使用aiohttp / asyncio发出1百万个请求-从字面上看 [英] Making 1 milion requests with aiohttp/asyncio - literally

问题描述

我遵循了本教程： https： //pawelmhm.github.io/asyncio/python/aiohttp/2016/04/22/asyncio-aiohttp.html ，当我处理5万个请求时，一切正常。但是我需要进行1百万次API调用，然后此代码出现问题：

I followed up this tutorial: https://pawelmhm.github.io/asyncio/python/aiohttp/2016/04/22/asyncio-aiohttp.html and everything works fine when I am doing like 50 000 requests. But I need to do 1 milion API calls and then I have problem with this code:

    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

由于Python需要创建1百万个任务，因此它基本上只是滞后在终端中显示 Killed 消息。有什么方法可以使用由预设的url组（或列表）插入的生成器吗？谢谢。

Because Python needs to create 1 milion tasks, it basically just lags and then prints Killed message in terminal. Is there any way to use a generator insted of pre-made set (or list) of urls? Thanks.

推荐答案

一次安排所有一百万个任务

这是您正在谈论的代码。它最多需要3 GB的RAM，因此如果您的可用内存较少，很可能会被操作系统终止。

Schedule all 1 million tasks at once

This is the code you are talking about. It takes up to 3 GB RAM so it is easily possible that it will be terminated by the operating system if you have low free memory.

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())

使用队列来简化工作

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

Use queue to streamline the work

This is my suggestion how to use asyncio.Queue to pass URLs to worker tasks. The queue is filled as-needed, there is no pre-made list of URLs.

仅需30 MB RAM：）

It takes only 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())

这篇关于使用aiohttp / asyncio发出1百万个请求-从字面上看的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！