如何用最快的方式发送 10 万个 http 请求?

321次阅读  |  发布于3年以前

”假如有一个文件,里面有 10 万个 url,需要对每个 url 发送 http 请求,并打印请求结果的状态码,如何编写代码尽可能快的完成这些任务呢?

Python 并发编程有很多方法,多线程的标准库 threading,concurrency,协程 asyncio,当然还有 grequests 这种异步库,每一个都可以实现上述需求,下面一一用代码实现一下,本文的代码可以直接运行,给你以后的并发编程作为参考:

队列+多线程

定义一个大小为 400 的队列,然后开启 200 个线程,每个线程都是不断的从队列中获取 url 并访问。

主线程读取文件中的 url 放入队列中,然后等待队列中所有的元素都被接收和处理完毕。代码如下:

from threading import Thread
import sys
from queue import Queue
import requests

concurrent = 200


def doWork():
    while True:
        url = q.get()
        status, url = getStatus(url)
        doSomethingWithResult(status, url)
        q.task_done()


def getStatus(ourl):
    try:
        res = requests.get(ourl)
        return res.status_code, ourl
    except:
        return "error", ourl


def doSomethingWithResult(status, url):
    print(status, url)


q = Queue(concurrent * 2)
for i in range(concurrent):
    t = Thread(target=doWork)
    t.daemon = True
    t.start()

try:
    for url in open("urllist.txt"):
        q.put(url.strip())
    q.join()
except KeyboardInterrupt:
    sys.exit(1)

运行结果如下: 有没有 get 到新技能?

线程池

如果你使用线程池,推荐使用更高级的 concurrent.futures 库:

import concurrent.futures
import requests

out = []
CONNECTIONS = 100
TIMEOUT = 5

urls = []
with open("urllist.txt") as reader:
    for url in reader:
        urls.append(url.strip())

def load_url(url, timeout):
    ans = requests.get(url, timeout=timeout)
    return ans.status_code

with concurrent.futures.ThreadPoolExecutor(max_workers=CONNECTIONS) as executor:
    future_to_url = (executor.submit(load_url, url, TIMEOUT) for url in urls)
    for future in concurrent.futures.as_completed(future_to_url):
        try:
            data = future.result()
        except Exception as exc:
            data = str(type(exc))
        finally:
            out.append(data)
            print(data)

协程 + aiohttp

协程也是并发非常常用的工具了:

import asyncio
from aiohttp import ClientSession, ClientConnectorError

async def fetch_html(url: str, session: ClientSession, **kwargs) -> tuple:
    try:
        resp = await session.request(method="GET", url=url, **kwargs)
    except ClientConnectorError:
        return (url, 404)
    return (url, resp.status)

async def make_requests(urls: set, **kwargs) -> None:
    async with ClientSession() as session:
        tasks = []
        for url in urls:
            tasks.append(
                fetch_html(url=url, session=session, **kwargs)
            )
        results = await asyncio.gather(*tasks)

    for result in results:
        print(f'{result[1]} - {str(result[0])}')

if __name__ == "__main__":
    import sys
    assert sys.version_info >= (3, 7), "Script requires Python 3.7+."
    with open("urllist.txt") as infile:
        urls = set(map(str.strip, infile))
    asyncio.run(make_requests(urls=urls))

grequests[1]

这是个第三方库,目前有 3.8K 个星,就是 Requests + Gevent[2],让异步 http 请求变得更加简单。Gevent 的本质还是协程。

使用前:

pip install grequests

使用起来那是相当的简单:

import grequests

urls = []
with open("urllist.txt") as reader:
    for url in reader:
        urls.append(url.strip())

rs = (grequests.get(u) for u in urls)

for result in grequests.map(rs):
    print(result.status_code, result.url)

注意 grequests.map(rs) 是并发执行的。运行结果如下: 也可以加入异常处理:

>>> def exception_handler(request, exception):
...    print("Request failed")

>>> reqs = [
...    grequests.get('http://httpbin.org/delay/1', timeout=0.001),
...    grequests.get('http://fakedomain/'),
...    grequests.get('http://httpbin.org/status/500')]
>>> grequests.map(reqs, exception_handler=exception_handler)
Request failed
Request failed
[None, None, <Response [500]>]

最后的话

今天分享了并发 http 请求的几种实现方式,有人说异步(协程)性能比多线程好,其实要分场景看的,没有一种方法适用所有的场景,笔者就曾做过一个实验,也是请求 url,当并发数量超过 500 时,协程明显变慢。

参考资料

[1] grequests: https://github.com/spyoungtech/grequests

[2] Gevent: http://www.gevent.org/

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