XOR on two lists in Python [duplicate]

Question

This question already has answers here:

Comparing two lists and only printing the differences? (XORing two lists) (6 answers)

Closed 9 years ago.

I'm a beginner in Python, and I have to do the XOR between two lists (the first one with the length : 600 and the other 60)

I really don't know how to do that, if somebody can explain me how, it will be a pleasure.

I have to do that to find the BPSK signal module, and I'm wondering on how doing that with two lists that haven't the same length. I saw this post : Comparing two lists and only printing the differences? (XORing two lists) but the length of lists is the same

Thanks for your help, and sry for my bad english Rom

Answer 1

Given sequences seq1 and seq2, you can calculate the symmetric difference with

set(seq1).symmetric_difference(seq2)

For example,

In [19]: set([1,2,5]).symmetric_difference([1,2,9,4,8,9])
Out[19]: {4, 5, 8, 9}

Tip: Generating the set with the smaller list is generally faster:

In [29]: %timeit set(range(60)).symmetric_difference(range(600))
10000 loops, best of 3: 25.7 µs per loop

In [30]: %timeit set(range(600)).symmetric_difference(range(60))
10000 loops, best of 3: 41.5 µs per loop

The reason why you may want to use symmetric difference instead of ^ (despite the beauty of its syntax) is because the symmetric difference method can take a list as input. ^ requires both inputs be sets. Converting both lists into sets is a little more computation than is minimally required.

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This question has been marked of as a duplicate of this question That question, however, is seeking a solution to this problem without using sets.

The accepted solution,

[a for a in list1+list2 if (a not in list1) or (a not in list2)]

is not the recommended way to XOR two lists if sets are allowed. For one thing, it's over 100 times slower:

In [93]: list1, list2 = range(600), range(60)

In [94]: %timeit [a for a in list1+list2 if (a not in list1) or (a not in list2)]
100 loops, best of 3: 3.35 ms per loop

Answer 2

There's the XOR operator on set. Assuming you have no duplicates (and you don't care about checking if an element appears more than 1 times or not in the second list), you can use the ^ operator on set:

>>> set([1, 2, 3, 4, 5]) ^ set([1, 3, 4, 5, 6])
set([2, 6])
>>> set(range(80)) ^ set(range(60))
set([60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79])