How to kill an asyncio coroutine (not First_completed case)

Question

There is an example: main coroutine creates coroutines that take a long time to complete, what means FIRST_COMPLETED case is unreachable. Problem: considering that await asyncio.wait(tasks) line blocks everything under itself, how to get access to pending futures set?

import asyncio

async def worker(i):
    #some big work
    await asyncio.sleep(100000)

async def main():
    tasks = [asyncio.create_task(worker(i), name=str(i)) for i in range(5)]
    done, pending = await asyncio.wait(tasks) #or asyncio.as_completed(tasks, return_when=FIRST_COMPLETED) no matter
    # everything below is unreachable until tasks are in process
    # we want to kill certain task
    for future in pending:
        if future.get_name == "4":
            future.close()

loop = asyncio.get_event_loop()
loop.run_until_complete(main())

How to avoid await block and kill certain coroutine? for example 4?

I don't really understand your question, if you don't want to wait for even one task to complete why using `asyncio.wait` at all? — Łukasz Kwieciński, May 24 '21 at 14:00
it doesnt work to me coz without await all child koroutines will break when parent koroutine is done — kostya, May 25 '21 at 11:52

user4815162342 · Accepted Answer · 2021-05-25T07:15:19.253

You can create another task to monitor your tasks:

async def monitor(tasks):
    # monitor the state of tasks and cancel some
    while not all(t.done() for t in tasks):
        for t in tasks:
            if not t.done() and t.get_name() == "4":
                t.cancel()
            # ...
        # give the tasks some time to make progress
        await asyncio.sleep(1)

async def main():
    tasks = [asyncio.create_task(worker(i), name=str(i)) for i in range(5)]
    tasks.append(asyncio.create_task(monitor(tasks[:])))
    done, pending = await asyncio.wait(tasks)
    # ...

How to kill an asyncio coroutine (not First_completed case)

1 Answers1