【问题标题】:Parallel processing in MatlabMatlab中的并行处理
【发布时间】:2013-01-05 19:01:34
【问题描述】:

我创建了两个函数:生成数据和处理数据。数据处理很耗时,所以我想在并行线程中处理它们。但我对他们有一些问题。首先,这是我的程序:

result = zeros(1, 10);

matlabpool open local 2
spmd
    for a = 1:5
        data = generate_data();
        display(sprintf('Received data on CPU%d: %d', labindex, data));
        result(end + 1) = process_data(data);
    end
    display(sprintf('All done on CPU%d', labindex));
end
matlabpool close

以及它返回的日志:

Starting matlabpool using the 'local' profile ... connected to 2 workers.
Lab 1: 
  Received data on CPU1: 100
Lab 2: 
  Received data on CPU2: 100
Lab 1: 
  Received data on CPU1: 101
  Received data on CPU1: 102
  Received data on CPU1: 103
  Received data on CPU1: 104
  All done on CPU1
Lab 2: 
  Received data on CPU2: 101
  Received data on CPU2: 102
  Received data on CPU2: 103
  Received data on CPU2: 104
  All done on CPU2
Sending a stop signal to all the workers ... stopped.

有问题,我有:

  1. generate_data 返回的值对于 两个线程。我应该与众不同。线程应该处理不同的 数据,而不是两次相同的数据。我无法生成全部数据 立即设置并使用 getLocalPart。

  2. 变量结果不是 1x10 的双精度矩阵,而是 1x2 的双精度矩阵 复合材料。我读到了(共同)分布式数组,但没有帮助 我。我应该怎么做才能收到 1x10 的双打矩阵?

  3. 我应该对 CPU1 做些什么来处理 CPU2 的数据,当完成时 处理自己的数据?一般来说,我不知道该怎么做。

  4. 可以删除“实验室 1:”和“实验室 2:”吗?他们在捣乱 我的日志:)

考虑到上述情况,日志(对于较大的数据集)应该是这样的:

Starting matlabpool using the 'local' profile ... connected to 2 workers.
Received data on CPU1: 100
Received data on CPU2: 101
Received data on CPU1: 102
Received data on CPU1: 103
Received data on CPU1: 104
Received data on CPU1: 105
Received data on CPU2: 106
Received data on CPU1: 107
Received data on CPU1: 108
Received data on CPU2: 109
All done on CPU1
All done on CPU2
Sending a stop signal to all the workers ... stopped.

【问题讨论】:

    标签: matlab parallel-processing distributed


    【解决方案1】:

    为什么不使用更简单的parfor?目前,您正在每个工作人员上运行循环,我假设您希望并行运行循环的迭代。

    nIter = 10;
    result = zeros(1, nIter);
    
    matlabpool open local 2
    
        parfor a = 1:nIter
            data = generate_data();
            fprintf('%s: processing set %i/%i\n',datestr(now),a,nIter)
            result(a) = process_data(data);
        end
    end
    matlabpool close
    

    【讨论】:

    • spmd .. end 结构意味着在多个线程上并行运行一个块。在他的代码中,for循环不是运行两次吗? Parfor 只是打破了 for 循环,对吗?
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