Python generator to yield everything from another generator call

Question

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:])

I don't quite get what this is supposed to do, I originally thought it was just a naive way of converting a list to a generator but I was mistaken. Could you enlighten me? — Davy8, Jun 06 '11 at 20:45
Correct me if I'm wrong, but that code looks like it would always return an empty list. I assume there must be other yields in the comment? — Dunes, Jun 06 '11 at 21:25
@Davy8: it's not a wrapper or converter, it's a custom generator I wrote. It takes a list of things and generates another list, which may contain some of the same things, or not. A "filter" of sorts. — John Zwinck, Jun 07 '11 at 01:28

score 26 · Answer 1 · answered Jun 06 '11 at 21:47

26

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.

answered Jun 06 '11 at 21:47

kindall

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10

Note: Python 3.3 has been released, along with PEP 380 (yield from). – jomido Sep 30 '12 at 12:34

score 17 · Accepted Answer · answered May 03 '18 at 18:29

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]

score 5 · Answer 3 · answered Jun 06 '11 at 20:33

5

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.

answered Jun 06 '11 at 20:33

Alexander Gessler

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1

Node generators have [a syntax](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/yield*) for delegating generation to another generator: `yield* my_generator(...)`. I personally find that more concise. And things become idiomatic with a PEP and adoption... – Carl G Mar 19 '18 at 18:41

score 4 · Answer 4 · answered Jun 06 '11 at 20:31

4

Your first example is the proper solution.

answered Jun 06 '11 at 20:31

mikerobi

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Python generator to yield everything from another generator call

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