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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Walter Roberson
Walter Roberson 2013-11-11
Your lambda is too low or your bw is too low.
  6 个评论
Walter Roberson
Walter Roberson 2013-11-11
Sorry, my bed is calling me now. I have already shut down my system for the evening.
vidya
vidya 2013-11-11
yes sir.!!! sorry for the disturbance..if possible reply me tomo sir!!..gud night :-)

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更多回答(1 个)

Image Analyst
Image Analyst 2013-11-10
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 个评论
Image Analyst
Image Analyst 2013-11-12
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.
vidya
vidya 2013-11-13
yes sir...thanks a lot for youa valuable solutions..i study and worked it out with the different method ..

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