在多处理/映射函数中返回计数器对象

Question

我有一个 python 脚本正在运行，它在多个线程中启动相同的函数。这些函数创建并处理 2 个计数器（c1 和 c2）。来自分叉进程的所有 c1 计数器的结果应合并在一起。与所有 c2 计数器的结果相同，由不同的分叉返回。

我的（伪）代码如下所示：

def countIt(cfg)
   c1 = Counter
   c2 = Counter
   #do some things and fill the counters by counting words in an text, like
   #c1= Counter({'apple': 3, 'banana': 0})
   #c2= Counter({'blue': 3, 'green': 0})    

   return c1, c2

if __name__ == '__main__':
        cP1 = Counter()
        cP2 = Counter()
        cfg = "myConfig"
        p = multiprocessing.Pool(4)  #creating 4 forks
        c1, c2 = p.map(countIt,cfg)[:2]
        # 1.) This will only work with [:2] which seams to be no good idea
        # 2.) at this point c1 and c2 are lists, not a counter anymore,
        # so the following will not work:
        cP1 + c1
        cP2 + c2

按照上面的示例，我需要如下结果： cP1 = Counter({'apple': 25, 'banana': 247, 'orange': 24}) cP2 = Counter({'red': 11, 'blue': 56, 'green': 3})

所以我的问题是：如何计算分叉进程中的事物，以便聚合父进程中的每个计数器（所有 c1 和所有 c2）？

Answer 1

您需要使用例如 for-each 循环来“解压缩”您的结果。您将收到一个元组列表，其中每个元组是一对计数器：(c1, c2)。
使用您当前的解决方案，您实际上将它们混合在一起。您将[(c1a, c2a), (c1b, c2b)] 分配给c1, c2 意味着c1 包含(c1a, c2a) 并且c2 包含(c1b, c2b)。

试试这个：

if __name__ == '__main__':
        from contextlib import closing

        cP1 = Counter()
        cP2 = Counter()

        # I hope you have an actual list of configs here, otherwise map will
        # will call `countIt` with the single characters of the string 'myConfig'
        cfg = "myConfig"

        # `contextlib.closing` makes sure the pool is closed after we're done.
        # In python3, Pool is itself a contextmanager and you don't need to
        # surround it with `closing` in order to be able to use it in the `with`
        # construct.
        # This approach, however, is compatible with both python2 and python3.
        with closing(multiprocessing.Pool(4)) as p:
            # Just counting, no need to order the results.
            # This might actually be a bit faster.
            for c1, c2 in p.imap_unordered(countIt, cfg):
                cP1 += c1
                cP2 += c2