What's the idiomatic way to waste the output of a generator in Python? [duplicate]

Question

Sometimes I need to consume everything that a generator outputs but don't actually need the output. In one case, some callers need progress updates from a coroutine, but some callers don't; in another case, some callers need to iterate through the results but some callers only need the side effects.

What's the idiomatic way to do this?

I currently use list():

def do_work_and_yield():
    for x in ...:
        y = do_work_on(x)
        yield y

def only_cares_about_side_effects():
    list(do_work_and_yield())

but of course this is an abuse of list() and a waste of memory. I could write a loop:

def only_cares_about_side_effects():
    for _ in do_work_and_yield():
        pass

but what I Really Mean is just "run do_work_and_yield to completion", which the loop obfuscates.

wjandrea · Accepted Answer · 2024-10-14 22:56:19Z

0

Naturally, the solution was hiding in the standard library documentation. The itertools docs give a recipe for a function called consume that does just what I want.

The "recipes" in the itertools documentation are conveniently available on PyPI in the more-itertools package.

So now I can just write

from more_itertools import consume

def do_work_and_yield():
    for x in ...:
        y = do_work_on(x)
        yield y

def only_cares_about_side_effects():
    consume(do_work_and_yield())

edited Oct 14, 2024 at 22:56

wjandrea

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answered Oct 14, 2024 at 22:27

zmccord

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