Asyncio - repeatedly invoke multiple functions and merge results

Question

I have two functions:

async def f(i):
    await asyncio.sleep(1)
    return f'f{i}'


async def g(i):
    await asyncio.sleep(2)
    return f'g{i}'

I want to write a loop that calls them repeatedly and prints the results as they come. Like this imaginary code:

for c in amerge(amap(f, itertools.count()), 
                amap(g, itertools.count())):
    x = await c
    print(x)

And the result should be approx f0, f1, g1, f2, f3, g2, f4, f5, g3, ...

My attempt was this:

async def run():
    """bad, always wait for f, then for g"""
    for i in itertools.count():
        for c in asyncio.as_completed([f(i), g(i)]):
            res = await c
            print(res)

asyncio.run(run())

But this is not correct, it prints f0, g0, f1, g1, ...

Any ideas?

larsks · Accepted Answer · 2022-08-02 00:45:45Z

1

The aiostream library provides a aiostream.stream.merge, which can be used to combine multiple async generators. If we rewrite your code like this:

import asyncio
import itertools

from aiostream import stream


async def f(i):
    await asyncio.sleep(1)
    return f"f{i}"


async def g(i):
    await asyncio.sleep(2)
    return f"g{i}"


async def amap(func, iterable):
    for i in iterable:
        res = await func(i)
        yield res


async def run():
    async with stream.merge(
        amap(f, itertools.count()), amap(g, itertools.count())
    ).stream() as streamer:
        async for x in streamer:
            print(x)


asyncio.run(run())

We get as output:

f0
f1
g0
f2
f3
g1
f4
f5
g2
f6
f7
g3
.
.
.

You'll note that the code here looks pretty much exactly like your pseudocode, except that amerge is provided by aiostream.stream.merge.

edited Aug 2, 2022 at 0:45

answered Aug 2, 2022 at 0:38

larsks

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Comments

Andrej Kesely · Accepted Answer · 2022-08-02 00:42:06Z

One way to do this is to use asyncio.wait with parameter return_when=asyncio.FIRST_COMPLETED (but each iteration you must schedule next task list to run):

import asyncio
import itertools


async def f(i):
    await asyncio.sleep(1)
    return f"f{next(i)}", f(i)


async def g(i):
    await asyncio.sleep(2)
    return f"g{next(i)}", g(i)


async def run():
    tasks = [
        asyncio.create_task(f(itertools.count())),
        asyncio.create_task(g(itertools.count())),
    ]

    while True:
        # wait for first completed task:
        done, pending = await asyncio.wait(
            tasks, return_when=asyncio.FIRST_COMPLETED
        )

        # print results:
        next_tasks = []
        for finished_task in done:
            s, nxt = finished_task.result()
            next_tasks.append(asyncio.create_task(nxt))
            print(s)

        # create next task list to run:
        tasks = [*pending, *next_tasks]


asyncio.run(run())

Prints:

f0
g0
f1
f2
g1
f3
f4
g2
f5

...

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