如何比较两个列表中的项目并根据结果创建一个新列表？

Question

As an exercise, I'm trying to code a simple XOR decipher.

I want to take a binary input of ASCII and store it in a list. Then take a binary key and also store it in a list.

A function should compare items from both lists respectively and check whether two items that are being compared are a 0 or 1. If both are 0 or both are 1 then output item = 0. If one item is 0 and the other 1 or vice versa, then output item = 1.

For each comparison that the function does, it should store the output binary item in a result list. Once the function iterates over each item in both lists, it should print the resulting binary output of result .

My logic:

def xor():

    binary_plaintext = [1,0,1,0,1,0,0,1, 0,1,0,0,0,1,1,0, 1,1,1,0,0,0,0,1]
    binary_key =       [1,1,1,1,1,0,1,1, 0,0,0,0,1,1,1,1, 1,0,1,1,0,1,0,1]
    result =           []

    for plaintext_item in binary_plaintext and key_item in binary_key:

        if plaintext_item == 1 and key_item == 1: # XOR 1|1 = 0
            output = 0
            output.append(result)

        elif plaintext_item == 0 and key_item == 0: # XOR 0|0 = 0
            output = 0
            output.append(result)

        elif plaintext_item == 1 and key_item == 0: # XOR 1|0 = 1
            output = 1
            output.append(result)

        elif plaintext_item == 0 and key_item == 1: # XOR 0|1 = 1
            output = 1
            output.append(result)

        print(result)

xor()

My example doesn't work. What would be the correct way of comparing binary_plaintext list and binary_key list and iterating their items so that an output value can be stored in a new list?

Answer 1

Basically you can implement an XOR with the addition of both inputs and then modulo 2.

For example that would work:

result = [(x+y) % 2 for x,y in zip(binary_plaintext,binary_key)]

print(result)

Output

[0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0]

Answer 2

你永远不会改变result ，尽管你反复将它附加到output ，当我认为你有倒退时（即附加output到result ）。

Answer 3

This should help:

result = [int(binary_plaintext[entry]!=binary_key[entry]) for entry in range(len(binary_plaintext))]
# [0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0]

Answer 4

A couple of mistakes in your function - firstly, to iterate over the lists in parallel, you can use zip() - so we can replace

for plaintext_item in binary_plaintext and key_item in binary_key:

with

for plaintext_item, key_item in zip(binary_plaintext, binary_key):

Secondly, you need to use

result.append(output)

instead of

output.append(result)

And lastly, unindent print(result) to take it out of the for loop.

The corrected function then prints

[0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0]

Answer 5

For reference you can do this with numpy.

import numpy as np

x = np.asarray([0,0,1,1])
y = np.asarray([0,1,0,1])

xor = x ^ y

In your own implementation, use zip to get matching pairs from two lists. You've also interchanged output and results by mistake:

def xor():

    binary_plaintext = [1,0,1,0,1,0,0,1, 0,1,0,0,0,1,1,0, 1,1,1,0,0,0,0,1]
    binary_key =       [1,1,1,1,1,0,1,1, 0,0,0,0,1,1,1,1, 1,0,1,1,0,1,0,1]
    result =           []

    for plaintext_item, key_item in zip(binary_plaintext, binary_key):

        if plaintext_item == 1 and key_item == 1: # XOR 1|1 = 0
            output = 0
            result.append(output)

        elif plaintext_item == 0 and key_item == 0: # XOR 0|0 = 0
            output = 0
            result.append(output)

        elif plaintext_item == 1 and key_item == 0: # XOR 1|0 = 1
            output = 1
            result.append(output)

        elif plaintext_item == 0 and key_item == 1: # XOR 0|1 = 1
            output = 1
            result.append(output)

    print(result)

xor()

Answer 6

We can use the python ^ operator to do this quite simply:

def xor():
    binary_plaintext = [1,0,1,0,1,0,0,1, 0,1,0,0,0,1,1,0, 1,1,1,0,0,0,0,1]
    binary_key =       [1,1,1,1,1,0,1,1, 0,0,0,0,1,1,1,1, 1,0,1,1,0,1,0,1]
    return [a ^ b for (a, b) in zip(binary_plaintext, binary_key]

We zip together binary_plaintext and binary_key and XOR the contents of each tuple in a list comprehension.

This assumes binary_plaintext and binary_key are the same length.

Answer 7

You can use zip. I'm using python3.

for text,key in  zip(binary_plaintext, binary_key)
   result.append(text ^ key)

Answer 8

this should work, edited your code for correct result

def xor():

    binary_plaintext = [1,0,1,0,1,0,0,1, 0,1,0,0,0,1,1,0, 1,1,1,0,0,0,0,1]
    binary_key =       [1,1,1,1,1,0,1,1, 0,0,0,0,1,1,1,1, 1,0,1,1,0,1,0,1]
    result =           []

    for i in range(0,len(binary_plaintext)):

        if binary_plaintext[i] == binary_key[i]: # XOR 1|1 = 0 # XOR 0|0 = 0
            output = 0
            result.append(output)

        elif binary_plaintext[i] == 1 and binary_key[i] == 0: # XOR 1|0 = 1
            output = 1
            result.append(output)

        elif binary_plaintext[i] == 0 and binary_key[i] == 1: # XOR 0|1 = 1
            output = 1
            result.append(output)

    print(result)

xor()