Compare matrix element without loops
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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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dpb
2019년 5월 4일
편집: dpb
2019년 5월 4일
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" :)
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dpb
2019년 5월 4일
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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