具有多个线程的'N'项的进程列表

Saj*_*jad 6 java concurrency multithreading threadpoolexecutor

ListN项目,我想这分List以连续的方式固定数目之间threads.

顺序我的意思是,我想传递1 to N/4给第一个thread,N/4 + 1 to N/2第二个线程和N/2+1 to N第三个thread,现在一旦所有的threads工作完成了,我想通知主要thread发送一些消息,说明所有处理都已完成.

到目前为止我所做的是我已经实施了 ExecutorService

我做了这样的事

ExecutorService threadPool = Executors.newFixedThreadPool(Number_of_threads); 
                //List of items List
                List <items>itemList = getList(); 
                  for (int i = 0 i < Number_of_threads ;i++ ) { 
                    //how to divide list here sequentially and pass it to some processor while will process those items.
                    Runnable processor = new Processor(Start, End)
                    executor.execute(process);
                   }
                  if(executor.isTerminated()){
                    logger.info("All threads completed");
                  }
Run Code Online (Sandbox Code Playgroud)
  • 如何在顺序块中划分列表?
  • 有没有更好的方法来实现这样的功能?

sta*_*off 15

如果您想要的是让所有线程尽快完成处理并且项目数量不是很大,那么只需将Runnable每个项目发布一个newFixedThreadPool(NUMBER_OF_THREADS):

    ExecutorService exec = Executors.newFixedThreadPool(NUMBER_OF_THREADS);
    List<Future<?>> futures = new ArrayList<Future<?>>(NUMBER_OF_ITEMS);
    for (Item item : getItems()) {
        futures.add(exec.submit(new Processor(item)));
    }
    for (Future<?> f : futures) {
        f.get(); // wait for a processor to complete
    }
    logger.info("all items processed");
Run Code Online (Sandbox Code Playgroud)

如果你真的想给每个线程一个连续的列表部分(但仍然希望它们尽可能快地完成,并且还希望处理每个项目花费大约相同的时间),那么将项目拆分为"均匀"你可以这样,每个线程的最大项目数不同于最小数量不超过一个(例如:14项目,4线程,然后你想要拆分,而[4,4,3,3]不是例如[3,3,3,5]).为此,您的代码将是例如

    ExecutorService exec = Executors.newFixedThreadPool(NUMBER_OF_THREADS);
    List<Item> items = getItems();
    int minItemsPerThread = NUMBER_OF_ITEMS / NUMBER_OF_THREADS;
    int maxItemsPerThread = minItemsPerThread + 1;
    int threadsWithMaxItems = NUMBER_OF_ITEMS - NUMBER_OF_THREADS * minItemsPerThread;
    int start = 0;
    List<Future<?>> futures = new ArrayList<Future<?>>(NUMBER_OF_ITEMS);
    for (int i = 0; i < NUMBER_OF_THREADS; i++) {
        int itemsCount = (i < threadsWithMaxItems ? maxItemsPerThread : minItemsPerThread);
        int end = start + itemsCount;
        Runnable r = new Processor(items.subList(start, end));
        futures.add(exec.submit(r));
        start = end;
    }
    for (Future<?> f : futures) {
        f.get();
    }
    logger.info("all items processed");
Run Code Online (Sandbox Code Playgroud)