# a simple clasification of matrix elements

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Jules Ray il 18 Ott 2011
dear coleages, i got a problem with some elements, i`m working with a regular rectangle vertices, so, that's what i get:
%this are the table values of a regular rectangle vertices. are showed as row and cols %as the typical notation, as you see is formed by 4 rows and two cols which %correspond to coordinates X and Y (that`s why i use the name x1, x2,... xi and also %y1, y2, ...yi for each one)
x1=p(1,1); x2=p(2,1); x3=p(3,1); x4=p(4,1); y1=p(1,2); y2=p(2,2); y3=p(3,2); y4=p(4,2);
%continuing i have got the distances between each one. As you remember this is a %rectangle, and i have just three resulting distances.
A=sqrt((x2-x1)^2+(y2-y1)^2); B=sqrt((x3-x2)^2+(y3-y2)^2); C=sqrt((x3-x4)^2+(y3-y4)^2); D=sqrt((x4-x2)^2+(y4-y2)^2); E=sqrt((x3-x1)^2+(y3-y1)^2); F=sqrt((x4-x1)^2+(y4-y1)^2); G=sqrt((x3-x1)^2+(y3-y1)^2);
%i have make some order of the obtained distances values
V=[A B C D E F G]; W=sort(V); H=W'; %max to min ordering of obtained results and transposing
%the resulting H is a table which contains the distances between each coordinates, i have charged an example rectangle and i have obtained this values:
H= 102,08 102,08 928,45 928,45 934,05 934,05 934,05
OK, HEre comes my question:
I NEED TO CREATE A SCRIPT TO:
1)ELIMINATE THE REPEATED VALUES 2)REMOVE THE MAXIMUM AND MINIMUM VALUE /OR/ JUST TAKE THE CETRAL VALUE (928,45)
THANKS FOR YOUR HELP
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### Risposta accettata

Naz il 18 Ott 2011
There is one quick way to do it, however it is based on the assumption that the initial array will always have only 3 distinct values:
minV=min(H); % find min value in H
H(H==minV)=0; % set all min values to zero
maxV=max(H); % find max value in H
H(H==maxV)=0; % set all max values to zero
At this time you should have intermediate values somewhere in H. To separate them in the new matrix, you do the following:
Intermediate=H(H~=0);
This will give you the matrix with non-zero values, that is max and min values are eliminated.
If you want to get the actual value from the new H matrix you can do the following:
ind = find(H~=0, 1, 'first');
finds the address of the first non-zero value;
yourValue=H(ind)
gives the first non-zero value that is assumed to be the intermediate value between min and max.
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### Più risposte (1)

Andrei Bobrov il 19 Ott 2011
idxs = nchoosek(1:4,2)
H = sqrt(sum(abs(diff(reshape(p(idxs,:),size(idxs,1),[],2),1,2)).^2,3));
1.
[out1,a,b] = unique(H,'first')
2.
out2 = median(out1);
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