Select first result from two coroutines in asyncio

Simple solution, by using asyncio.wait and its FIRST_COMPLETED parameter:

import asyncio

async def something_to_wait():
    await asyncio.sleep(1)
    return "something_to_wait"

async def something_else_to_wait():
    await asyncio.sleep(2)
    return "something_else_to_wait"


async def wait_first():
    done, pending = await asyncio.wait(
        [something_to_wait(), something_else_to_wait()],
        return_when=asyncio.FIRST_COMPLETED)
    print("done", done)
    print("pending", pending)

asyncio.get_event_loop().run_until_complete(wait_first())

gives:

done {<Task finished coro=<something_to_wait() done, defined at stack.py:3> result='something_to_wait'>}
pending {<Task pending coro=<something_else_to_wait() running at stack.py:8> wait_for=<Future pending cb=[Task._wakeup()]>>}
Task was destroyed but it is pending!
task: <Task pending coro=<something_else_to_wait() running at stack.py:8> wait_for=<Future pending cb=[Task._wakeup()]>>

In the case of wanting to apply a timeout to a task, there is a standard library function that does exactly this: asyncio.wait_for(). Your example can be written like this:

try:
  result = await asyncio.wait_for(queue.get(), timeout=1)
except asyncio.TimeoutError:
  # This block will execute if queue.get() takes more than 1s.
  result = ...

But this only works for the specific case of a timeout. The other two answers here generalize to any arbitrary set of tasks, but neither of those answers shows how to clean up the tasks which don't finish first. This is what causes the "Task was destroyed but it is pending" messages in the output. In practice, you should do something with those pending tasks. Based on your example, I'll assume that you don't care about the other tasks' results. Here's an example of a wait_first() function that returns the value of the first completed task and cancels the remaining tasks.

import asyncio, random

async def foo(x):
    r = random.random()
    print('foo({:d}) sleeping for {:0.3f}'.format(x, r))
    await asyncio.sleep(r)
    print('foo({:d}) done'.format(x))
    return x

async def wait_first(*futures):
    ''' Return the result of the first future to finish. Cancel the remaining
    futures. '''
    done, pending = await asyncio.wait(futures,
        return_when=asyncio.FIRST_COMPLETED)
    gather = asyncio.gather(*pending)
    gather.cancel()
    try:
        await gather
    except asyncio.CancelledError:
        pass
    return done.pop().result()

async def main():
    result = await wait_first(foo(1), foo(2))
    print('the result is {}'.format(result))

if __name__ == '__main__':
    loop = asyncio.get_event_loop()
    loop.run_until_complete(main())
    loop.close()

Running this example:

# export PYTHONASYNCIODEBUG=1
# python3 test.py
foo(1) sleeping for 0.381
foo(2) sleeping for 0.279
foo(2) done
the result is 2
# python3 test.py
foo(1) sleeping for 0.048
foo(2) sleeping for 0.515
foo(1) done
the result is 1
# python3 test.py
foo(1) sleeping for 0.396
foo(2) sleeping for 0.188
foo(2) done
the result is 2

There are no error messages about pending tasks, because each pending task has been cleaned up correctly.

In practice, you probably want wait_first() to return the future, not the future's result, otherwise it will be really confusing trying to figure out which future finished. But in the example here, I returned the future's result since it looks a little cleaner.

Here's a more robust solution based upon earlier examples that deals with the following:

• Uses recursion to return the first non-null result (earlier examples will return first result regardless if null or non-null)
• Returns the first non-null result even if another task raises an exception
• In the event only non-null results are returned and an exception is raised, the last exception is raised
• Deals with multiple tasks completing at the same time - in practice this would be very rare but it can pop up in unit tests where fake async tasks complete immediately.

Note this example requires Python 3.8 due to use of the assignment operator.

async def wait_first(*tasks):
    """Return the result of first async task to complete with a non-null result"""
    # Get first completed task(s)
    done, pending = await asyncio.wait(tasks, return_when=asyncio.FIRST_COMPLETED)

    # Tasks MAY complete at same time e.g. in unit tests :)
    # Coalesce the first result if present
    for task in done:
        exception = task.exception()
        if exception is None and (result := task.result()):
            break
        else:
            result = None
        
    # Gather remaining tasks without raising exceptions
    gather = asyncio.gather(*pending, return_exceptions=True)

    # Cancel remaining tasks if result is non-null otherwise await next pending tasks
    if result:
        gather.cancel()
    elif pending:
        result = await wait_first(*pending)
    
    # Await remaining tasks to ensure they are cancelled
    try:
        await gather
    except asyncio.CancelledError:
        pass
    
    # Return result or raise last exception if no result was returned
    if exception and result is None:
        raise exception
    else:
        return result

For the examples using asyncio.FIRST_COMPLETED, they do not seem to work in Python 3.11.

A "TypeError: Passing coroutines is forbidden, use tasks explicitly." is thrown

However, the asyncio.as_completed(futures) does work properly.