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Compare matrix element without loops

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shdotcom shdotcom
shdotcom shdotcom on 4 May 2019
Commented: dpb on 4 May 2019
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

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Accepted Answer

dpb
dpb on 4 May 2019
Edited: dpb on 4 May 2019
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" :)

  2 Comments

dpb
dpb on 4 May 2019
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...

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