How in asyncio run independence co-routine in other co-routine from run loop.Python 3.5

Question

Necessary , with the main loop run coroutine that something will be done. And from it ,start another coroutine, which will not block. Use Python 3.5

import asyncio,time  
async def cor1():
    for i in range(10):
        await asyncio.sleep(0)
        print("cor1",i)

async def cor2():
    for i in range(10):
        await asyncio.sleep(0)
        time.sleep(1)
        print("cor2",i)

async def main():
    asyncio.ensure_future(cor1())
    asyncio.ensure_future(cor2())
    print("cor3")

loop = asyncio.get_event_loop()
asyncio.ensure_future(main())
loop.run_forever()

Now the main loop creates two coroutines.But they do not run parallel,one after another. Until someone one complete, the other will not start its work.Of course you can run them on different threads and to establish communication using queues. But is it possible to do this with the help of asyncio in python 3.5

Answer 1

this is a way how you can run them in parallel:

import asyncio

async def cor1():
    for i in range(10):
        await asyncio.sleep(1)
        print("cor1", i)

async def cor2():
    for i in range(10):
        await asyncio.sleep(1)
        print("cor2", i)

loop = asyncio.get_event_loop()
cors = asyncio.wait([cor1(), cor2()])
loop.run_until_complete(cors)

note that time.sleep(1) (unlike asyncio.sleep(1)) is a blocking call and will not run concurrently.

Luciano Ramalho's book Fluent Python has an excellent chapter about coroutines and asyncio... just in case.

Answer 2

This decision helped me

import asyncio,time
from concurrent.futures import ProcessPoolExecutor
def cor1():
    for i in range(10):
        print("cor1", i)
        time.sleep(2)

def cor2():
    for i in range(10):
        print("cor2", i)
        time.sleep(1)

executor = ProcessPoolExecutor(2)
loop = asyncio.get_event_loop()

asyncio.ensure_future(loop.run_in_executor(executor, cor1))
asyncio.ensure_future(loop.run_in_executor(executor, cor2))

loop.run_forever()