使用concurrent.futures一次消耗许多出列消息

Question

我正在使用来自RabbitMQ频道的消息,我希望我一次可以消耗n个元素.我想我可以使用ProcessPoolExecutor(或ThreadPoolExecutor).我只是想知道是否可以知道池中是否有免费执行程序.

这就是我想写的:

executor = futures.ProcessPoolExecutor(max_workers=5)
running = []
def consume(message):
    print "actually consuming a single message"

def on_message(channel, method_frame, header_frame, message):
    # this method is called once per incoming message
    future = executor.submit(consume, message)
    block_until_a_free_worker(executor, future)

def block_until_a_free_worker(executor, future):
    running.append(future) # this grows forever!
    futures.wait(running, timeout=5, return_when=futures.FIRST_COMPLETED)

[...]
channel.basic_consume(on_message, 'my_queue')
channel.start_consuming()

我需要编写函数block_until_a_free_worker.此方法应该能够检查是否所有正在运行的工作程序都在使用中.

在替代方案中,我可以使用任何阻塞executor.submit选项(如果可用).

我尝试了一种不同的方法,并在完成后改变期货清单.我试图从列表中明确添加和删除期货,然后像这样等待:

futures.wait(running, timeout=5, return_when=futures.FIRST_COMPLETED)

这似乎不是解决方案.

我可以设置future.add_done_callback,并可能计算正在运行的实例...

任何提示或想法？谢谢.

Answer 1

我在这里给出了类似的答案。

信号量的目的是限制一组工作人员对资源的访问。

from threading import Semaphore
from concurrent.futures import ProcessPoolExecutor 

class TaskManager:
    def __init__(self, workers):
        self.pool = ProcessPoolExecutor(max_workers=workers)
        self.workers = Semaphore(workers)

    def new_task(self, function):
        """Start a new task, blocks if all workers are busy."""
        self.workers.acquire()  # flag a worker as busy

        future = self.pool.submit(function, ... )

        future.add_task_done(self.task_done)

    def task_done(self, future):
        """Called once task is done, releases one worker."""
        self.workers.release()