来自Executor的concurrent.futures.Future的asyncio yield

Question

我有一个long_task运行大量cpu绑定计算的函数,我希望通过使用新的asyncio框架使其异步.生成的long_task_async函数使用a ProcessPoolExecutor将工作卸载到不同的进程,不受GIL约束.

麻烦的是,由于某种原因,concurrent.futures.Future实例ProcessPoolExecutor.submit从投掷时返回a TypeError.这是设计的吗？那些期货与asyncio.Future班级不兼容吗？什么是解决方法？

我也注意到发电机不是可拣选的,所以提交一个couroutine ProcessPoolExecutor就会失败.这个也有任何清洁的解决方案吗？

import asyncio
from concurrent.futures import ProcessPoolExecutor

@asyncio.coroutine
def long_task():
    yield from asyncio.sleep(4)
    return "completed"

@asyncio.coroutine
def long_task_async():
    with ProcessPoolExecutor(1) as ex:
        return (yield from ex.submit(long_task)) #TypeError: 'Future' object is not iterable
                                                 # long_task is a generator, can't be pickled

loop = asyncio.get_event_loop()

@asyncio.coroutine
def main():
    n = yield from long_task_async()
    print( n )

loop.run_until_complete(main())

Answer 1

您想要使用loop.run_in_executor,它使用concurrent.futures执行程序,但将返回值映射到asyncio未来.

最初的asyncioPEP 表明, concurrent.futures.Future有朝一日可能会增加一种__iter__方法,因此它也可以使用yield from,但是现在这个库的设计只需要yield from支持而已.(否则一些代码在3.3中实际上不起作用.)

Answer 2

我们可以通过调用包装concurrent.futures.Future成一个.我用下面的代码试了一下.工作良好asyncio.futureasyncio.wrap_future(Future)

from asyncio import coroutine
import asyncio
from concurrent import futures


def do_something():
    ls = []
    for i in range(1, 1000000):
        if i % 133333 == 0:
            ls.append(i)
    return ls


@coroutine
def method():
    with futures.ProcessPoolExecutor(max_workers=10) as executor:
        job = executor.submit(do_something)
        return (yield from asyncio.wrap_future(job))

@coroutine
def call_method():
    result = yield from method()
    print(result)


def main():
    loop = asyncio.get_event_loop()
    try:
        loop.run_until_complete(call_method())
    finally:
        loop.close()


if __name__ == '__main__':
    main()