Sampling pixel intensities according to distance matrix...

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 ! ! !

Answer 1

[Voss]

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

[DGM]

Might also want to round X so that the equality test works reliably. If other binning methods are used, it might be good to test for equality with tolerance.

[Chaz Pelas]

Benjamin thank you, I was able to get much closer to what I was looking for with this!

Answer 2

[Image Analyst]

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

[Chaz Pelas]

splitapply is not working well in this particular instance (the values in Mat(Y) are not integers and are very small) I am receiving this error:

**********************************

Error using splitapply (line 87)

Group numbers must be a vector of positive integers, and cannot be a sparse vector.

Error in (line 430)

theMeans = splitapply(@mean, A, B+1)

**********************************

Is there a workaround for this issue using the splitapply function?

[Image Analyst]

I simply used the data you supplied, and it worked. If you supplied the wrong data, or have different data that it does not work with, then attach those variables in a .mat file.

Alternatives include the functions groupsummary() and grpstats().

Also check out the function findgroups().

[Chaz Pelas]

I apologize for the erroneous example, I am near completely unaware of the limitations and uses of this platform. I greatly appreciate the hastey reply and the few suggestions, the stats toolbox had some neat things in it and the other functions had some interesting uses too.

[Image Analyst]

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.

Answer 3

[DGM]

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.