Find center of mass of parts of a matrix

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Hi, Let us say that I have a matrix as shown below:
Then I apply a treshold to this matrix and am left with a region of interest. As such:
How can I find the center of mass for this region of interest (using values from the corresponding indexes in the first matrix) and the location of this center of mass in the original matrix?
Thanks for any help

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Guillaume
Guillaume 2016-11-28
编辑:Guillaume 2016-11-28
Well, go back to the definition of the centre of mass, which is just a weighted mean:
%inputs:
%originalmatrix: the original matrix
%binarisedmatrix = originalmatrix > threshold; %the thresholded matrix, a logical array
[rows, cols] = ndgrid(1:size(originalmatrix, 1), 1:size(originalmatrix, 2));
rowcentre = sum(rows(binarisedmatrix) .* originalmatrix(binarisedmatrix)) / sum(originalmatrix(binarisedmatrix));
colcentre = sum(cols(binarisedmatrix) .* originalmatrix(binarisedmatrix)) / sum(originalmatrix(binarisedmatrix));
  4 个评论
Guillaume
Guillaume 2016-11-28
Works fine for me:
originalmatrix = ones(10);
originalmatrix(4:7, 4:7) = 2 %semicolon missing for display
binarisedmatrix = originalmatrix > 1 %semicolon missing for display
[rows, cols] = ndgrid(1:size(originalmatrix, 1), 1:size(originalmatrix, 2));
rowcentre = sum(rows(binarisedmatrix) .* originalmatrix(binarisedmatrix)) / sum(originalmatrix(binarisedmatrix)) %semicolon missing for display
colcentre = sum(cols(binarisedmatrix) .* originalmatrix(binarisedmatrix)) / sum(originalmatrix(binarisedmatrix)) %semicolon missing for display
returns 5.5 for both rowcentre and colcentre
OH
OH 2016-11-29
You are right, it works. I had just made a silly mistake. Thanks!

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KSSV
KSSV 2016-11-28
doc mean
  1 个评论
OH
OH 2016-11-28
Hey, thanks. I know how to get the mean value of the region of interest: let us call the first matrix A and the tresholded matrix, where the region of interest indexes = 1 and outside region indexes = 0,B. Then I find the mean value by: mean(A(B==1))
However, I need to find the location of the center of mass in A

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