hey guys,I built this function:
err=zeros(15,15,15);
for p=-7:7;
for r=-7:7;
for q=-7:7;
zR=gwl*p*1.12+xR;
zG=gwl*q+xG;
zB=gwl*r*0.88+xB;
h=0.33*(abs(zR-zB)+abs(zB-zG)+abs(zR-zG));
err(8+p,8+r,8+q)=h;
end
end
end
and now i need know the 3 indices of the min value of err, any ideas how to achieve it?
thanks in advance

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Roger Stafford
Roger Stafford 2014년 9월 17일

1 개 추천

[~,ix] = min(err(:));
[i1,i2,i3] = ind2sub(size(err),ix);
The minimum value is err(i1,i2,i3).

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