How to break time.sleep() in a python concurrent.futures

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渐次进展 2021-02-15 23:29

I am playing around with concurrent.futures.

Currently my future calls time.sleep(secs).

It seems that Future.cancel() does less than I thought.

3条回答
  •  滥情空心
    2021-02-16 00:22

    As it is written in its link, You can use a with statement to ensure threads are cleaned up promptly, like the below example:

    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)))
    

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