Python，调用进程池而不阻塞事件循环

Question

如果我运行以下代码：

import asyncio
import time
import concurrent.futures

def cpu_bound(mul):
    for i in range(mul*10**8):
        i+=1
    print('result = ', i)
    return i

async def say_after(delay, what):
    print('sleeping async...')
    await asyncio.sleep(delay)
    print(what)

# The run_in_pool function must not block the event loop
async def run_in_pool():
    with concurrent.futures.ProcessPoolExecutor() as executor:
        result = executor.map(cpu_bound, [1, 1, 1])

async def main():
    task1 = asyncio.create_task(say_after(0.1, 'hello'))
    task2 = asyncio.create_task(run_in_pool())
    task3 = asyncio.create_task(say_after(0.1, 'world'))

    print(f"started at {time.strftime('%X')}")
    await task1
    await task2
    await task3
    print(f"finished at {time.strftime('%X')}")

if __name__ == '__main__':
    asyncio.run(main())

输出是：

started at 18:19:28
sleeping async...
result =  100000000
result =  100000000
result =  100000000
sleeping async...
hello
world
finished at 18:19:34

这表明事件循环会阻塞，直到 cpu 绑定作业 ( task2) 完成，然后继续执行task3.

如果我只运行一项 cpu 密集型作业（即run_in_pool以下一项）：

async def run_in_pool():
    loop = asyncio.get_running_loop()
    with concurrent.futures.ProcessPoolExecutor() as executor:
        result = await loop.run_in_executor(executor, cpu_bound, 1)

那么事件循环似乎不会阻塞，因为输出是：

started at 18:16:23
sleeping async...
sleeping async...
hello
world
result =  100000000
finished at 18:16:28

如何task2在进程池中运行许多 cpu 绑定作业（in ）而不阻塞事件循环？

Answer 1

正如您所发现的，您需要使用 asyncio 自己的run_in_executor来等待提交的任务完成而不阻塞事件循环。Asyncio 没有提供 的等效项map，但模拟它并不难：

async def run_in_pool():
    with concurrent.futures.ProcessPoolExecutor() as executor:
        futures = [loop.run_in_executor(executor, cpu_bound, i)
                   for i in (1, 1, 1)]
        result = await asyncio.gather(*futures)