2

我想做类似以下的事情：

import asyncio

async def g():
    print('called g')
    return 'somevalue'

async def f():
    x = g()

loop = asyncio.get_event_loop()
loop.run_until_complete(f())
loop.close()

没有输出的地方。请注意，我没有await. g()这将产生一个g was not awaited异常，但我正在寻找g最肯定没有运行的行为。

这对我来说很有用，因为我有一个长时间运行的操作和复杂的设置，但我只在某些情况下需要它的结果，所以为什么在不需要它时还要运行它。一种“按需”的情况。

我怎样才能做到这一点？

4

1 回答 1

0

一种选择是使用简单的标志来指示任务：

import asyncio

import random


async def g(info):
    print('> called g')

    if not info['skip']:
        print('* running g', info['id'])
        await asyncio.sleep(random.uniform(1, 3))
    else:
        print('- skiping g', info['id'])

    print('< done g', info['id'])

    return info['id']


async def main():
    data = [{
                'id': i,
                'skip': False
            } for i in range(10)]

    # schedule 10 tasks to run later
    tasks = [asyncio.ensure_future(g(info)) for info in data]

    # tell some tasks to skip processing
    data[2]['skip'] = True
    data[5]['skip'] = True
    data[6]['skip'] = True

    # wait for all results
    results = await asyncio.gather(*tasks)

    print(results)
    print("Done!")


loop = asyncio.get_event_loop()
loop.run_until_complete(main())
loop.close()

一个不同的选择是使用task.cancel：

import asyncio


async def coro(x):
    print('coro', x)
    return x


async def main():
    task1 = asyncio.ensure_future(coro(1))
    task2 = asyncio.ensure_future(coro(2))
    task3 = asyncio.ensure_future(coro(3))

    task2.cancel()

    for task in asyncio.as_completed([task1, task2, task3]):
        try:
            result = await task
            print("success", result)
        except asyncio.CancelledError as e:
            print("cancelled", e)


loop = asyncio.get_event_loop()
loop.run_until_complete(main())
loop.close()
于 2017-01-15T09:25:40.627 回答