sub2ind多维-元素访问

问题描述

我有一个三维矩阵，例如u.我还有另一个三维尺寸的矩阵，例如B.假设u和B的尺寸为CxTxW.对于u的每个元素，我希望执行以下操作:如果u为> 0，则报告回u的值以及其所在的相应T尺寸以及B/u的相应元素.如果u为0，则使用正u和T以及相应的B/u元素报告下一个T(对于相同的C，W).在下面的代码中，我将B/u称为benefit.

I have a matrix of three dimensions, say u. I have another matrix of three dimensions, say B. Say the dimensions of u and B are CxTxW. For every element of u, I wish to do the following:If u is >0, then report back both the value of u as well as the corresponding T dimension it is in, and the corresponding element of B/u. If the u is 0, then report back the next T (for the same C,W) with a positive u and the T and the corresponding B/u element. In the code below I call B/u as benefit.

以下链接的问题以非常有效的方式回答了C = 1(即二维)的问题.我可以在c周围编写一个for循环，并为C的每个元素重复链接的答案，但我希望在没有for循环的情况下执行.

The following linked question answers this in a very efficient manner for C=1 (i.e, two dimensions). I can write a for loop around the c and repeat the linked answer for every element of C but I wish to do it without the for loop.

带有if语句MATLAB的可向量化FIND函数

以下是带有循环的解决方案:

Here is the solution with a loop:

for c=1:C
    dummy_u = reshape(u(c,:,:),[T+1 W]) ;
    dummy_B = reshape(B(c,:,:),[T+1 W]) ;

    dummy_i = nan(size(dummy_u)) ;  % Start with all nan values
    [dum_row, ~] = find(dummy_u);  % Get row indices of non-zero values
    dummy_i(dummy_u ~= 0) = dum_row;     % Place row indices in locations of non-zero values
    dummy_i = cummin(dummy_i, 1, 'reverse');  % Column-wise cumulative minimum, starting from bottomi
    dum_i = dummy_i+(T+1).*repmat(0:(W-1), T+1, 1);  % Create linear index
    dummy_benefit = dummy_B(dum_i)./dummy_u(dum_i);
    i(c,:,:)       = dummy_i ;
    benefit(c,:,:) = dummy_benefit ;
    clear dum_i dum_row dummy_i dummy_benefit
end

没有for循环怎么办?

How can I do it without a for loop?

推荐答案

您可以将整个行置换为T x CW矩阵，而不是将一行重新整形为T x W矩阵:

Instead of reshaping one row into a T x W matrix, you can permute the whole thing into a T x CW matrix:

>> C = 4; T = 5; W = 3;
>> u = reshape(1:C*T*W, C, T, W)   % generate sample u
u =

ans(:,:,1) =

    1    5    9   13   17
    2    6   10   14   18
    3    7   11   15   19
    4    8   12   16   20

ans(:,:,2) =

   21   25   29   33   37
   22   26   30   34   38
   23   27   31   35   39
   24   28   32   36   40

ans(:,:,3) =

   41   45   49   53   57
   42   46   50   54   58
   43   47   51   55   59
   44   48   52   56   60

>> du = reshape(permute(u, [2 1 3]), T, [])
du =

    1    2    3    4   21   22   23   24   41   42   43   44
    5    6    7    8   25   26   27   28   45   46   47   48
    9   10   11   12   29   30   31   32   49   50   51   52
   13   14   15   16   33   34   35   36   53   54   55   56
   17   18   19   20   37   38   39   40   57   58   59   60

du上的其余代码应保持不变(添加虚拟行除外).

The rest of your code should work unchanged (except for adding the dummy row) on du.

倒数是:

>> permute(reshape(du, T, C, W), [2 1 3])
ans =

ans(:,:,1) =

    1    5    9   13   17
    2    6   10   14   18
    3    7   11   15   19
    4    8   12   16   20

ans(:,:,2) =

   21   25   29   33   37
   22   26   30   34   38
   23   27   31   35   39
   24   28   32   36   40

ans(:,:,3) =

   41   45   49   53   57
   42   46   50   54   58
   43   47   51   55   59
   44   48   52   56   60

我不知道这种解决方案是否真的更快.

I have no idea if this solution is actually faster.

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