下面是一个“异步”解决方案，它不使用asyncio，而是使用asyncio使用的低级机制(在Linux上):select()。(或者asyncio可能使用poll或epoll，但这是类似的原理。)

它是对PyCurl示例的稍微修改版本。

(为了简单起见，它多次请求相同的URL，但您可以轻松地修改它以检索一系列不同的URL。)

(另一个轻微的修改可以使这个检索相同的URL作为一个无限循环。提示:将while url和句柄更改为while句柄，将while nprocessed<nurls更改为while 1。)

import pycurl,io,gzip,signal, time, random
signal.signal(signal.SIGPIPE, signal.SIG_IGN)  # NOTE! We should ignore SIGPIPE when using pycurl.NOSIGNAL - see the libcurl tutorial for more info

NCONNS = 2  # Number of concurrent GET requests
url    = 'example.com'
urls   = [url for i in range(0x7*NCONNS)]  # Copy the same URL over and over

# Check args
nurls  = len(urls)
NCONNS = min(NCONNS, nurls)
print("\x1b[32m%s \x1b[0m(compiled against 0x%x)" % (pycurl.version, pycurl.COMPILE_LIBCURL_VERSION_NUM))
print(f'\x1b[37m{nurls} \x1b[91m@ \x1b[92m{NCONNS}\x1b[0m')

# Pre-allocate a list of curl objects
m         = pycurl.CurlMulti()
m.handles = []
for i in range(NCONNS):
  c = pycurl.Curl()
  c.setopt(pycurl.FOLLOWLOCATION,  1)
  c.setopt(pycurl.MAXREDIRS,       5)
  c.setopt(pycurl.CONNECTTIMEOUT,  30)
  c.setopt(pycurl.TIMEOUT,         300)
  c.setopt(pycurl.NOSIGNAL,        1)
  m.handles.append(c)

handles    = m.handles  # MUST make a copy?!
nprocessed = 0
while nprocessed<nurls:

  while urls and handles:  # If there is an url to process and a free curl object, add to multi stack
    url   = urls.pop(0)
    c     = handles.pop()
    c.buf = io.BytesIO()
    c.url = url  # store some info
    c.t0  = time.perf_counter()
    c.setopt(pycurl.URL,        c.url)
    c.setopt(pycurl.WRITEDATA,  c.buf)
    c.setopt(pycurl.HTTPHEADER, [f'user-agent: {random.randint(0,(1<<256)-1):x}', 'accept-encoding: gzip, deflate', 'connection: keep-alive', 'keep-alive: timeout=10, max=1000'])
    m.add_handle(c)

  while 1:  # Run the internal curl state machine for the multi stack
    ret, num_handles = m.perform()
    if ret!=pycurl.E_CALL_MULTI_PERFORM:  break

  while 1:  # Check for curl objects which have terminated, and add them to the handles
    nq, ok_list, ko_list = m.info_read()
    for c in ok_list:
      m.remove_handle(c)
      t1 = time.perf_counter()
      reply = gzip.decompress(c.buf.getvalue())
      print(f'\x1b[33mGET  \x1b[32m{t1-c.t0:.3f}  \x1b[37m{len(reply):9,}  \x1b[0m{reply[:32]}...')  # \x1b[35m{psutil.Process(os.getpid()).memory_info().rss:,} \x1b[0mbytes')
      handles.append(c)
    for c, errno, errmsg in ko_list:
      m.remove_handle(c)
      print('\x1b[31mFAIL {c.url} {errno} {errmsg}')
      handles.append(c)
    nprocessed = nprocessed + len(ok_list) + len(ko_list)
    if nq==0: break

  m.select(1.0)  # Currently no more I/O is pending, could do something in the meantime (display a progress bar, etc.). We just call select() to sleep until some more data is available.

for c in m.handles:
  c.close()
m.close()

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)。

(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])

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。