Please help, logical indexing of 3D matrices
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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!!!
10 个评论
James Tursa
2015-3-6
What's wrong with using the last 2-line form that works?
Miguel Lopes
2015-3-6
Sad Grad Student
2015-3-6
I am unable to understand what exactly you want.. Can you give an example of the sample input of A and B matrices and sample output where what content of A is combined with that of B ?
James Tursa
2015-3-6
编辑：James Tursa
2015-3-6
@Miguel: Do you mean that the statement D = A(:,:,1) is in an of itself causing an out of memory error? Where, exactly, are you getting an out of memory error? Sizes you show above are not that large.
Miguel Lopes
2015-3-6
编辑：Miguel Lopes
2015-3-6
Miguel Lopes
2015-3-6
编辑：Miguel Lopes
2015-3-6
Miguel Lopes
2015-3-6
James Tursa
2015-3-6
编辑：James Tursa
2015-3-6
Here is what I get ... as you can see the memory usage is not that much:
>> A = rand(480,848,3);
>> B = rand(480,848)<0.5;
>> C = zeros(480,848);
>> D = A(:,:,1);
>> C(B) = D(B);
>> whos
Name Size Bytes Class Attributes
A 480x848x3 9768960 double
B 480x848 407040 logical
C 480x848 3256320 double
D 480x848 3256320 double
What am I missing? Apparently I don't understand what you mean by the phrase "work with indexes in 3D".
Miguel Lopes
2015-3-6
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?
回答（5 个）
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);
2 个评论
Miguel Lopes
2015-3-6
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.
(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)
4 个评论
James Tursa
2015-3-6
This does not give a 480 x 848 result C as Miguel stated he wants.
David Young
2015-3-6
Note the solution above gives a vector - C contains only the elements of A corresponding to the non-zero elements of B. I think what is requested is the matrix given by
C = zeros(size(B));
ind2d = find(B);
C(ind2d) = A(ind2d);
where I have used ind2d to fit in with the comment about mulitplying to get planes other than the first.
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
2015-3-6
Wow, that is odd...we both corrected it to the same solution even to the naming of the intermediate variable!!
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
2015-3-6
0 个投票
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
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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2015-3-6
2019-11-6
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