Matlab中的并行处理

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，所以我想在并行线程中处理它们。但是我和他们有一些问题。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

以及它返回的内容的日志：

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.

有一个问题，我有：

它通过生成_数据返回的值与 两条线。我应该与众不同。线程应该处理不同的 数据，而不是相同的数据两次。我无法生成整个数据 立即设置并使用getLocalPart

可变结果不是双精度的1x10矩阵，而是双精度的1x2矩阵 复合材料。我读过关于（联合）分布式阵列的文章，但没有帮助 我我应该如何接收1x10双工矩阵

我应该对CPU1做什么处理CPU2的数据，当完成 处理自己的数据？一般来说，我不知道该怎么做

可以删除“实验室1:”和“实验室2:”？他们在捣乱 我的日志：）

考虑到上述因素，日志（对于较大的数据集）应如下所示：

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.

---

为什么不使用更简单的

parfor

？目前，您正在每个worker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

---

spmd。。端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循环不是运行了两次吗？Parfor只是打破了for循环，对吗？