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Size of Images varying

Asked by FIR on 28 Jul 2012
Input = imread('baby1.jpeg') ;
size(Input)
P=rgb2gray(Input);
image= im2double(P);
[m n ] = size(A);
Med = [];
%Modified filter 
for i=2:m-1
    for j=2:n-1
            Med(1) = image(i,j);
            Med(2) =image(i-1,j) ;
            Med(3) = image(i-1,j+1);
            Med(4) = image(i,j-1);
            Med(5) = image(i,j+1);
            Med(6) = image(i+1, j-1);
            Med(7) = image(i+1,j);
            Med(8) = image(i+1,j+1);
            Afilteres(i, j) = median(Med(:));
    end
end 
imshow(Af);

In this the size of image varies

   size(P)=201   300
size(Af)
   200   299

PLease help

2 Comments

Sean de Wolski on 30 Jul 2012

In reponse to the flags and requests for deletion. I will not delete this question since Andrei, Wayne and IA all put effort into helping you.

FIR on 31 Jul 2012

ok Sean thank u

FIR

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3 Answers

Answer by Andrei Bobrov on 28 Jul 2012
Edited by Andrei Bobrov on 28 Jul 2012
Accepted answer
m = size(A);
A1 = zeros([m(1:2) + 2,m(3)]);
A1(2:end-1,2:end-1,:) = A;
Af = zeros(m);
for ii = 1:m(1)
    for jj = 1:m(2)
        k = reshape(A1(ii:ii+2,jj:jj+2,:),[],1,3);
        Af(ii,jj,:) = median(k([1:4,6:9],:,:)); % EDIT
    end
end

2 Comments

Image Analyst on 28 Jul 2012

You forgot to exclude the middle pixel when taking the median, like he did. Though I don't know what the use of this strange type of median filter would be. Maybe he can explain that.

Andrei Bobrov on 28 Jul 2012

Thank you Image, I corrected my answer.

Andrei Bobrov
Answer by Wayne King on 28 Jul 2012

The Image Processing Toolbox has medfilt2

3 Comments

FIR on 28 Jul 2012

wayne that function must not be used ,is there any other way ,please help

Wayne King on 28 Jul 2012

The file exchange appears to have median filter implementations, but I'm guessing if you're not allowed to use medfilt2, then you're not supposed to use those either

Image Analyst on 28 Jul 2012

If the built-in function must not be used, then it must be some kind of homework assignment, which means that code solutions provided by Answers must not be used either. Too bad because I have a nice demo using blockproc that I could have given you. However it's strange that median(), the 1D version, can be used while medfilt2(), the sliding 2D version, cannot be used.

Wayne King
Answer by Image Analyst on 28 Jul 2012

Of course the image sizes are different, because you're sliding a window along, and you have it so that when the edge of the window touches the edge of the image, it stops. So of course the output image will not be as big as the input image.

Again, I have demo code but since you said you're not allowed to use built in functions, and probably not code handed over to you in Answers or code you got from the File Exchange, I won't post it. Yeah, sometimes instructors/professors are picky about doing your own work.

4 Comments

Image Analyst on 30 Jul 2012

"median" like he used is also a single word function, as would "sort()" which is what I would have suggested as a second method. Sort and take the middle value. By the way, why are you excluding the pixel itself when you are taking the median?

FIR on 30 Jul 2012

any code is suggested without using medfilt2

Image Analyst on 30 Jul 2012

Well then you're all set, because Andrei gave you code that used median() (and you've already accepted his answer), and I suggested using sort() and taking the middle element. So I'm assuming that you're done now.

Image Analyst

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