Compare matrix element without loops

Hi,

Is there any way to get same result without using loops?

G = [5 8; 8 5; 3 9; 7 3; 1 4; 5 10; 6 7; 4 10; 4 7; 1 6];
n = 2;
nG = size(G,1);   
for ii=1:nG
   zz =1;
   isDom = [];
   for kk=1:n
        for jj=1:nG
            if ii ~= jj
                isDom(zz) = G(ii,kk) <  G(jj,kk);
                zz = zz +1;
            end
        end
   end        
  R(ii) = sum(isDom==1);
end

Answer 1

[dpb]

Not w/o zero loops, think not, but can reduce to one...

idx=1:nG;            % working index array for element logical lookup/exclusion
R=zeros(nG,1);       % preallocate
for i=1:nG
  isDom=(G(i,:)-G(idx~=i,:));
  R(i)=sum(isDom(:)<0);
end

You can eliminate the intermediate isDom temporary if desired..."exercise for the student" :)

[shdotcom shdotcom]

Thank you

[dpb]

You can also, of course, remove the explicit loop via arrayfun, but the loop is still there and resulting code is somewhat obfuscated and may well be slower, besides...