Question

我有一个pool_map函数，可以用来限制同时执行函数的数量。

我们的想法是让coroutine function接受一个映射到可能参数列表的参数，但也要将所有函数调用包装到信号量采集中，这样一次只运行一个有限的数字：

from typing import Callable, Awaitable, Iterable, Iterator
from asyncio import Semaphore

A = TypeVar('A')
V = TypeVar('V')

async def pool_map(
    func: Callable[[A], Awaitable[V]],
    arg_it: Iterable[A],
    size: int=10
) -> Generator[Awaitable[V], None, None]:
    """
    Maps an async function to iterables
    ensuring that only some are executed at once.
    """
    semaphore = Semaphore(size)

    async def sub(arg):
        async with semaphore:
            return await func(arg)

    return map(sub, arg_it)

为了示例，我修改了并且没有测试上面的代码，但我的变体运行良好。例如。你可以像这样使用它：

from asyncio import get_event_loop, coroutine, as_completed
from contextlib import closing

URLS = [...]

async def run_all(awaitables):
    for a in as_completed(awaitables):
        result = await a
        print('got result', result)

async def download(url): ...


if __name__ != '__main__':
    pool = pool_map(download, URLS)

    with closing(get_event_loop()) as loop:
        loop.run_until_complete(run_all(pool))

但是，如果在等待未来时抛出异常，则会出现问题。我无法看到如何取消所有已安排或仍在运行的任务，也无法等待仍在等待获取信号量的任务。

我不知道是否有图书馆或优雅的构建块，或者我是否必须自己构建所有部件？（即Semaphore可以访问其服务员，as_finished可以访问其正在运行的任务队列，...）

Answer 1

这是一个天真的解决方案，基于如果任务已经完成cancel是无操作的事实：

async def run_all(awaitables):
    futures = [asyncio.ensure_future(a) for a in awaitables]
    try:
        for fut in as_completed(futures):
            result = await fut
            print('got result', result)
    except:
        for future in futures:
            future.cancel()
        await asyncio.wait(futures)

Answer 2

使用ensure_future获取Task而不是协程：

import asyncio
from contextlib import closing


def pool_map(func, args, size=10):
    """
    Maps an async function to iterables
    ensuring that only some are executed at once.
    """
    semaphore = asyncio.Semaphore(size)

    async def sub(arg):
        async with semaphore:
            return await func(arg)

    tasks = [asyncio.ensure_future(sub(x)) for x in args]

    return tasks


async def f(n):
    print(">>> start", n)

    if n == 7:
        raise Exception("boom!")

    await asyncio.sleep(n / 10)

    print("<<< end", n)
    return n


async def run_all(tasks):
    exc = None
    for a in asyncio.as_completed(tasks):
        try:
            result = await a
            print('=== result', result)
        except asyncio.CancelledError as e:
            print("!!! cancel", e)
        except Exception as e:
            print("Exception in task, cancelling!")
            for t in tasks:
                t.cancel()
            exc = e
    if exc:
        raise exc


pool = pool_map(f, range(1, 20), 3)

with closing(asyncio.get_event_loop()) as loop:
    loop.run_until_complete(run_all(pool))

如何使asyncio池可取消？

2 个答案: