我们可以用请求库替换此代码中的urlopen吗？

Question

在这个例子中，我们可以用python 2.7中的请求库替换urlopen库吗？

import concurrent.futures 
import urllib.request 

URLS = ['http://www.foxnews.com/', 
     'http://www.cnn.com/', 
     'http://europe.wsj.com/', 
     'http://www.bbc.co.uk/', 
     'http://some-made-up-domain.com/'] 

# Retrieve a single page and report the URL and contents 
def load_url(url, timeout): 
    with urllib.request.urlopen(url, timeout=timeout) as conn: 
     return conn.read() 

# We can use a with statement to ensure threads are cleaned up promptly 
with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor: 
    # Start the load operations and mark each future with its URL 
    future_to_url = {executor.submit(load_url, url, 60): url for url in URLS} 
    for future in concurrent.futures.as_completed(future_to_url): 
     url = future_to_url[future] 
     try: 
      data = future.result() 
     except Exception as exc: 
      print('%r generated an exception: %s' % (url, exc)) 
     else: 
      print('%r page is %d bytes' % (url, len(data))) 

谢谢！

Answer 1

是的，你可以。

你的代码似乎做超时一个简单的HTTP GET，所以与请求等价物是：

import requests 

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