python - 使用 aiohttp/asyncio 发出 100 万次请求 - 字面意思

Question

我跟进了本教程:https://pawelmhm.github.io/asyncio/python/aiohttp/2016/04/22/asyncio-aiohttp.html当我处理 50 000 个请求时，一切正常。但是我需要进行 100 万次 API 调用，然后我遇到了这段代码的问题:

    url = "http://some_url.com/?id={}"
    tasks = set()

    sem = asyncio.Semaphore(MAX_SIM_CONNS)
    for i in range(1, LAST_ID + 1):
        task = asyncio.ensure_future(bound_fetch(sem, url.format(i)))
        tasks.add(task)

    responses = asyncio.gather(*tasks)
    return await responses

因为 Python 需要创建 100 万个任务，它基本上只是滞后，然后在终端打印 Killed 消息。有什么方法可以使用预制的 url 集(或列表)的生成器？谢谢。

Answer 1

一次安排所有 100 万个任务

这是您正在谈论的代码。它最多占用 3 GB RAM，因此如果您的可用内存不足，它很可能会被操作系统终止。

import asyncio
from aiohttp import ClientSession

MAX_SIM_CONNS = 50
LAST_ID = 10**6

async def fetch(url, session):
    async with session.get(url) as response:
        return await response.read()

async def bound_fetch(sem, url, session):
    async with sem:
        await fetch(url, session)

async def fetch_all():
    url = "http://localhost:8080/?id={}"
    tasks = set()
    async with ClientSession() as session:
        sem = asyncio.Semaphore(MAX_SIM_CONNS)
        for i in range(1, LAST_ID + 1):
            task = asyncio.create_task(bound_fetch(sem, url.format(i), session))
            tasks.add(task)
        return await asyncio.gather(*tasks)

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

使用队列简化工作

这是我的建议如何使用 asyncio.Queue将 URL 传递给工作任务。队列按需填充，没有预制的 URL 列表。

它只需要 30 MB RAM :)

import asyncio
from aiohttp import ClientSession

MAX_SIM_CONNS = 50
LAST_ID = 10**6

async def fetch(url, session):
    async with session.get(url) as response:
        return await response.read()

async def fetch_worker(url_queue):
    async with ClientSession() as session:
        while True:
            url = await url_queue.get()
            try:
                if url is None:
                    # all work is done
                    return
                response = await fetch(url, session)
                # ...do something with the response
            finally:
                url_queue.task_done()
                # calling task_done() is necessary for the url_queue.join() to work correctly

async def fetch_all():
    url = "http://localhost:8080/?id={}"
    url_queue = asyncio.Queue(maxsize=100)
    worker_tasks = []
    for i in range(MAX_SIM_CONNS):
        wt = asyncio.create_task(fetch_worker(url_queue))
        worker_tasks.append(wt)
    for i in range(1, LAST_ID + 1):
        await url_queue.put(url.format(i))
    for i in range(MAX_SIM_CONNS):
        # tell the workers that the work is done
        await url_queue.put(None)
    await url_queue.join()
    await asyncio.gather(*worker_tasks)

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