Replace Values that Meet a Condition - 3D Matrix

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Flo
Flo 2018 年 6 月 28 日
コメント済み: Flo 2018 年 6 月 28 日
Hi, i have two 3D-Matrices each with a dimension of 3. I work with images and i want to use the rgb-code to create conditions for my needs. The color red got the values of 1 0 0. I want to replace the values of Matrix Z with the values of Matrix X, if Matrix X got the value 1 in dimension 1, the value 0 in dimension 2 and the value 0 in dimension 3. So 1 0 0 for red.
My code for 3D-Matrices X and Z:
Z ( X(:,:,1) == 1 & X(:,:,2) == 0 & X(:,:,3) == 0) = X ( X(:,:,1) == 1 & X(:,:,2) == 0 & X(:,:,3) == 0);
The result is not satisfying. The code changes the values in Z(:,:,1) as i want them to do, but dimension 2 and 3 are left unchanged... I hope someone can solve my problem. I guess i got the wrong syntax...
Thanks in advance Flo

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Guillaume
Guillaume 2018 年 6 月 28 日
mask = X(:, :, 1) == 1 & X(:, :, 2) == 0 & X(:, :, 3) == 0; %fully saturated pure red
mask = repmat(mask, 1, 1, 3); %replicate mask across all 3 colour channels
Z(mask) = X(mask);
  4 件のコメント
Stephen23
Stephen23 2018 年 6 月 28 日
"But it would be great to know why my code isn't working."
Because you used a 2D logical index with a 3D array. Unfortunately there is no simple way to use a 2D mask as you want to, without splitting the array or duplicating the mask. Read this for a short discussion on this topic:
Flo
Flo 2018 年 6 月 28 日
Thx!

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