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필터 지우기

constructiong 3x3 matrix window

조회 수: 7 (최근 30일)
FIR
FIR 2012년 12월 19일
I am designing a filter removing impulses noises from an RGB IMAGE
For identifying the noise pixels in the image I need a 3x3 window to slide over the image starting from the first Pixel to the Last.
If the corrupted Pixel is found i have to do some calculation to correct it.
To find this Corrupted Pixel I need a 3x3 Window to slide over my Image.
please send me the code for this

채택된 답변

Matt J
Matt J 2012년 12월 19일
Hint - the CONV command

추가 답변 (2개)

Walter Roberson
Walter Roberson 2012년 12월 19일
blockproc()
  댓글 수: 7
FIR
FIR 2012년 12월 19일
or
I = imread(...)
kernel = ones(3, 3) / 9; % 3x3 mean kernel
J = conv2(I, kernel, 'same');
Walter Roberson
Walter Roberson 2012년 12월 19일
Matt J is referring to the second of those forms, where you create your own kernel.

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Image Analyst
Image Analyst 2012년 12월 19일
FIR, I've ALREADY posted code to get rid of impulse noise. In fact it was to your very own question just a few days ago! Perhaps you've forgotten, or just ignored my answer. Here is the link: http://www.mathworks.com/matlabcentral/answers/56515#answer_68417 If you want to replace noise pixels with blurred pixels, just replace the medfilt2() with conv2() like Matt said, though I don't know why you'd do that because you'd be designing a worse filter.
  댓글 수: 6
FIR
FIR 2012년 12월 20일
no walter my question is without separating the RGB image to ,R,G,B plane can we directly proceed on RGB IMAGE,i dont want to separate into planes ,is it possible
Walter Roberson
Walter Roberson 2012년 12월 20일
Not using the code posted by Image Analyst. That code could be altered to effectively pull apart the planes without looking like it was pulling apart the planes. You can for example create a routine similar to medfilt2() but which accepts an RGB image and does plane-by-plane filtering internally.
You need to ask yourself, though, what it means to do a median() with respect to values that have three components (R, G, B), and how that differs from [median®, median(G), median(B)] applied individually. The problem becomes rather similar to that of comparing two complex numbers: just as there is no defined sorting order for all P < Q when P and Q are complex, there is also no defined sorting order for all (R1,G1,B1) < (R2,G2,B2) pixels. Without a defined sorting order, you cannot determine median() as median() requires logically fully ordering the values to find which value is in the "middle" of the fully ordered list.

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