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
le 4 Mai 2019
Modifié(e) : dpb
le 4 Mai 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 commentaires
dpb
le 4 Mai 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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