# Sampling pixel intensities according to distance matrix...

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Chaz Pelas2021-10-12

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 ! ! !

### 采纳的回答

Benjamin 2021-10-12
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 个评论显示隐藏 1更早的评论
Chaz Pelas 2021-10-15
Benjamin thank you, I was able to get much closer to what I was looking for with this!

### 更多回答（2 个）

Image Analyst 2021-10-12
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
##### 4 个评论显示隐藏 3更早的评论
Image Analyst 2021-10-18
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 2021-10-12

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.

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