Indexing 3-D matrix with 2-D matrix of indexes

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I have a 3-D matrix that represents data for XYZ. I would like to pull the values from data given a 2-D matrix reprenting the Z indices of the values I want to pull. I thought there would be an easy way to index this without using a loop since my data is rather large, but I am stumped.
Below I have an example of the output I want using a loop...
% Sample 3-D Matrix of Data
data = ones(3,3,3);
data(:,:,1) = 0.1*data(:,:,1);
data(:,:,2) = 0.2*data(:,:,2);
data(:,:,3) = 0.3*data(:,:,3);
% Sample 2-D matrix of Z indices
idx = randi(3,3,3);
% Code for Loop, would like to do this without loop
sz = size(idx);
output = nan(sz);
for i = 1:numel(idx)
[j,k] = ind2sub(sz,i);
output(i) = data(j,k,idx(i));
end
  2 个评论
meghannmarie
meghannmarie 2019-5-31
So, I am not applying the a 3D logical matrix to all the data. I have a 4D matrix with indexes to the points in 3rd dimension which I want to keep and store in a 3D matrix (obviously this value is not static, it changes from point to point). I am still struggling with correct answer. I wonder if trying to turn 3D matrix into a 4D logical matrix would be easier, but not sure....
To put this in perspective, I have a very large 4 dimensional matrix (time, depth, lat, and lon) and I am trying to pull just the data from the bottom layer which the depth changes. My output would be a 3 dimensional matrix (time, lat, and lon) - all the data is from the bottom.
I could easily do this in a loop but would take lots of time...

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采纳的回答

Matt J
Matt J 2019-5-31
编辑:Matt J 2019-5-31
I don't think your example works because idx wasn't 2D as the comments say it should be. Here is what I think you want, though.
[m,n,p]=size(data);
[I,J]=ndgrid(1:m,1:n);
ouput = data( sub2ind([m,n,p], I,J,idx) );
  3 个评论
Matt J
Matt J 2019-5-31
No, I see nothing untoward in the first 4 lines, at least.
meghannmarie
meghannmarie 2019-5-31
I was super tired yesterday, probably something earlier in code I screwed up...

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