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pythonrecursiongeneratoryield
John Zwinck
John Zwinck

Reputation: 249223

Python generator to yield everything from another generator call

I have a Python generator that can call itself to get more elements to yield. It looks like this:

def gen(list):
    # ...
    if list:
        for x in gen(list[1:]):
            yield x

My question is about the last two lines: is there a more concise way to express this? I am hoping for something like this (understanding this isn't valid Python as-is):

def gen(list):
    # ...
    if list:
        yield each in gen(list[1:])

Upvotes: 26

Views: 12931

Answers (4)

Carl G
Carl G

Reputation: 18260

Python 3.3 added the yield from keyword. Here's a comparison between what you currently have and code using the new keyword:

yield_from_test.py:

def gen_for(a_list):
    if a_list:
        yield a_list[0]
        for x in gen_for(a_list[1:]):
            yield x

def gen_yield(a_list):
    if a_list:
        yield a_list[0]
        yield from gen_yield(a_list[1:])

if __name__ == '__main__':
    assert list(gen_for([1,2,3])) == list(gen_yield([1,2,3]))
    print(list(gen_yield([1,2,3])))

» python3 yield_from_test.py [1, 2, 3]

Upvotes: 21

kindall
kindall

Reputation: 184211

There's been some call for a yield from or the like that "passes through" all the values returned by a subgenerator. See PEP 380 for some ideas that have been bounced around. However, nothing has been implemented yet. Your first example is correct.

Upvotes: 28

Alexander Gessler
Alexander Gessler

Reputation: 46607

Your code sample is very idiomatic and concise, no need and no real chance for further improvements and especially not when it comes at readability.

Upvotes: 6

mikerobi
mikerobi

Reputation: 20878

Your first example is the proper solution.

Upvotes: 4

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