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Single line for loop over iterator with an “if” filter?

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清歌不尽
清歌不尽 2020-12-13 12:55

Silly question:
I have a simple for loop followed by a simple if statement: 

for airport in airports:
    if airport.is_important:

and

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  • 轻奢々
    2020-12-13 13:05

    I'd use a negative guard on the loop. It's readable, and doesn't introduce an extra level of indentation.

    for airport in airports:
    if not airport.is_important: continue
    <body of loop>
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  • 日久生厌
    2020-12-13 13:06

    Here's an alternative to some of the other filter versions:

    from operator import attrgetter as attr
for airport in filter(attr('is_important'), airports):
    ...

    This has the advantages of being pretty concise and also letting you use dot notation attr('first_class.is_full').

    You could also put something like that (or a version using a list comprehension) into a utility function like filter_by_attr. Then you could do:

    for airport in filter_by_attr(airports, 'is_important'):
    ...

    I still think e-satis is right to put it in a new variable no matter the method you use, though. It is just clearer that way, especially if the use doesn't exactly match the name of the attribute in question (or the the criteria is more complex).

    My only note on that would be that if you find yourself using this in several places, perhaps you should make airports a special collection with 'important_airports' being a @property which returns the filtered collection. Or some sort other abstraction to hide away the filtering (like a service call).

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  • 遥遥无期
    2020-12-13 13:09

    You could do

    for airport in filter(lamdba x: x.is_important, airports):
    # do stuff...
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  • 不知归路
    2020-12-13 13:14

    I found the usage of the filter & lambda to be the easiest in a single line.

    for filtered_airport in filter(lambda airport: airport.is_important, airports)):
    print(filtered_airport.some_property)

    If you wish to return a list from output of this filter object you can do something like this.

    filtered_airport_list = list(filter(lambda airport: airport.is_important, airports)))
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  • 野性不改
    2020-12-13 13:16

    No, there is no shorter way. Usually, you will even break it into two lines :

    important_airports = (airport for airport in airports if airport.is_important)
for airport in important_airports:
    # do stuff

    This is more flexible, easier to read and still don't consume much memory.

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  • 时光说笑
    2020-12-13 13:17

    This is a design philosophy of python. If it takes you too many words to put it on one line, it should be broken into a few lines to help the person who comes after you. List and generator expressions are more for transforming iterables in-place -- making more readable forms of map and filter.

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