如何使异步池可取消？

Question

我有一个pool_map函数可以用来限制同时执行函数的数量

想法是 coroutine function 接受映射到可能参数列表的单个参数，但也将所有函数调用包装到信号量获取中，因此只有有限的数量是运行一次：

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，我修改了上面的代码，但没有测试它，但我的变体运行良好。例如。你可以这样使用它：

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，就会出现问题。我看不到如何取消所有计划的或仍然-运行任务，那些仍在等待获取信号量的任务。

是否有我不知道的库或优雅的构建块，或者我必须自己构建所有部分？（即 Semaphore 可以访问其服务员，as_finished 可以访问其运行任务队列，...）

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 no-op 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得到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))

如何使异步池可取消？

How to make a asyncio pool cancelable?

python

pool

python-asyncio