Please help, logical indexing of 3D matrices

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Miguel Lopes
Miguel Lopes 2015-3-6
回答： Tongyao Pu 2019-11-6
Hello
I'm stuck on this stupid thing and i know is obvious.
I have this matrice A: 480x848x3
And this logical matrice B: 480x848
And I want a matrice C=zeros(480, 848) like this: C(B)=A(B,1)
It should work but there's something wrong. I thought it might be like this:
C(B)=A([B 1])
or like this:
C(B)=A([B, 1])
or like this:
C(B)=A([B; 1])
But nothing works.
I know it works if i make:
D=A(:,:,1)
C(B)=D(B)
But how do i make it work with the indexes in 3D???
Thanks!!!
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Miguel Lopes
Miguel Lopes 2015-3-6
yes but I have many fotos like that, not just one!!! I also need the HSV channel for each one (i have to do the same), so I get no memory at a point!
James Tursa
James Tursa 2015-3-6
So, it sounds like the memory problem isn't this calculation for a single photo, but the fact that you have very many of these photos in memory at the same time?

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回答（5 个）

Star Strider
Star Strider 2015-3-6
编辑：Star Strider 2015-3-6
Your ‘A’ matrix looks like it could be an image. To collapse the third dimension, use the rgb2gray function in the Image Processing Toolbox.
For example:
A = uint8(randi([0 255], 480, 848, 3));
figure(1)
imshow(A)
G = rgb2gray(A);
figure(2)
imshow(G)
Another option is to expand ‘C’ to three dimensions by replicating it three times in the third dimension:
C = zeros(480,848);
Cx = repmat(C, 1, 1, 3);
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Miguel Lopes
Miguel Lopes 2015-3-6
What you mean "to collapse"??
I need the 3 channels and I have different masks for each channel! I need the data from the 3 channels!!
Star Strider
Star Strider 2015-3-6
If you need all three channels, then don’t collapse them into grayscale.
If you have different masks for each channel, concatenate them:
C1 = uint8(randi([0 255],480,848)); % Red Mask
C2 = uint8(randi([0 255],480,848)); % Green Mask
C3 = uint8(randi([0 255],480,848)); % Blue Mask
Cc= cat(3, C1, C2, C3); % All Masks
That creates a (480x848x3) matrix in ‘Cc’ that you may be able to use.

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Adam
Adam 2015-3-6
编辑：Adam 2015-3-6
(Edited version as in comments below - original edit was incorrect!)
C = zeros( size(B) );
ind2d = find(B);
C( ind2d ) = A( ind2d );
works without needing to create another matrix. Note that it works specifically because it is the 1st index of the 3rd dimension that you want. If you wanted the 2nd then you would need to add the x*y size to the ind2d array and for the 3rd index of the 3rd dimension you would need to add twice the x*y size to the indices.
In case you aren't aware, this works because you can use 1d (linear) indexing into an array of any dimension in addition to the standard subscript access of e.g. (300,200,3)
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Adam
Adam 2015-3-6
Ah yes, I tried to simplify my original answer and messed up the result:
C = zeros( size(B) );
ind2d = find(B);
C( ind2d ) = A( ind2d );
was what I meant.
I should stick to the way I usually program stuff with logical matrices assigned to their own variable rather than trying to simplify it in-place without checking it first in Matlab!
Adam
Adam 2015-3-6
Wow, that is odd...we both corrected it to the same solution even to the naming of the intermediate variable!!

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David Young
David Young 2015-3-6
编辑：David Young 2015-3-6
One way, slightly fiddly but the simplest I can think of now:
[rowsB, colsB] = find(B);
C = zeros(size(B));
ind3 = 1; % index of the plane of A that is wanted
C(B) = A(sub2ind(size(A), rowsB, colsB, ind3*ones(size(rowsB))));
Sad Grad Student
Sad Grad Student 2015-3-6
Does changing from D = A(:,:,1) to D = A(:,:,1).*B and then assign C(B) = D help? Or does it still give memory problems?
Tongyao Pu
Tongyao Pu 2019-11-6
Hello,
I am a bit confused that you would like C to be 480*848.
I wonder if you would like to index each layer of A with logical matrix B? If that is the case, that is how I do it:
C = A;
B2 = repmat(B, [1,1,3]); % just copy B into a 480*848*3 array
C(B2) = 'something you would like it to be, e.g. NaN'

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