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。

其他回答

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)。

最简单的方法是使用Python的内置线程库。它们不是“真正的”/内核线程。它们有问题(比如序列化),但足够好了。你需要一个队列和线程池。这里有一个选项,但是编写自己的选项很简单。您无法并行处理所有100,000个调用,但可以同时发出100个(或左右)调用。

(工具)

Apache Bench是您所需要的全部。—用于测量HTTP web服务器性能的命令行计算机程序

给你一篇不错的博客文章:https://www.petefreitag.com/item/689.cfm(来自Pete Freitag)

我发现使用tornado包是最快和最简单的方法来实现这一点:

from tornado import ioloop, httpclient, gen


def main(urls):
    """
    Asynchronously download the HTML contents of a list of URLs.
    :param urls: A list of URLs to download.
    :return: List of response objects, one for each URL.
    """

    @gen.coroutine
    def fetch_and_handle():
        httpclient.AsyncHTTPClient.configure(None, defaults=dict(user_agent='MyUserAgent'))
        http_client = httpclient.AsyncHTTPClient()
        waiter = gen.WaitIterator(*[http_client.fetch(url, raise_error=False, method='HEAD')
                                    for url in urls])
        results = []
        # Wait for the jobs to complete
        while not waiter.done():
            try:
                response = yield waiter.next()
            except httpclient.HTTPError as e:
                print(f'Non-200 HTTP response returned: {e}')
                continue
            except Exception as e:
                print(f'An unexpected error occurred querying: {e}')
                continue
            else:
                print(f'URL \'{response.request.url}\' has status code <{response.code}>')
                results.append(response)
        return results

    loop = ioloop.IOLoop.current()
    web_pages = loop.run_sync(fetch_and_handle)

    return web_pages

my_urls = ['url1.com', 'url2.com', 'url100000.com']
responses = main(my_urls)
print(responses[0])

Twistedless解决方案:

from urlparse import urlparse
from threading import Thread
import httplib, sys
from Queue import Queue

concurrent = 200

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

def getStatus(ourl):
    try:
        url = urlparse(ourl)
        conn = httplib.HTTPConnection(url.netloc)   
        conn.request("HEAD", url.path)
        res = conn.getresponse()
        return res.status, 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)

这个方案比twisted方案稍微快一点,并且使用更少的CPU。