我知道这已经关闭了一段时间，但我认为推广另一种基于请求库的异步解决方案可能是有用的。

list_of_requests = ['http://moop.com', 'http://doop.com', ...]

from simple_requests import Requests
for response in Requests().swarm(list_of_requests):
    print response.content

文档在这里:http://pythonhosted.org/simple-requests/

我测试了两个请求——未来请求和请求请求。Grequests速度更快，但会带来猴子补丁和依赖关系的其他问题。请求-期货比请求慢几倍。我决定编写自己的请求，并简单地将请求包装到ThreadPoolExecutor中，它几乎和grequest一样快，但没有外部依赖。

import requests
import concurrent.futures

def get_urls():
    return ["url1","url2"]

def load_url(url, timeout):
    return requests.get(url, timeout = timeout)

with concurrent.futures.ThreadPoolExecutor(max_workers=20) as executor:

    future_to_url = {executor.submit(load_url, url, 10): url for url in     get_urls()}
    for future in concurrent.futures.as_completed(future_to_url):
        url = future_to_url[future]
        try:
            data = future.result()
        except Exception as exc:
            resp_err = resp_err + 1
        else:
            resp_ok = resp_ok + 1

Async现在是一个独立的模块:grequests。

请看这里:https://github.com/kennethreitz/grequests

还有:通过Python发送多个HTTP请求的理想方法?

安装:

$ pip install grequests

用法:

建立一个堆栈:

import grequests

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

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

发送堆栈

grequests.map(rs)

结果如下所示

[<Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>]

grequest似乎没有设置并发请求的限制，即当多个请求被发送到同一个服务器时。

也许请求-期货是另一种选择。

from requests_futures.sessions import FuturesSession

session = FuturesSession()
# first request is started in background
future_one = session.get('http://httpbin.org/get')
# second requests is started immediately
future_two = session.get('http://httpbin.org/get?foo=bar')
# wait for the first request to complete, if it hasn't already
response_one = future_one.result()
print('response one status: {0}'.format(response_one.status_code))
print(response_one.content)
# wait for the second request to complete, if it hasn't already
response_two = future_two.result()
print('response two status: {0}'.format(response_two.status_code))
print(response_two.content)

办公文档中也有建议。如果你不想卷入gevent，这是一个不错的选择。

我已经使用python请求异步调用github的gist API有一段时间了。

举个例子，请看下面的代码:

https://github.com/davidthewatson/flasgist/blob/master/views.py#L60-72

这种风格的python可能不是最清晰的例子，但我可以向您保证代码是有效的。如果这让你感到困惑，请告诉我，我会记录下来。

from threading import Thread

threads=list()

for requestURI in requests:
    t = Thread(target=self.openURL, args=(requestURI,))
    t.start()
    threads.append(t)

for thread in threads:
    thread.join()

...

def openURL(self, requestURI):
    o = urllib2.urlopen(requestURI, timeout = 600)
    o...