I am opening a file which has 100,000 URL's. I need to send an HTTP request to each URL and print the status code. I am using Python 2.6, and so far looked at the many confusing ways Python implements threading/concurrency. I have even looked at the python concurrence library, but cannot figure out how to write this program correctly. Has anyone come across a similar problem? I guess generally I need to know how to perform thousands of tasks in Python as fast as possible - I suppose that means 'concurrently'.


当前回答

使用grequests,它是requests + Gevent模块的组合。

GRequests允许您使用带有Gevent的Requests来轻松地生成异步HTTP请求。

用法很简单:

import grequests

urls = [
   'http://www.heroku.com',
   'http://tablib.org',
   'http://httpbin.org',
   'http://python-requests.org',
   'http://kennethreitz.com'
]

创建一组未发送的请求:

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

同时发送:

>>> grequests.map(rs)
[<Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>]

其他回答

(下一个项目的自我提示)

Python 3解决方案只使用请求。它是最简单且快速的,不需要多处理或复杂的异步库。

最重要的方面是重用连接,特别是对于HTTPS (TLS需要额外的往返才能打开)。注意,连接是特定于子域的。如果在多个域上抓取多个页面,则可以对url列表进行排序,以最大化连接重用(它有效地按域进行排序)。

当给定足够的线程时,它将与任何异步代码一样快。(请求在等待响应时释放python GIL)。

[带有日志记录和错误处理的生产等级代码]

import logging
import requests
import time
from concurrent.futures import ThreadPoolExecutor, as_completed

# source: https://stackoverflow.com/a/68583332/5994461

THREAD_POOL = 16

# This is how to create a reusable connection pool with python requests.
session = requests.Session()
session.mount(
    'https://',
    requests.adapters.HTTPAdapter(pool_maxsize=THREAD_POOL,
                                  max_retries=3,
                                  pool_block=True)
)

def get(url):
    response = session.get(url)
    logging.info("request was completed in %s seconds [%s]", response.elapsed.total_seconds(), response.url)
    if response.status_code != 200:
        logging.error("request failed, error code %s [%s]", response.status_code, response.url)
    if 500 <= response.status_code < 600:
        # server is overloaded? give it a break
        time.sleep(5)
    return response

def download(urls):
    with ThreadPoolExecutor(max_workers=THREAD_POOL) as executor:
        # wrap in a list() to wait for all requests to complete
        for response in list(executor.map(get, urls)):
            if response.status_code == 200:
                print(response.content)

def main():
    logging.basicConfig(
        format='%(asctime)s.%(msecs)03d %(levelname)-8s %(message)s',
        level=logging.INFO,
        datefmt='%Y-%m-%d %H:%M:%S'
    )

    urls = [
        "https://httpstat.us/200",
        "https://httpstat.us/200",
        "https://httpstat.us/200",
        "https://httpstat.us/404",
        "https://httpstat.us/503"
    ]

    download(urls)

if __name__ == "__main__":
    main()

考虑使用风车,虽然风车可能不能做那么多线程。

您可以在5台机器上使用手卷Python脚本,每台机器使用端口40000-60000连接出站,打开100,000个端口连接。

另外,使用一个线程良好的QA应用程序(如OpenSTA)做一个示例测试可能会有所帮助,以了解每个服务器可以处理多少。

另外,试着在LWP::ConnCache类中使用简单的Perl。这样您可能会获得更好的性能(更多的连接)。

使用线程池是一个很好的选择,这将使这相当容易。不幸的是,python并没有一个标准库来简化线程池。但这里有一个不错的图书馆,你应该开始: http://www.chrisarndt.de/projects/threadpool/

来自他们网站的代码示例:

pool = ThreadPool(poolsize)
requests = makeRequests(some_callable, list_of_args, callback)
[pool.putRequest(req) for req in requests]
pool.wait()

希望这能有所帮助。

对于您的情况,线程可能会做的技巧,因为您可能会花费大部分时间等待响应。标准库中有一些有用的模块,如Queue,可能会有所帮助。

我以前做过类似的并行下载文件的事情,对我来说已经足够好了,但它不是你所说的那种规模。

如果您的任务对cpu的限制更大,您可能需要考虑multiprocessing模块,它将允许您利用更多的cpu /内核/线程(更多的进程不会相互阻塞,因为锁定是每个进程)

自从2010年这篇文章发布以来,事情发生了很大的变化,我还没有尝试过所有其他的答案,但我尝试了一些,我发现使用python3.6对我来说这是最好的。

在AWS上运行时,我每秒可以获取大约150个独特的域名。

import concurrent.futures
import requests
import time

out = []
CONNECTIONS = 100
TIMEOUT = 5

tlds = open('../data/sample_1k.txt').read().splitlines()
urls = ['http://{}'.format(x) for x in tlds[1:]]

def load_url(url, timeout):
    ans = requests.head(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)
    time1 = time.time()
    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(str(len(out)),end="\r")

    time2 = time.time()

print(f'Took {time2-time1:.2f} s')