如何用最快的方式发送 10 万个 http 请求
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2021-08-30 09:08
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。
关于 Python 并发编程,推荐阅读文章:
如果你有更好的实现异步 I/O 的操作,不妨留言分享哦。
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参考资料
grequests: https://github.com/spyoungtech/grequests
[2]Gevent: http://www.gevent.org/