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。

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])

我知道这是一个老问题，但在Python 3.7中，您可以使用asyncio和aiohttp来做到这一点。

import asyncio
import aiohttp
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 pathlib
    import sys

    assert sys.version_info >= (3, 7), "Script requires Python 3.7+."
    here = pathlib.Path(__file__).parent

    with open(here.joinpath("urls.txt")) as infile:
        urls = set(map(str.strip, infile))

    asyncio.run(make_requests(urls=urls))

你可以阅读更多关于它的内容，并在这里看到一个例子。

解决这个问题的一个好方法是首先编写获得一个结果所需的代码，然后合并线程代码来并行化应用程序。

In a perfect world this would simply mean simultaneously starting 100,000 threads which output their results into a dictionary or list for later processing, but in practice you are limited in how many parallel HTTP requests you can issue in this fashion. Locally, you have limits in how many sockets you can open concurrently, how many threads of execution your Python interpreter will allow. Remotely, you may be limited in the number of simultaneous connections if all the requests are against one server, or many. These limitations will probably necessitate that you write the script in such a way as to only poll a small fraction of the URLs at any one time (100, as another poster mentioned, is probably a decent thread pool size, although you may find that you can successfully deploy many more).

您可以遵循以下设计模式来解决上述问题:

Start a thread which launches new request threads until the number of currently running threads (you can track them via threading.active_count() or by pushing the thread objects into a data structure) is >= your maximum number of simultaneous requests (say 100), then sleeps for a short timeout. This thread should terminate when there is are no more URLs to process. Thus, the thread will keep waking up, launching new threads, and sleeping until your are finished. Have the request threads store their results in some data structure for later retrieval and output. If the structure you are storing the results in is a list or dict in CPython, you can safely append or insert unique items from your threads without locks, but if you write to a file or require in more complex cross-thread data interaction you should use a mutual exclusion lock to protect this state from corruption.

我建议您使用threading模块。您可以使用它来启动和跟踪正在运行的线程。Python的线程支持是完全的，但是对问题的描述表明它完全满足了您的需求。

最后，如果您希望看到用Python编写的并行网络应用程序的相当简单的应用程序，请查看ssh.py。它是一个小型库，使用Python线程并行处理许多SSH连接。该设计非常接近您的需求，您可能会发现它是一个很好的资源。

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)

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。