I substracted two 3D matrix and get a 2D matrix instead of 3D matrix, why?

To facilitate the understanding of indexing behavior with number of index less than the dimension of original matrix, in other word how MATLAB reshape implicitly, try this examples

A = rand(2,3,5);
size(A)
ans = 1x3
     2     3     5
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size(A(:,:,:))
ans = 1x3
     2     3     5
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size(A(:,:))
ans = 1x2
     2    15
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size(A(:))
ans = 1x2
    30     1
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size(A(:,:,:,:))
ans = 1x3
     2     3     5
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size(A(:,:,:,:,:)) % yes all trailing dimensions are 1s but not displayed
ans = 1x3
     2     3     5
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Answer 2

Vinay 2024-9-5，9:46

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2150319-i-substracted-two-3d-matrix-and-get-a-2d-matrix-instead-of-3d-matrix-why#answer_1511334

在 MATLAB Online 中打开

Hii Wang,

The original matrix p has dimensions 3264x4912x3. When using p(1:Y, 1:X-d) and p(1:Y, 1+d:X), you are slicing the first two dimensions and ignoring the third dimension.

Therefore the result of the subtraction is a 2D matrix of size 3264x4911

ans = p(1:Y, 1:X-d)- p(1:Y, 1+d:X)

The result can be obtained as 3D matrix by using the below code.

result = zeros(Y, X-d, 3);
result = p(1:Y, 1:X-d, :) - p(1:Y, 1+d:X, :);

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Stephen23 2024-9-5，23:23

编辑：Stephen23 2024-9-5，23:26

"you are slicing the first two dimensions and ignoring the third dimension."

No, that is incorrect.

In fact all 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. Other discussions on this topic:

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

https://www.mathworks.com/matlabcentral/answers/2015866-why-does-accessing-a-multi-dimensional-array-with-fewer-indices-than-its-total-dimensions-not-result

DGM 2024-9-6，12:42

在 MATLAB Online 中打开

Similarly, this helps explain why size() behaves the way it does when using an underspecified set of scalar outputs:

A = rand(2,3,5);
sz1 = size(A(:,:,:))
sz1 = 1x3
     2     3     5
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[d1,d2,d3] = size(A)
d1 = 2
d2 = 3
d3 = 5
sz2 = size(A(:,:)) % the input to size() is reshaped first
sz2 = 1x2
     2    15
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[d1,d2] = size(A) % the input to size() is not reshaped
d1 = 2
d2 = 15