如何等待Python asyncial call_稍后完成所有挂起的操作

你可以用 asyncio.all_tasks() 但这样做可能是错误的。

call_later 将它们标记为已完成:

import asyncio
from functools import partial
START_RATE = 1


def say_hi(who):
    print(f'hello {who}')


def run_on_finish(callback):
    def wrapper(func, *args, **kwargs):
        try:
            return func(*args, **kwargs)
        finally:
            callback()
    return wrapper


def release_waiter(waiter, *args):
    """Taken from standard library"""
    if not waiter.done():
        waiter.set_result(None)


def run():
    # Let's go!
    incr = START_RATE
    loop = asyncio.get_event_loop()
    tasks = []
    for x in range(5):
        wait_delta = 5 - x

        # Create a waiter
        waiter = loop.create_future()
        release_cb = partial(release_waiter, waiter)

        # Schedule the function, making sure we release the waiter on finish
        handle = loop.call_later(incr * wait_delta, run_on_finish(release_cb),
                                 say_hi, x)

        # If waiter is cancelled, cancel the handle too.
        waiter.add_done_callback(lambda *args: handle.cancel)

        tasks.append(waiter)

    loop.run_until_complete(asyncio.gather(*tasks))


run()

待会儿打电话给你 用于正常函数而不是协程。如果 say_hi ensure_future 或 loop.create_task 加入混音。

添加它们确实会带来更多的复杂性-您需要添加更多的函数 确保未来 的结果,并以类似的方式与你的服务员连锁 futures._chain_future .

我强烈建议在这种情况下使用你自己的协同程序,就像你已经做的那样。