How to make asyncio task / coroutine return values to be assigned

Question

TL;DR: I want to create asyncio task/coroutine and get the return values to be assigned in a var.

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I found this question Getting values from functions that run as asyncio tasks Which seems to talk about a similar issue, but the sintax has changed a lot in the asyncio module that I not even sure if it's related. [I'm on Python 3.7.2]

Sample code for me to explain what I'm trying to do:

async def s(f):
    func = {
        1: op1,
        2: op2,
        3: op3
      }.get(f,False)
    var = func(f)
    return var

def op1(f):
    print('Op1',f+f)
    return f+f

def op2(f):
    print('Op2', f*f)
    return f*f

def op3(f):
    print('Op3', f**f)
    return f**f

async def main():
    task1 = asyncio.create_task(s(1))
    task2 = asyncio.create_task(s(3))
    print(f"started at {time.strftime('%X')}")
    await task1
    await task2
    print(task1,task2)
    print(f"finished at {time.strftime('%X')}")

The main() is a example coroutine from docs https://docs.python.org/3/library/asyncio-task.html#coroutines

The s() is supposed to be a switch which will select the proper function (one of op1,op2,op3) for the scenario, run the function and RETURN the result, (from what I expected) to be assigned to task1/task2.

What is actually assigned to the task1/task2 is:

<Task finished coro=<s() done, defined at C:\...\test.py:17> result=2> 
<Task finished coro=<s() done, defined at C:\...\test.py:17> result=27>

As you can see, the return is stored in 'result' (attribute? var?) but not directly assigned to the var.

I need either: Assign the return directly to the var OR a way to access the 'result' and assign it to a var where I can further manipulate.

BTW: print(task1.result) returns this: <built-in method result of _asyncio.Task object at 0x0000028592DDA048>

Sounds like you need to write: print(task1.result()).

Brian Rodriguez
– Brian Rodriguez

2019-04-21 01:58:21 +00:00
Commented Apr 21, 2019 at 1:58 — Brian Rodriguez
– Brian Rodriguez, Commented Apr 21, 2019 at 1:58

Tom Lubenow · Accepted Answer · 2019-04-21 02:04:29Z

13

async def main():
    task1 = asyncio.create_task(s(1))
    task2 = asyncio.create_task(s(3))
    print(f"started at {time.strftime('%X')}")
    result_of_task1 = await task1
    result_of_task2 = await task2
    print(result_of_task1,result_of_task2)
    print(f"finished at {time.strftime('%X')}")

is one way to do it. I prefer this, as it makes awaiting tasks more like normal function calls, but also equivalent is

await task1
# some time later
result_of_task1 = task1.result()

You just missed the (), because result is a method, not a member variable.

edited Apr 21, 2019 at 2:04

answered Apr 21, 2019 at 1:58

Tom Lubenow

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3 Comments

Justcurious Over a year ago

I was under the impression you couldn't assign await expression. Anyway, both ways solved my issue. Thanks!

Anurag A S Over a year ago

How can I change the above such that function s can return multiple values?

Tom Lubenow Over a year ago

@AnuragAS you can simply return a tuple and receive it as a tuple, just like any other python function

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