Python 并发 executor.map() 和 submit()

Question

我正在学习如何使用并发 executor.map()和 executor.submit() .
我有一个包含 20 个 url 的列表，想同时发送 20 个请求，问题是 .submit()从头开始以与给定列表不同的顺序返回结果。我读过 map()做我需要的，但我不知道如何用它编写代码。
下面的代码对我来说很完美。
问题:有没有map()的代码块？相当于下面的代码，或者任何可以从 submit() 中对结果列表进行排序的排序方法按给定列表的顺序？

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)))

Answer 1

这是您现有代码的 map 版本。请注意，回调现在接受一个元组作为参数。我在回调中添加了一个 try\except ，因此结果不会引发错误。结果根据输入列表排序。

from concurrent.futures import ThreadPoolExecutor
import urllib.request

URLS = ['http://www.foxnews.com/',
        'http://www.cnn.com/',
        'http://www.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(tt):  # (url,timeout)
    url, timeout = tt
    try:
      with urllib.request.urlopen(url, timeout=timeout) as conn:
         return (url, conn.read())
    except Exception as ex:
        print("Error:", url, ex)
        return(url,"")  # error, return empty string

with ThreadPoolExecutor(max_workers=5) as executor:
    results = executor.map(load_url, [(u,60) for u in URLS])  # pass url and timeout as tuple to callback
    executor.shutdown(wait=True) # wait for all complete
    print("Results:")
for r in results:  # ordered results, will throw exception here if not handled in callback
    print('   %r page is %d bytes' % (r[0], len(r[1])))

输出

Error: http://www.wsj.com/ HTTP Error 404: Not Found
Results:
   'http://www.foxnews.com/' page is 320028 bytes
   'http://www.cnn.com/' page is 1144916 bytes
   'http://www.wsj.com/' page is 0 bytes
   'http://www.bbc.co.uk/' page is 279418 bytes
   'http://some-made-up-domain.com/' page is 64668 bytes