How to check if one of the following items is in a list?

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Track title: Cosmic Puzzle

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Chapters
00:00 Question
00:30 Accepted answer (Score 345)
00:47 Answer 2 (Score 306)
01:02 Answer 3 (Score 34)
01:15 Answer 4 (Score 20)
01:59 Thank you

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Full question
https://stackoverflow.com/questions/7402...

Answer 1 links:
[Tobias' solution]: https://stackoverflow.com/a/740359/45183...

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Tags
#python

#avk47

ACCEPTED ANSWER

Score 360

>>> L1 = [2,3,4]
>>> L2 = [1,2]
>>> [i for i in L1 if i in L2]
[2]


>>> S1 = set(L1)
>>> S2 = set(L2)
>>> S1.intersection(S2)
set([2])

Both empty lists and empty sets are False, so you can use the value directly as a truth value.

ANSWER 2

Score 331

I was thinking of this slight variation on Tobias' solution:

>>> a = [1,2,3,4]
>>> b = [2,7]
>>> any(x in a for x in b)
True

ANSWER 3

Score 36

Maybe a bit more lazy:

a = [1,2,3,4]
b = [2,7]

print any((True for x in a if x in b))

ANSWER 4

Score 20

Think about what the code actually says!

>>> (1 or 2)
1
>>> (2 or 1)
2

That should probably explain it. :) Python apparently implements "lazy or", which should come as no surprise. It performs it something like this:

def or(x, y):
    if x: return x
    if y: return y
    return False

In the first example, x == 1 and y == 2. In the second example, it's vice versa. That's why it returns different values depending on the order of them.