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317 points est | 1 comments | 03 Jul 18 08:30 UTC | HN request time: 0.2s | source

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BerislavLopac ◴[03 Jul 18 10:24 UTC] No.17448998[source]▶

I would like to see adding a comprehension-like filtering clause to for-statements:

    for n in range(100) if n%2:
        print(f'{n} is odd number')

Does anyone know if there is a PEP covering that?

replies(5): >>17449022 #>>17449029 #>>17449035 #>>17449036 #>>17452328 #

theSage ◴[03 Jul 18 10:34 UTC] No.17449029[source]▶

>>17448998 #

Agreed these are a little verbose but they get the job done no?

    for n in filter(is_even, range(100)):
        print(f'{n} is odd number')

    for n in (i for i in range(100) if i % 2 == 0):
        print(f'{n} is odd number')

Are there any points against these solutions other than verbosity?

replies(1): >>17452040 #

1. BerislavLopac ◴[03 Jul 18 17:27 UTC] No.17452040[source]▶

>>17449029 #

Yes, that's what I've been using so far, especially filter, which works quite well with lambda. But if you have a separate function anyway it's better to make it into a generator:

    def odd_range(count):
        return (x for x in range(count) if x%2)
        
    for n in odd_range(100):
        ...

As for the second one, I'm just not too happy with the implied two loops (even if it amounts to only one in practice).

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PEP 572: Python assignment expressions has been accepted