Any ideas how i can reduce the execution time?
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Dennis_Pana
le 19 Juil 2016
Réponse apportée : Dennis_Pana
le 20 Juil 2016
newdes=[0.6250 0.8750; 0.2500 0.7500; 0.3750 0.1250;
0.8750 0.3750; 0.1250 0.6250; 0 0; 0.5000 0.5000; 0.7500 0.2500;
0.6875 0.0625; 0.0625 0.8125; 0.1875 0.6875; 0.3125 0.3125;
0.4375 0.9375; 0.5625 0.5625; 0.9375 0.1875; 0.8125 0.4375;] ;
dis=zeros(v,1);
v=16;
m=2;
sum=0
s=2;
for j=1:v
for k=1:m
ddii=(newdes(s,k)-newdes(j,k))^2;%Distance calculation of point s at its new location with all the other points.
sum=sum+ddii;
end
dis(j,1)=sum; %Store all the distances.
sum=0;
end
I am trying to use bxsfun, but i cannot solve it. I want reduce the time effort. Thank you in advance.
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Azzi Abdelmalek
le 19 Juil 2016
newdes=[0.6250 0.8750; 0.2500 0.7500; 0.3750 0.1250;
0.8750 0.3750; 0.1250 0.6250; 0 0; 0.5000 0.5000; 0.7500 0.2500;
0.6875 0.0625; 0.0625 0.8125; 0.1875 0.6875; 0.3125 0.3125;
0.4375 0.9375; 0.5625 0.5625; 0.9375 0.1875; 0.8125 0.4375;] ;
[nn,mm]=size(newdes)
v=16;
m=2;
s=2;
ay=repmat(1:m,1,v)
ax=s*ones(1,numel(ay))
idx1=sub2ind([nn mm],ax',ay')
[xx,yy]=meshgrid(1:v,1:m)
idx2=sub2ind([nn mm],xx(:),yy(:))
ww=(newdes(idx1)-newdes(idx2)).^2 %Distance calculation of point s at its new location with all the other points.
dis1=sum(reshape(ww,2,[]))'
5 commentaires
Azzi Abdelmalek
le 19 Juil 2016
Modifié(e) : Azzi Abdelmalek
le 19 Juil 2016
I made some improvement, but it's less faster. Sometimes for loop is more efficient!
[nn,mm]=size(newdes);
s=2;
ay=repmat(1:mm,nn,1);
ax=s*ones(size(ay));
idx1=reshape(sub2ind([nn mm],ax',ay'),mm,[]);
[xx,yy]=meshgrid(1:v,1:m);
idx2=sub2ind([nn mm],xx,yy);
ww=(newdes(idx1)-newdes(idx2)).^2 ; %Distance calculation of point s at its new location with all the other points.
dis1=sum(ww)';
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