Question

我创建了两个函数：生成数据和处理数据。数据处理非常耗时，因此我想在并行线程中处理它们。但是我遇到了一些问题。首先，这是我的计划：

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

并记录它返回的内容：

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：

generate_data返回的值与之相同两个线程。我应该是不同的。线程应该处理不同数据，而不是两次相同的数据。我无法生成整个数据立即设置并使用getLocalPart。
变量结果不是1x10的双精度矩阵，而是1x2矩阵的复合材料。我读了（共）分布式阵列，但没有帮助我。我应该怎样做才能获得1x10的双打矩阵？
完成后，我应该对CPU1处理CPU2的数据做些什么处理自己的数据？一般来说，我不知道该怎么做。
可以删除“实验1：”和“实验2：”吗？他们搞乱了我的日志：）

考虑到上述情况，log（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.

Answer 1

为什么不使用更简单的parfor？目前你在每个worker上运行循环，我假设你想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

Matlab中的并行处理

1 个答案: