How to speed up this code consisting of 6 nested FOR loops?
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Hi All,
I am trying to speed up the code below which consists of 6 nested FOR loops.
The purpose of this code is to fill a large, sparse matrix A with ones, and a large vector b with elements from a matrix dd of size TxTxN.
I have access to the Parallel Computing Toolbox, and ultimately this code will run on a cluster.
Any comments/suggestions would be greatly appreciated.
N= big number (e.g. 10000)
T=24;
A=sparse(N*T*(T+1)/2,(N-1)*T+T); %matrix to be filled
b=sparse(N*T*(T+1)/2,1); %vector to be filled
tempT = (T+1)*T/2; % constant used in the loop
for j=1:N
for t1=1:T
for t2=t1:T
for t3=1:t2-t1+1
rw = (j-1)*tempT + (t1-1)*(T-t1/2)+t2;
cl = (j-1)*T+t1-1+t3;
A(rw,cl) = 1;
for t4=t1:t2
for t5=t4:t2
b(rw) = b(rw) + dd(t4,t5,j); % dd is given, size(dd)=TxTxN
end
end
end
end
end
end
An earlier discussion on a simplified version of this with only 3 nested loops can be found here: http://www.mathworks.com/matlabcentral/answers/52005-how-to-transform-these-three-nested-for-loops-into-a-parfor-loop
8 Kommentare
Akzeptierte Antwort
Matt J
am 30 Okt. 2012
Bearbeitet: Matt J
am 30 Okt. 2012
Here's one proposal,
N=1e4;
T=24;
%dd=rand(T,T,N); %fake
tempT = (T+1)*T/2;
[T1,T2,T3]=ndgrid(1:T,1:T,1:T);
idx = (T2>=T1) & (T3<=T2-T1+1);
T1=T1(idx);
T2=T2(idx);
T3=T3(idx);
nt=numel(T1);
%Make row/col data
RW=(T1-1).*(T-T1/2)+T2;
CL=T1-1+T3;
RW=bsxfun(@plus,RW,(0:N-1)*tempT);
CL=bsxfun(@plus,CL,(0:N-1)*T);
%For building b
Z=zeros(nt,T^2);
for ii=1:nt % A very small loop, no need to optimize
[T4,T5]=ndgrid(T1(ii):T2(ii),T1(ii):T2(ii));
idx=T5>=T4;
T4=T4(idx);
T5=T5(idx);
ind=sub2ind([T,T],T4,T5);
Z(ii,ind)=1;
end
ddd=Z*reshape(dd,[],N);
A=sparse(RW,CL,1, N*T*(T+1)/2,N*T);
b=sparse(RW(:),1,ddd(:),N*T*(T+1)/2,1);
4 Kommentare
Matt J
am 1 Nov. 2012
The values I get for max(b1-b2) are tiny ones on the order of 1e-13. This is easily attributable to floating point precision issues. Remember, the two approaches are not adding up the numbers in the same order. Are you not getting similarly small values?
A stronger check is to look at the percent error, for which I also get tiny values:
>> percentError = full( sum(abs(b1-b2))/sum(abs(b1))*100 )
percentError =
2.5104e-14
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Performance and Memory finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!