Converting a 3D Matrix to multiple 2D matrices

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Aldo Hernandez
Aldo Hernandez 2021 年 1 月 17 日
編集済み: Bruno Luong 2021 年 1 月 17 日
I have a 3D matrix, 1000 x 1000 x 40
I want to convert this matrix to 40 different matrices, 1000 x 1000 each. I suppose I am asking for the inverse of repmat, however searches into that have not led to progress.
Any help is appreciated.
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Athrey Ranjith Krishnanunni
Athrey Ranjith Krishnanunni 2021 年 1 月 17 日
編集済み: Athrey Ranjith Krishnanunni 2021 年 1 月 17 日
Something like this?
largeMatrix = randn(1000,1000,40);
smallMatrixArray = cell(40,1);
for iOuterDim = 1:40
smallMatrixArray{iOuterDim} = largeMatrix(:,:,iOuterDim);
end
Steven Lord
Steven Lord 2021 年 1 月 17 日
I want to convert this matrix to 40 different matrices
Why? What benefit do you hope to gain by doing that instead of indexing into the array to retrieve the appropriate slice when needed?
A = reshape(1:(3*4*5), [3 4 5]);
A4 = A(:, :, 4)
A4 = 3×4
37 40 43 46 38 41 44 47 39 42 45 48
What happens when or if you expand your problem sizes to an array of size 10000 in the third dimension? Your workspace would become awfully cluttered with ten thousand and one arrays in it.

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Bruno Luong
Bruno Luong 2021 年 1 月 17 日
編集済み: Bruno Luong 2021 年 1 月 17 日
Assuming A is array 1000 x 1000 x 40
C = num2cell(A,[1 2]);
  2 件のコメント
Athrey Ranjith Krishnanunni
Athrey Ranjith Krishnanunni 2021 年 1 月 17 日
You'll have to write
C = shiftdim(num2cell(A,[1 2]));
to get rid of the leading singleton dimensions, though.
Bruno Luong
Bruno Luong 2021 年 1 月 17 日
Well it is nicer but it is not requirement by OP.
C{k} still give the the kth matrix as requested.

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