Effacer les filtres
Effacer les filtres

How to improve the code for vector calculation

1 vue (au cours des 30 derniers jours)
krishnamurthy
krishnamurthy le 6 Juin 2013
N=2;
MM=4;
AA=rand(1,12);AA1=AA(1:3);AA2=AA(4:6);AA3=AA(7:9);AA4=AA(10:12);
BB=rand(1,12);BB1=BB(1:3);BB2=BB(4:6);BB3=BB(7:9);BB4=BB(10:12);
DD=rand(1,12);DD1=DD(1:3);DD2=DD(4:6);DD3=DD(7:9);DD4=DD(10:12);
EE=rand(1,12);EE1=EE(1:3);EE2=EE(4:6);EE3=EE(7:9);EE4=EE(10:12);
HH=rand(1,12);HH1=HH(1:3);HH2=HH(4:6);HH3=HH(7:9);HH4=HH(10:12);
BBB=rand(12,3);BBB1=BBB((1:3),:);BBB2=BBB((4:6),:);BBB3=BBB((7:9),:);BBB4=BBB((10:12),:);
LAD=rand(1,4);LAD1=LAD(1);LAD2=LAD(2);LAD3=LAD(3);LAD4=LAD(4)
for iter=1:N
iter
for i=1:MM
if i==1
AA=AA1;BB=BB1;DD=DD1;EE=EE1;HH=HH1;BBB=BBB1;LAD=LAD1;
elseif i==2
AA=AA2;BB=BB2;DD=DD2;EE=EE2;HH=HH2;BBB=BBB2;LAD=LAD2;
elseif i==3
AA=AA3;BB=BB3;DD=DD3;EE=EE3;HH=HH3;BBB=BBB3;LAD=LAD3;
elseif i==4
AA=AA4;BB=BB4;DD=DD4;EE=EE4;HH=HH4;BBB=BBB4;LAD=LAD4;
end
Eeta=diag((AA+HH.*DD)./LAD)+BBB;
Delta=0.5*(1-(BB+HH.*EE)./LAD)';
Penta=Eeta\Delta;
Penta1(i,:)=Penta
end
PPP=reshape(Penta1',1,12)
end

Réponses (1)

Jan
Jan le 6 Juin 2013
It is a bad idea to pack indices into the name of the variables like in "AA4". If you use a cell instead, the FOR-loop get's much nicer:
AA = rand(1,12);
AAC {AA(1:3), AA(4:6), AA(7:9), AA(10:12)};
...
for i = 1:MM
AA = AAC{i};
...
I do not assume that a completely vectorized version is nicer or faster.

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