pixel by pixel comparison of images
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hello I am working with gabor filter. I have 12 images with different orientations which is output by gabor filter.I want to perform pixel by pixel comparison of images and find the maximum response image. maximum response of image is the image which has the highest pixel value other than all the images .can anybody tell me how do i do it. I have a small piece of code.
for n=1:N
count = 1;
gb = gabor_fn(bw,gamma,psi(1),lambda,theta);%...
figure;
imshow(gb);
theta = theta + pi/N;
end
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Accepted Answer
Walter Roberson
on 11 Nov 2013
Your lambda is too low or your bw is too low.
You appear to be using gabor_fn from http://www.mathworks.com/matlabcentral/fileexchange/23253-gabor-filter
6 Comments
Walter Roberson
on 11 Nov 2013
Sorry, my bed is calling me now. I have already shut down my system for the evening.
More Answers (1)
Image Analyst
on 10 Nov 2013
Something like this perhaps:
theMaxValues = zeros(1, N);
theta = 0; % whatever...
for n=1:N
gb = gabor_fn(bw,gamma,psi(1),lambda,theta(n));
theMaxValues(n) = max(gb(:));
figure;
imshow(gb);
theta(n+1) = theta(n) + pi/N;
end
[overallMax, index] = max(theMaxValues);
thetaOfMax = theta(index);
final_gb = gabor_fn(bw,gamma,psi(1),lambda,thetaOfMax);
5 Comments
Image Analyst
on 12 Nov 2013
I don't have your code so I can't go much more. Even if I did, I probably wouldn't have time. Please review this: http://blogs.mathworks.com/videos/2012/07/03/debugging-in-matlab/ and you will be able to solve it yourself. Also please make yourself aware how dot operators work - when and why you use them. There is a difference between matrix multiplication of two matrices, and multiplication of those two matrices element-by-element. You can matrix multiply a n*m matrix by a m*n matrix, but you can't do an element by element multiplication if n is not equal to m. Similarly you can dot multiply element-by-element an m*n matrix by another n*m matrix, but you can't do a matrix multiplication of those matrices. Make very sure you understand those concepts and operators very well.
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