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

我也尝试过使用python中的异步方法做一些事情，然而我使用twisted进行异步编程的运气要好得多。它的问题较少，并且有良好的文档记录。这里有一个类似于你在twisted中尝试的东西的链接。

http://pythonquirks.blogspot.com/2011/04/twisted-asynchronous-http-request.html

上面的答案都没有帮助我，因为他们假设你有一个预定义的请求列表，而在我的情况下，我需要能够侦听请求和异步响应(类似于它在nodejs中的工作方式)。

def handle_finished_request(r, **kwargs):
    print(r)


# while True:
def main():
    while True:
        address = listen_to_new_msg()  # based on your server

        # schedule async requests and run 'handle_finished_request' on response
        req = grequests.get(address, timeout=1, hooks=dict(response=handle_finished_request))
        job = grequests.send(req)  # does not block! for more info see https://stackoverflow.com/a/16016635/10577976


main()

handle_finished_request回调函数将在收到响应时被调用。注意:由于某些原因，超时(或无响应)在这里不会触发错误

这个简单的循环可以触发异步请求，类似于它在nodejs服务器中的工作方式

Note

下面的答案不适用于v0.13.0+请求。在写完这个问题之后，异步功能被移到了请求中。但是，您可以用下面的请求替换请求，它应该可以工作。

我保留这个答案，以反映最初的问题，即使用请求< v0.13.0。

异步完成多个任务。异步映射你必须:

为每个对象(任务)定义一个函数 将该函数作为事件钩子添加到请求中 调用异步。映射到所有请求/操作的列表上

例子:

from requests import async
# If using requests > v0.13.0, use
# from grequests import async

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

# A simple task to do to each response object
def do_something(response):
    print response.url

# A list to hold our things to do via async
async_list = []

for u in urls:
    # The "hooks = {..." part is where you define what you want to do
    # 
    # Note the lack of parentheses following do_something, this is
    # because the response will be used as the first argument automatically
    action_item = async.get(u, hooks = {'response' : do_something})

    # Add the task to our list of things to do via async
    async_list.append(action_item)

# Do our list of things to do via async
async.map(async_list)

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/