# Sampling pixel intensities according to distance matrix...

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Chaz Pelas on 12 Oct 2021
Commented: Image Analyst on 18 Oct 2021
I have Mat(X) that consists of distances; lets say
Mat(X) = [2 1 2; 1 0 1; 2 1 2]
And an image, let's call it Mat(Y), containing an intensity value in every i, j element; for example
Mat(Y) = [9 5 6; 7 1 3; 2 8 4]
I would like a vector describing the average pixel intensity at the distances described by Mat(X) such that
y(0) = 1
y(1) = (5+3+7+8)/4
y(2) = (9+6+4+2)/4
I am not a very saavy coder as I imagine that this should not be very difficult to do, yet I am struggling to make it happen; ANY POINTERS ARE GREATLY APPRECIATED ! ! !

Benjamin on 12 Oct 2021
Let X be your matrix of distances and Y be your matrix of intensities. Then the following code makes use of logical indexing to calculate the average value of Y at each unique value of X:
uX = unique(X(:)); % vector of unique distances
n_uX = numel(uX); % number of unique distances
uY = zeros(1,n_uX); % initialize average intensity vector
for i = 1:n_uX % for each unique distance
uY(i) = mean(Y(X == uX(i))); % average intensity is the mean of the intensities where distance == that unique distance
end
##### 2 CommentsShowHide 1 older comment
Chaz Pelas on 15 Oct 2021
Benjamin thank you, I was able to get much closer to what I was looking for with this!

### More Answers (2)

Image Analyst on 12 Oct 2021
Use splitapply() which was meant for this kind of thing:
MatX = [2 1 2; 1 0 1; 2 1 2]
% And an image, let's call it Mat(Y), containing an intensity value in every i, j element; for example
MatY = [9 5 6; 7 1 3; 2 8 4]
theMeans = splitapply(@mean, MatY(:), MatX(:)+1)
theMeans =
1
5.75
5.25
Image Analyst on 18 Oct 2021
So did my code work for you like it did for me? Are we done here? If not, attach your nonworking code and nonworking data.

DGM on 12 Oct 2021
Edited: DGM on 12 Oct 2021
Disregarding splitapply() for a moment, the issue of working with non-integers can be avoided by using the histogram tools to bin the distance array as desired.
X = [2 1 2; 1 0 1; 2 1 2]/100;
Y = [9 5 6; 7 1 3; 2 8 4];
nbins = 3; % you probably want more than 3
[~,~,idx] = histcounts(X,nbins);
binmeans = zeros(nbins,1);
for b = 1:nbins
binmeans(b) = mean(Y(idx == b));
end
binmeans
binmeans = 3×1
1.0000 5.7500 5.2500
If you want to use splitapply instead of the loop, you can do that too:
X = [2 1 2; 1 0 1; 2 1 2]/100;
Y = [9 5 6; 7 1 3; 2 8 4];
nbins = 3; % you probably want more than 3
[~,~,idx] = histcounts(X,nbins);
binmeans2 = splitapply(@mean,Y(:),idx(:))
binmeans2 = 3×1
1.0000 5.7500 5.2500
I'm sure findgroups would work too.
X = [2 1 2; 1 0 1; 2 1 2]/100;
Y = [9 5 6; 7 1 3; 2 8 4];
nbins = 3; % you probably want more than 3
idx = findgroups(X(:));
binmeans2 = splitapply(@mean,Y(:),idx(:))
binmeans2 = 3×1
1.0000 5.7500 5.2500
Findgroups may be simpler to use than assuming that groups are uniformly distributed (as with histogram tools). Depends on what you want, I guess.