Question

我几乎没有阻止功能foo，bar而且我无法更改这些功能（我无法控制某些内部库。与一个或多个网络服务进行对话）。我如何将其用作异步？例如。我不想做以下事情。

results = []
for inp in inps:
    val = foo(inp)
    result = bar(val)
    results.append(result)

这样效率很低，因为我在等待foo的第一个输入时可以调用bar进行第二次输入。如何将它们包装起来以便它们可以与asyncio一起使用（即新的async，await语法）？

让我们假设这些功能是可重入的。也就是说，只有在前一个foo正在处理时再次致电foo。

更新

使用可重复使用的装饰器扩展答案。例如，点击here。

def run_in_executor(f):
    @functools.wraps(f)
    def inner(*args, **kwargs):
        loop = asyncio.get_running_loop()
        return loop.run_in_executor(None, functools.partial(f, *args, **kwargs))

    return inner

Answer 1

这里有两个问题：第一，如何异步运行阻塞代码，第二，如何并行运行异步代码（asyncio是单线程的，所以GIL仍然适用，所以它不是真正并发，但我离题了。）

可以使用asyncio.ensure_future创建并行任务，如文档here所述。

要运行同步代码，您需要run the blocking code in an executor。例如：

import concurrent.futures
import asyncio
import time

def blocking(delay):
    time.sleep(delay)
    print('Completed.')

async def non_blocking(loop, executor):
    # Run three of the blocking tasks concurrently. asyncio.wait will
    # automatically wrap these in Tasks. If you want explicit access
    # to the tasks themselves, use asyncio.ensure_future, or add a
    # "done, pending = asyncio.wait..." assignment
    await asyncio.wait(
        fs={
            # Returns after delay=12 seconds
            loop.run_in_executor(executor, blocking, 12),

            # Returns after delay=14 seconds
            loop.run_in_executor(executor, blocking, 14),

            # Returns after delay=16 seconds
            loop.run_in_executor(executor, blocking, 16)
        },
        return_when=asyncio.ALL_COMPLETED
    )

loop = asyncio.get_event_loop()
executor = concurrent.futures.ThreadPoolExecutor(max_workers=5)
loop.run_until_complete(non_blocking(loop, executor))

如果您想使用for循环计划这些任务（如您的示例所示），您有几种不同的策略，但基本方法是使用for循环（或列表）计划任务理解等），用asyncio.wait等待它们，然后检索结果。例如：

done, pending = await asyncio.wait(
    fs=[loop.run_in_executor(executor, blocking_foo, *args) for args in inps],
    return_when=asyncio.ALL_COMPLETED
)

# Note that any errors raise during the above will be raised here; to
# handle errors you will need to call task.exception() and check if it
# is not None before calling task.result()
results = [task.result() for task in done]

Answer 2

扩展接受的答案以实际解决相关问题。

注意：需要python 3.7 +

import functools

from urllib.request import urlopen
import asyncio


def legacy_blocking_function():  # You cannot change this function
    r = urlopen("https://example.com")
    return r.read().decode()


def run_in_executor(f):
    @functools.wraps(f)
    def inner(*args, **kwargs):
        loop = asyncio.get_running_loop()
        return loop.run_in_executor(None, lambda: f(*args, **kwargs))

    return inner


@run_in_executor
def foo(arg):  # Your wrapper for async use
    resp = legacy_blocking_function()
    return f"{arg}{len(resp)}"


@run_in_executor
def bar(arg):  # Another wrapper
    resp = legacy_blocking_function()
    return f"{len(resp)}{arg}"


async def process_input(inp):  # Modern async function (coroutine)
    res = await foo(inp)
    res = f"XXX{res}XXX"
    return await bar(res)


async def main():
    inputs = ["one", "two", "three"]
    input_tasks = [asyncio.create_task(process_input(inp)) for inp in inputs]
    print([await t for t in asyncio.as_completed(input_tasks)])
    # This doesn't work as expected :(
    # print([await t for t in asyncio.as_completed([process_input(inp) for inp in input_tasks])])


if __name__ == '__main__':
asyncio.run(main())

单击here获取此示例的最新版本并发送请求请求。

Answer 3

import asyncio
from time import sleep
import logging

logging.basicConfig(
    level=logging.DEBUG, format="%(asctime)s %(thread)s %(funcName)s %(message)s")


def long_task(t):
    """Simulate long IO bound task."""
    logging.info("2. t: %s", t)
    sleep(t)
    logging.info("4. t: %s", t)
    return t ** 2


async def main():
    loop = asyncio.get_running_loop()
    inputs = range(1, 5)
    logging.info("1.")
    futures = [loop.run_in_executor(None, long_task, i) for i in inputs]
    logging.info("3.")
    results = await asyncio.gather(*futures)
    logging.info("5.")
    for (i, result) in zip(inputs, results):
        logging.info("6. Result: %s, %s", i, result)


if __name__ == "__main__":
    asyncio.run(main())

输出：

2020-03-18 17:13:07,523 23964 main 1.
2020-03-18 17:13:07,524 5008 long_task 2. t: 1
2020-03-18 17:13:07,525 21232 long_task 2. t: 2
2020-03-18 17:13:07,525 22048 long_task 2. t: 3
2020-03-18 17:13:07,526 25588 long_task 2. t: 4
2020-03-18 17:13:07,526 23964 main 3.
2020-03-18 17:13:08,526 5008 long_task 4. t: 1
2020-03-18 17:13:09,526 21232 long_task 4. t: 2
2020-03-18 17:13:10,527 22048 long_task 4. t: 3
2020-03-18 17:13:11,527 25588 long_task 4. t: 4
2020-03-18 17:13:11,527 23964 main 5.
2020-03-18 17:13:11,528 23964 main 6. Result: 1, 1
2020-03-18 17:13:11,528 23964 main 6. Result: 2, 4
2020-03-18 17:13:11,529 23964 main 6. Result: 3, 9
2020-03-18 17:13:11,529 23964 main 6. Result: 4, 16

如何在现有的阻塞库中使用asyncio？

3 个答案: