Fill up submatrix along third dimension with two dimensional matrix

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Michael Stollenwerk
Michael Stollenwerk 2022 年 8 月 29 日
編集済み: Bruno Luong 2022 年 8 月 29 日
Given a p by p by N array, e.g. p=3 and N=4,
x = NaN(3,3,4);
I can easily fill up the 1 by 1 by N submatrix given a "1 by 1 matrix"doing
x(1,1,:) = 1;
But what if I want to fill up the 2 by 2 by N submatrix like this
x(1:2,1:2,:) = [1 2; 3 4];
This does not work.
How could I do it elegantly?

採用された回答

Bruno Luong
Bruno Luong 2022 年 8 月 29 日
編集済み: Bruno Luong 2022 年 8 月 29 日
x(1:2,1:2,:) = repmat( [1 2; 3 4], 1, 1, size(x,3));
I would love to have MATLAB indexing to behave with auto expansion, but probably many users would be unhappy and complain because of coding error becoming silence.

その他の回答 (1 件)

Torsten
Torsten 2022 年 8 月 29 日
x = NaN(3,3,4);
x(1:2,1:2,:) = repmat([1 2; 3 4],1,1,4);
x
x =
x(:,:,1) = 1 2 NaN 3 4 NaN NaN NaN NaN x(:,:,2) = 1 2 NaN 3 4 NaN NaN NaN NaN x(:,:,3) = 1 2 NaN 3 4 NaN NaN NaN NaN x(:,:,4) = 1 2 NaN 3 4 NaN NaN NaN NaN

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