Program to find the median of the matrix (need to use for loop)
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Hi,
I need to make a progrqam for the following logic:
1. Consider a matrix of size 120x120. Now consider a section of 3x3 and find the median of the 3x3 section, replacing the values of the matrix to make a new matrix (of a smaller size). The new 3x3 section will be taking the next element as the central element.
2. Please see that there is a command in MATLAB, called medfilt, medfilt2, which does the same, but I need to develop a program which does this manually i.e. it first sorts our the elements in ascending or descending order and then selects the middle value as the median thereby forming a new matrix with the median values.
The program that I am building is more complex than the section which I need help on. Any help in deleveloping this program will be highly appreciated.
Thanks and Regards
Manpreet Kaur
2 Comments
bym
on 17 Apr 2011
what have you tried so far?
Paulo Silva
on 17 Apr 2011
interesting homework!
Answers (2)
Andrei Bobrov
on 17 Apr 2011
variant % size(Min) = [m1 n1], size(Msection) = [m2 n2]
mn1 = size(Min)+2;
Min1 = zeros(mn1);
Min1(2:end-1,2:end-1) = Min;
Mout = zeros(mn1-2);
IJ = arrayfun(@(x)subsref(buffer(1:x,3,2),struct('type','()','subs',{{':',3:x}})),mn1,'UniformOutput', false);
[I,J] = IJ{:};
for ii = 1:length(I)
for jj = 1:length(J)
Mout(ii,jj) = median(median(Min1(I(:,ii),J(:,jj))));
end
end
as in the previous variant without a loop (as in medfilt2)
mn1 = size(Min)+2;
Min1 = zeros(mn1);
Min1(2:end-1,2:end-1) = Min;
IJ = arrayfun(@(x)subsref(buffer(1:x,3,2),struct('type','()','subs',{{':',3:x}})),mn1,'Un', false);
[I,J] = IJ{:};
[j3,i3] = meshgrid((1:mn1(2)-2),(1:mn1(1)-2));
Mout = arrayfun(@(x,y)median( reshape(Min1(I(:,x),J(:,y)),[],1)),i3,j3);
2 Comments
Manpreet Kapoor
on 18 Apr 2011
Andrei Bobrov
on 18 Apr 2011
analogy to your task: m1 = 120, n1 = 120, m2 = 3, n2 = 3
Manpreet Kapoor
on 21 Apr 2011
0 votes
3 Comments
Matt Fig
on 22 Apr 2011
Perhaps (I hope) Anderei left some things out because this is YOUR homework problem and you should do at least some of it yourself.
This site is not a homework solution service....
Show some work of your own....
Manpreet Kapoor
on 27 Apr 2011
Sean de Wolski
on 27 Apr 2011
I wouldn't want to take a class taught by someone who: A) doesn't know how to follow instructions (this post, and your identical duplicate post) and B) Can't read documentation. $0.02
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