Making async for loops in Python

Question

The following code outputs as follows:

1 sec delay, print \"1\", 
1 sec delay, print \"2\", 
1 sec delay, print \"1\", 
1 sec delay, print \"2\"

Answer 1

Looking at the desired output, it seems that the goal is to leave the individual iteration as it is - i.e. run first and second sequentially - but execute both loop iterations in parallel.

Assuming you only want to modify main(), it could be achieved like this:

async def main():
    async def one_iteration():
        result = await first()
        print(result)
        result2 = await second()
        print(result2)
    coros = [one_iteration() for _ in range(2)]
    await asyncio.gather(*coros)

Instead of iterating in sequence, the above creates a coroutine for each iteration task, and uses asyncio.gather to execute all the iterations in parallel.

Note that simply creating a coroutine doesn't start executing it, so a large number of coros won't block the event loop.

Answer 2

To run the two functions simultaneously you can use gather. However, the results will be provided to you in the order you provide them. So for example if you do

results = await asyncio.gather(first(), second())

Then you will get [the result of first(), the result of second()] back. If you want to do something whenever each one returns then you should use Tasks explicitly and add callbacks.

Answer 3

With the aysncio library you can use aysncio.gather()

loop.run_until_complete(asyncio.gather(
  first(),
  second()
))

This can come in handy if you are also sending HTTP requests in parallel:

loop.run_until_complete(asyncio.gather(
  request1(),
  request2()
))