当 te coro 在睡眠超时之前完成时，如何在 coro 中清理 asyncio.sleep()

Question

当此计数器被另一个计时器事件（stop_future）中断并完成时，我无法清理下面的 count_timer（睡眠）。

import asyncio
import datetime
from concurrent.futures import FIRST_COMPLETED

DISPLAY_DT_TIMEOUT = 7
MAX_DISPLAY_COUNT = 3
COUNT_TIMEOUT = 4


def stop(stop_future, display_count):

    print('stop', display_count, datetime.datetime.now())
    stop_future.set_result('Done!')

async def display_dt1():

    display_count = 0
    stop_future = asyncio.Future()
    stop_handle = loop.call_later(DISPLAY_DT_TIMEOUT, stop, stop_future, display_count)

    # loop until
    while not stop_future.done() and display_count < MAX_DISPLAY_COUNT:
        print('dt1-1', display_count, datetime.datetime.now())
        count_timer = asyncio.sleep(COUNT_TIMEOUT)            # cleanup ??
        await asyncio.wait([count_timer, stop_future], return_when=FIRST_COMPLETED)
        print('dt-2', display_count, datetime.datetime.now())
        display_count += 1

    if not stop_future.done():
        # cleanup stop future
        stop_handle.cancel()
        stop_future.cancel()

async def wait_before_loop_close():

    await asyncio.sleep(10)


loop = asyncio.get_event_loop()
coro = display_dt1()
loop.run_until_complete(coro)

# this print shows the count_timer is still pending
print('tasks-1', asyncio.Task.all_tasks())

# wait for the count_timer to finish     
loop.run_until_complete(wait_before_loop_close())

loop.close()
print('tasks-2', asyncio.Task.all_tasks())

print('finished', datetime.datetime.now())

结果：

dt1-1 0 2015-11-10 15:20:58.200107
dt-2 0 2015-11-10 15:21:02.201654
dt1-1 1 2015-11-10 15:21:02.201654
stop 0 2015-11-10 15:21:05.186800      # stop interrupt
dt-2 1 2015-11-10 15:21:05.186800      # results in stop counting
tasks-1 {<Task pending coro=<sleep() running at D:\Python\Python35-32\lib\asyncio\tasks.py:495> wait_for=<Future pending cb=[Task._wakeup()]>>}
>>> tasks-2 set()
finished 2015-11-10 15:21:15.193669

Update-1：取消 count_timer（包装在未来）

async def display_dt1():

    count_timer_task = None
    display_count = 0
    stop_future = asyncio.Future()
    stop_handle = loop.call_later(DISPLAY_DT_TIMEOUT, stop, stop_future, display_count)

    while not stop_future.done() and display_count < MAX_DISPLAY_COUNT:
        print('dt1-1', display_count, datetime.datetime.now())
        count_timer = asyncio.sleep(COUNT_TIMEOUT)            # cleanup ??
        count_timer_task = asyncio.ensure_future(count_timer)
        await asyncio.wait([count_timer_task, stop_future], return_when=FIRST_COMPLETED)
        print('dt-2', display_count, datetime.datetime.now())
        display_count += 1

    if count_timer_task and not count_timer_task.cancelled():
        count_timer_task.cancel()
        print('check-1', datetime.datetime.now(), count_timer_task)

    if not stop_future.done():
        stop_handle.cancel()
        stop_future.cancel()
        print('check-2', datetime.datetime.now(), stop_future)

结果：

dt1-1 0 2015-11-10 16:44:29.180372
dt-2 0 2015-11-10 16:44:33.180908
dt1-1 1 2015-11-10 16:44:33.180908
stop 0 2015-11-10 16:44:36.181062
dt-2 1 2015-11-10 16:44:36.181062
check-1 2015-11-10 16:44:36.181062 <Task pending coro=<sleep() running at D:\Python\Python35-32\lib\asyncio\tasks.py:495> wait_for=<Future cancelled>>
tasks-1 set()
>>> tasks-2 set()
finished 2015-11-10 16:44:46.181965

Answer 1

asyncio.wait_for 实现了您正在尝试做的大部分事情。将其与asyncio.shield 结合使用，以防止您的 Future 在超时到期时被取消：

await asyncio.wait_for(asyncio.shield(stop_future), COUNT_TIMEOUT)

编辑：如果不清楚，这将进入while 循环，替换对asyncio.wait 的调用。

更新 by voscausa：显示代码

async def display_dt1():

    display_count = 0
    stop_future = asyncio.Future()
    stop_handle = loop.call_later(DISPLAY_DT_TIMEOUT, stop, stop_future, display_count)

    while not stop_future.done() and display_count < MAX_DISPLAY_COUNT:
        print('dt-1', display_count, datetime.datetime.now())
        try:
            await asyncio.wait_for(asyncio.shield(stop_future), COUNT_TIMEOUT)
        except asyncio.TimeoutError:
            print('timeout', datetime.datetime.now())
        print('dt-2', display_count, datetime.datetime.now())
        display_count += 1

    if not stop_future.done():
        stop_handle.cancel()
        stop_future.cancel()
        print('check-2', datetime.datetime.now(), stop_future)

与：

DISPLAY_DT_TIMEOUT = 10
MAX_DISPLAY_COUNT = 3
COUNT_TIMEOUT = 4

结果：

dt-1 0 2015-11-10 21:43:04.549782
timeout 2015-11-10 21:43:08.551319   # count timeout
dt-2 0 2015-11-10 21:43:08.551319
dt-1 1 2015-11-10 21:43:08.551319
timeout 2015-11-10 21:43:12.552880   # count timeout
dt-2 1 2015-11-10 21:43:12.552880
dt-1 2 2015-11-10 21:43:12.552880
stop 0 2015-11-10 21:43:14.554649    # stop timeout
dt-2 2 2015-11-10 21:43:14.555650
tasks-1 set()
tasks-2 set()
finished 2015-11-10 21:43:24.558510

Answer 2

您已将wait() 配置为在计时器结束或用户取消循环中较早者返回。听起来您想取消count_timer，因为它不是导致wait 调用返回的未来。你可以通过询问两个future中的哪一个来了解这一点。

count_timer = asyncio.sleep(COUNT_TIMEOUT)
await asyncio.wait([count_timer, stop_future], return_when=FIRST_COMPLETED)
if not count_timer.done():
    count_timer.cancel()