Linear indexing over a subset of dimensions

Question

James Bowen 2022-11-9

2
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1846698-linear-indexing-over-a-subset-of-dimensions

编辑： Stephen23 2022-11-10

Linear indexing over the last two dimensions of a three dimensional array seems to work:

A = zeros([3 5 5]);
aIdx = [1 7 13 19 25];
aVal = [1 2 3 ; 4 5 6 ; 7 8 9 ; 10 11 12 ; 13 14 15].';
A(:,aIdx) = aVal;

This puts the triplets of aVal into the diagonal locations of the [5 5] part of A. That is

>> A(:,2,2)
ans =
4
5
6

I want to do the same thing but with first two dimension of an array. That is something like:

B = zeros([5 5 3]);
bIdx = [1 7 13 19 25];
bVal = [1 2 3 ; 4 5 6 ; 7 8 9 ; 10 11 12 ; 13 14 15];
B(bIdx,:) = bVal;

But this does not work. A couple of ways that do work are:

B([ bIdx bIdx + 25 bIdx + 50]) = bVal;

and

bLogIdx = false([5 5]);
bLogIdx(bIdx) = true;
B(repmat(bLogIdx, [1 1 3])) = bVal;

Both of these give

>> squeeze(B(2,2,:))
ans =
4
5
6

Is there a more clever way to accomplish this? More generally, is there a way to linear index over a subset of dimensions. That is use linear indexing in the y1,y2 dimensions of A(x1,x2,y1,y2,x3,x4)?

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Stephen23 2022-11-9

编辑：Stephen23 2022-11-10

"Is there a more clever way to accomplish this?"

RESHAPE or PERMUTE

"More generally, is there a way to linear index over a subset of dimensions."

By definition trailing dimensions collapse into the last subscript index. As Loren Shure wrote: "Indexing with fewer indices than dimensions If the final dimension i<N, the right-hand dimensions collapse into the final dimension."

Source: https://blogs.mathworks.com/loren/2006/08/09/essence-of-indexing/

If you think about it, linear indexing is really just a side-effect of this. An earlier discussion on this:

https://www.mathworks.com/matlabcentral/answers/1459369-determining-order-of-flattening-an-array-using-colons

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Answer 1

Matt J 2022-11-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1846698-linear-indexing-over-a-subset-of-dimensions#answer_1094938

编辑：Matt J 2022-11-9

在 MATLAB Online 中打开

If you are indexing into the initial dimensions of an array, reshape will be cheaper than permute:

B = zeros([25 3]);
bIdx = [1 7 13 19 25];
bVal = [1 2 3 ; 4 5 6 ; 7 8 9 ; 10 11 12 ; 13 14 15];
B(bIdx,:) = bVal;
B=reshape(B,5,5,[])
B = 
B(:,:,1) =

     1     0     0     0     0
     0     4     0     0     0
     0     0     7     0     0
     0     0     0    10     0
     0     0     0     0    13


B(:,:,2) =

     2     0     0     0     0
     0     5     0     0     0
     0     0     8     0     0
     0     0     0    11     0
     0     0     0     0    14


B(:,:,3) =

     3     0     0     0     0
     0     6     0     0     0
     0     0     9     0     0
     0     0     0    12     0
     0     0     0     0    15