l[i] is "B":
        l[i] = "y"
    elif l[i] is "C":
        l[i] = "X"
    elif l[i] is "D":
        l[i] = "W"
    elif l[i] is "E":
        l[i] = "V"
    elif l[i] is "F":
        l[i] = "U"
    elif l[i] is "G":
        l[i] = "T"
    elif l[i] is "H":
        l[i] = "S"
    elif l[i] is "I":
        l[i] = "R"
    elif l[i] is "J":
        l[i] = "Q"
    elif l[i] is "K":
        l[i] = "P"
    elif l[i] is "L":
        l[i] = "O"
    elif l[i] is "M":
        l[i] = "N"
    elif l[i] is "N":
        l[i] = "M"
    elif l[i] is "O":
        l[i] = "L"
    elif l[i] is "P":
        l[i] = "K"
    elif l[i] is "Q":
        l[i] = "J"
    elif l[i] is "R":
        l[i] = "I"
    elif l[i] is "S":
        l[i] = "H"
    elif l[i] is "T":
        l[i] = "G"
    elif l[i] is "U":
        l[i] = "F"
    elif l[i] is "V":
        l[i] = "E"
    elif l[i] is "W":
        l[i] = "D"
    elif l[i] is "X":
        l[i] = "C"
    elif l[i] is "Y":
        l[i] = "B"
    elif l[i] is "Z":
        l[i] = "A"
    print("".join(l))

我收到错误消息，说列表位置[i]无效，因为它是一个字符串。但是，我的意图是使用值作为计数器来遍历列表对象并更改字母。

Answer 1

from string import ascii_lowercase as lc,ascii_uppercase as uc,maketrans
transtab = maketrans(lc+uc,lc[::-1]+uc[::-1])
print("Hello".translate(transtab))

是一种方式（可能是首选方式）

from string import ascii_lowercase as lc,ascii_uppercase as uc,maketrans
translator = dict((letter1,letter2) for letter1,letter2 in zip(uc,uc[::-1]))
print("".join(translator.get(c,c) for c in "HELLO")

是使用字典的另一种方式

input_msg= "HELLO"
new_message = ""
for letter in input_msg:
    new_message += chr(ord("Z")-(ord(letter)-ord("A")))

print (new_message)

是使用数学的另一种方式...... 基本上有一百万种方法可以实现这个目标

关于您的代码i的

是您的实际字母，因此请更改为

if i=='A':...

如果您还想要索引（替换）

for index,letter in enumerate(l):
    if letter == "A":
        l[index] = "Z"

Answer 2

您可以使用ord功能将字符映射到其unicode代码点。对于26个大写英文字母，这些字母对应于介于65和90之间的整数值。从155中减去每个字符的整数代码将为其提供补码的字符代码。

w = input('Enter a word\n').upper()
print(''.join(map(lambda c: chr(155 - ord(c)), w)))

Answer 3

for i in l:遍历l中的字符，而不是数字索引。要使代码以最少的更改运行，您可以更改for循环，如下所示。您还需要在循环外移动print调用。

w = str.upper(input("Enter a word\n"))
l = list(w)
for i in range(len(l)):
    if l[i] is "A":
        l[i] = "Z"
    elif l[i] is "B":
        l[i] = "Y"
    elif l[i] is "C":
        l[i] = "X"
    elif l[i] is "D":
        l[i] = "W"
    elif l[i] is "E":
        l[i] = "V"
    elif l[i] is "F":
        l[i] = "U"
    elif l[i] is "G":
        l[i] = "T"
    elif l[i] is "H":
        l[i] = "S"
    elif l[i] is "I":
        l[i] = "R"
    elif l[i] is "J":
        l[i] = "Q"
    elif l[i] is "K":
        l[i] = "P"
    elif l[i] is "L":
        l[i] = "O"
    elif l[i] is "M":
        l[i] = "N"
    elif l[i] is "N":
        l[i] = "M"
    elif l[i] is "O":
        l[i] = "L"
    elif l[i] is "P":
        l[i] = "K"
    elif l[i] is "Q":
        l[i] = "J"
    elif l[i] is "R":
        l[i] = "I"
    elif l[i] is "S":
        l[i] = "H"
    elif l[i] is "T":
        l[i] = "G"
    elif l[i] is "U":
        l[i] = "F"
    elif l[i] is "V":
        l[i] = "E"
    elif l[i] is "W":
        l[i] = "D"
    elif l[i] is "X":
        l[i] = "C"
    elif l[i] is "Y":
        l[i] = "B"
    elif l[i] is "Z":
        l[i] = "A"
print("".join(l))

但是，有多更好的方法可以执行此操作，不需要26个if / elif语句。

顺便说一句，尽管它在这里有用，但最好是if l[i] == "A":而不是if l[i] is "A":。

这是一种改进的方法，希望不会太为您提升。它使用扩展切片来创建字母的反转版本。它使用zip函数迭代一对字符串; zip可用于任意数量的字符串，也可用于列表等其他容器。它使用字典来存储转换表。这个版本处理上部和上部小写字母。

letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
rev_letters = letters[::-1]

# Build translation dictionary
trans = {}

# Loop over both letter strings in parallel
for u, v in zip(letters, rev_letters):
    trans[u] = v
    trans[u.lower()] = v.lower()    

words = input("Enter some words: ")
lst = []
for ch in words:
    # Get the translated version of ch if it exists, 
    # otherwise return ch
    newch = trans.get(ch, ch)
    # Add it to the list
    lst.append(newch)
print("".join(lst))

测试

Enter some words: Wizard abc
Draziw zyx

Python中的基本交叉加密

3 个答案:

是您的实际字母，因此请更改为

测试