filter based on set membership

Question

I would love

small_evens = set([2, 4, 6, 8])
only_small_odds = filter(not in small_evens, [1, 3, 4, 1, 1, 2, 7, 6, 4, 5])

but of course this is a syntax error.

both filter(lambda x: x not in set, items) and [x for x in items if x not in set] feel too verbose.

Is there another way?

I suspect there might be because, for example, newer Python versions have made map(str.lower, strings) possible (used to have to map(lambda s: s.lower(), strings))

Why use `filter`? Since you're already working with sets, why not: `set([1, 2, 3, 4, 5, 6, 7, 8, 9]) - set([2, 4, 6, 8])` — Paul M., Oct 17 '20 at 15:01
'x and y feel to verbose' is opinion based. Both are pythonic solutions to your question. — Michael Szczesny, Oct 17 '20 at 15:08

Stuart · Answer 1 · 2020-10-17T15:39:37.277

0

You could write a function:

def not_in(it):
    def inner(x):
        return x not in it
    return inner

only_small_odds = filter(not_in(small_evens), [1, 3, 4, 1, 1, 2, 7, 6, 4, 5])

But the 'verbose' ways you don't like are more pythonic.

You could also subclass list to make a subtractable list class, and then do:

only_small_odds = SubtractableList([1, 3, 4, 1, 1, 2, 7, 6, 4, 5]) - small_evens

edited Oct 17 '20 at 15:39

answered Oct 17 '20 at 15:24

Stuart

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score 0 · Answer 2 · answered Oct 17 '20 at 15:49

0

You can do this, which is pretty close:

from itertools import filterfalse

only_small_odds = filterfalse(small_evens.__contains__, [1, 3, 4, 1, 1, 2, 7, 6, 4, 5])

answered Oct 17 '20 at 15:49

Brian McCutchon

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filter based on set membership

2 Answers2