Getting values from functions that run as asyncio tasks

Question

I was trying the following code:

import asyncio

@asyncio.coroutine
def func_normal():
        print("A")
        yield from asyncio.sleep(5)
        print("B")
        return 'saad'

@asyncio.coroutine
def func_infinite():
    i = 0
    while i<10:
        print("--"+str(i))
        i = i+1
    return('saad2')

loop = asyncio.get_event_loop()

tasks = [
    asyncio.async(func_normal()),
    asyncio.async(func_infinite())]

loop.run_until_complete(asyncio.wait(tasks))
loop.close()

I can't figure out how to get values in variables from these functions. I can't do this:

asyncio.async(a = func_infinite())

as this would make this a keyword argument. How do I go about accomplishing this?

jfs · Accepted Answer · 2015-09-08 18:09:31Z

87

The coroutines work as is. Just use the returned value from loop.run_until_complete() and call asyncio.gather() to collect multiple results:

#!/usr/bin/env python3
import asyncio

@asyncio.coroutine
def func_normal():
    print('A')
    yield from asyncio.sleep(5)
    print('B')
    return 'saad'

@asyncio.coroutine
def func_infinite():
    for i in range(10):
        print("--%d" % i)
    return 'saad2'

loop = asyncio.get_event_loop()
tasks = func_normal(), func_infinite()
a, b = loop.run_until_complete(asyncio.gather(*tasks))
print("func_normal()={a}, func_infinite()={b}".format(**vars()))
loop.close()

Output

--0
--1
--2
--3
--4
--5
--6
--7
--8
--9
A
B
func_normal()=saad, func_infinite()=saad2

answered Sep 8, 2015 at 18:09

jfs

417k210 gold badges1k silver badges1.7k bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Saad Aleem Over a year ago

This works. I did run functions for my django app concurrently using this approach but I didn't see a gain in performance. It takes the same time to process.

jfs Over a year ago

@AleemSaadullah: asyncio allows you to run 10 func_normal() tasks in 5 seconds instead of 50 seconds. All tasks are executed in the same thread unless you use an executor to offload some work to a thread/process pool. I'm not familiar with running django using asyncio (if it is even possible or whether it makes any sense without substancial restructuring in django).

Saad Aleem Over a year ago

Yes, the code works for functions such as the ones defined here. But I guess, the mongodb ORM(mongoengine) is supposed to support asynchronous queries. I think that is the problem here.

Saad Aleem Over a year ago

The functions in my actual project primarily build queries and execute them.

dano · Accepted Answer · 2015-09-08 15:31:05Z

loop.run_until_complete returns the value returned by the function you pass into it. So, it will return the output of asyncio.wait:

import asyncio

@asyncio.coroutine
def func_normal():
    print("A")
    yield from asyncio.sleep(5)
    print("B")
    return 'saad'

@asyncio.coroutine
def func_infinite():
    i = 0
    while i<10:
        print("--"+str(i))
        i = i+1
    return('saad2')

loop = asyncio.get_event_loop()

tasks = [
    asyncio.async(func_normal()),
    asyncio.async(func_infinite())]

done, _ = loop.run_until_complete(asyncio.wait(tasks))
for fut in done:
    print("return value is {}".format(fut.result()))
loop.close()

Output:

A
--0
--1
--2
--3
--4
--5
--6
--7
--8
--9
B
return value is saad2
return value is saad

You can also access the results directly from the tasks array:

print(tasks[0].result())
print(tasks[1].result())

Petr Javorik · Accepted Answer · 2017-05-15 21:02:59Z

If you want to use any value returned by coroutine as soon as coroutine ends you can pass future object into the coro and update this future by computed value. As soon as future is updated it passes its future.result() to the callback function which is bound with given future. See code below:

import asyncio


async def func_normal(future):
    print("A")
    await asyncio.sleep(5)
    print("B")
    # return 'saad'
    future.set_result('saad')


async def func_infinite(future):
    i = 0
    while i<10:
        print("--"+str(i))
        i = i+1
    # return('saad2')
    future.set_result('saad2')

def got_result(future):
    print(future.result())

loop = asyncio.get_event_loop()
future1 = asyncio.Future()
future2 = asyncio.Future()

future1.add_done_callback(got_result)
future2.add_done_callback(got_result)

# Coros are automatically wrapped in Tasks by asyncio.wait() 
coros = [
    func_normal(future1),
    func_infinite(future2)]

loop.run_until_complete(asyncio.wait(coros))
loop.close()

The callback function is called with a single argument - the future object which it is bound with. If you need to pass more arguments into the callback use partial from functools package:

future1.add_done_callback(functools.partial(print, "future:", argin))

will call

print("future:", argin)

Sergio Pulgarin · Accepted Answer · 2022-09-11 16:57:19Z

3

I found a solution that I like better:

    loop = asyncio.new_event_loop()
    task = loop.create_task(awaitable(*args, **kwargs))

    loop.run_until_complete(task)

    return task.result()

answered Sep 11, 2022 at 16:57

Sergio Pulgarin

9299 silver badges21 bronze badges

Collectives™ on Stack Overflow

Getting values from functions that run as asyncio tasks

4 Answers 4

Output

4 Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Output

4 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related