# Row wise indexing of 3d array

7 views (last 30 days)
Oscar Frick on 14 Jun 2018
Commented: Oscar Frick on 15 Jun 2018
I have a 3d array, for example:
(:,:,1)
-0.1468 -0.4846 -0.3310
0.3212 -0.4570 0.1491
(:,:,2)
-0.3110 -0.3165 0.1256
0.1868 -0.1315 0.2802
(:,:,3)
-0.4189 0.2757 -0.0641
0.4294 -0.0132 -0.0532
What I want to do is find all values of the last column (:,3,:) that is less than 0, and then set the corresponding rows to NaN. In the above case, the results should be:
(:,:,1)
NaN NaN NaN
0.3212 -0.4570 0.1491
(:,:,2)
-0.3110 0.3165 0.1256
0.1868 -0.1315 0.2802
(:,:,3)
NaN NaN NaN
NaN NaN NaN
I can achieve it in the 2d case using the following code:
array = rand(5,3)-0.5
array(find(array(:,3)<0),:) = NaN
But I struggle with the 3d case.

Guillaume on 14 Jun 2018
Note that in the 2D case you didn't need find (which only slows things done):
array(array(:, 3) < 0, :) = NaN;
For the ND case it's a bit more complicated since the result of the comparison is a matrix not a single dimension vector. One possibility:
array(repmat(array(:, 3, :) < 0, 1, size(array, 2), 1)) = NaN
##### 2 CommentsShow 1 older commentHide 1 older comment
Oscar Frick on 15 Jun 2018
This works perfectly as well as being effective. Some short testing I did seems to indicate that the version without the trailing argument has an extremely slight time advantage when having "small" arrays (on the format (n,3,4), small being size seems to be about n<10000000). For larger arrays the version with the trailing one starts getting ahead, getting a more significant efficiency advantage as the arrays grow bigger.

Mark Saad on 14 Jun 2018
You could try:
[I,J,K] = ind2sub(size(array), find(array(:,:,:) < 0));
ind = find(J == 3);
I = I(ind);
J = J(ind);
K = K(ind);
array(I,:,K) = NaN;
The first line gets the indices of every value that is less than 0. Lines 2-4 filter out any indices not in the third column. The last line sets the desired values to NaN.

### Categories

Find more on Multidimensional Arrays in Help Center and File Exchange

R2018a

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by