Mean of neighbor pixel

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Efstathios Kontolatis
Efstathios Kontolatis on 4 Oct 2016
Edited: Matt J on 25 Mar 2021
Hi guys,
I try to calculate the average intensity value of the neighbors of a pixel(either 8 or not). I am confused with the way matlab uses matrices(i-th,j-th position of a pixel is at j-th,i-th position of the matrix) and I have difficulties find the neighbors.
Thanks

Accepted Answer

Matt J
Matt J on 4 Oct 2016
Edited: Matt J on 4 Oct 2016
Just use convolution,
k=[1 1 1; 1 0 1; 1 1 1]/8;
averageIntensities = conv2(double(yourImage),k,'same');
  3 Comments
Arghya Pathak
Arghya Pathak on 25 Mar 2021
"same" stands for what?
Matt J
Matt J on 25 Mar 2021
Edited: Matt J on 25 Mar 2021

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More Answers (1)

Adam
Adam on 4 Oct 2016
If you are finding all 8 neighbours then it doesn't make a difference which is x,y and rows, columns, but your confusion and that comes from the fact that a matrix is indexed as (row,column), whereas an image is (x,y) more usually where x is horizontal and is therefore represented by columns of the image.
The neighbours of a pixel are simply
neighbours = yourMatrix( yLoc-1:yLoc+1, xLoc-1:xLoc+1 )
for a pixel at ( xLoc, yLoc ) in a matrix called yourMatrix.
This will also include the point itself, but that is easy to get rid of and you didn't state what you want to do with the neighbourhood so it may or may not be better to keep the central point for it to retain its 2d structure.
  2 Comments
Efstathios Kontolatis
Efstathios Kontolatis on 4 Oct 2016
Edited: Efstathios Kontolatis on 4 Oct 2016
I want to find the average intensity of the neighbors without using the central point. Also, I don't think that this solution will give me the neighbors on the borders of the image. I used this code
for i=1:height
for j=1:width
neighbours(i,j).ar=img(j-1:j+1, i-1:i+1);
end
end
and I take the following message 'Subscript indices must either be real positive integers or logicals'. Img is a uint16 image. As you can see I want to save the neighbors of each pixel.
Adam
Adam on 4 Oct 2016
Edited: Adam on 4 Oct 2016
Yes, you have to deal with the borders differently, but that is not difficult.
If you don't want the middle point then just throw it away and collapse the neighbours to a 1d array since you will just be taking the mean anyway - e.g.
neighbours(5) = [];

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