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。

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。

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')

(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()

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。