decreasing run time of a program -

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som
som 2012-5-3
Hi all;
I've written a program including foure loops like below:
clear ; clc ; tic ;
%== input data ================================================
cap= 30;
T=3;
q=[0,10,20; 40,50,60; 10,20,30;] ;
pr=ones(3)*0.33;
s=[0;10;20;30];
r=[10;20;30;40];
nq=size(q,2);
ns=length(s);
nr=length(r);
ts=20;
tr=25;
% THE MAIN PART OF PROGRAM WHICH MUST BE VECTORISED : the part
% below up to "n=n+1" line, must be vectorised using ommiting all
% three loops i.e "ss" , "qq", "rr"
n=1;
while n<10000
for t=T:-1:1
for ss=1:ns
for qq=1:nq
for rr=1:nr
s2{t,ss}(qq,rr)= s(ss)+q(t,qq)-r(rr);
if s2{t,ss}(qq,rr)<=cap && s2{t,ss}(qq,rr)>=0
tsd{t,ss}(qq,rr)=(ts-s(ss))^2+ (tr- r(rr))^2 ;
else
s2{t,ss}(qq,rr)=NaN; tsd{t,ss}(qq,rr)=NaN;
end
if n==1 && t==T
tsd_total{n,ss}(qq,rr)=tsd{t,ss}(qq,rr);
else
if s2{t,ss}(qq,rr)>=0
for kk=1:ns
if s2{t,ss}(qq,rr)==s(kk);
tsd_total{n,ss}(qq,rr)=tsd{t,ss}(qq,rr)+ pr(qq,:)*f_min{n-1,kk}(:,1);
end
end;
else;
tsd_total{n,ss}(qq,rr)=NaN ;
end
end
end
f_min{n,ss}(qq,1)= min(tsd_total{n,ss}(qq,:));
end
end
n=n+1;
end
%= below lines are criterias to stop program ==========
y_brk1=0;
if n>3*T
nT=fix(n/T);
for p=T:nT+1
if (n-1)==(p-1)*T;
c1 =(p-T-1)*T; c2=(p-T)*T; c3 =(p+1-T)*T;
f_1=cell2mat(tsd_total(c1+1:c2,:));
f_2=cell2mat(tsd_total(c2+1:c3,:));
f_3=cell2mat(tsd_total(c3+1:n-1,:));
dif_32=fix(f_3-f_2);
dif_21=fix(f_2-f_1);
if isequalwithequalnans(dif_32,dif_21)==1 ; y_brk1=1; end
end
if y_brk1==1; break; end
end
end
if y_brk1==1; break; end
end
toc end
end
if y_brk1==1; break; end
end
toc
sine s, r and q are predetemind in the form of inputs, their relevant loops i.e. 'ss', ' qq'and 'rr' can be removed from the program. this deletion make my program so faster in run especially when the dimension of s, r and q are big.
I've tried to remove 'rr' and 'qq' loops by following programe, but I need to remove "ss loop" as well. how can I do this?? Any help would be appreciated a lot.
clear ; clc; tic ;
%===== DATA ==============
s_max= 30;
s_min=0;
T=3;
q=[0,10,20; 40,50,60; 10,20,30;] ;
pr=ones(3)*0.33;
s=[0;10;20;30];
r=[10,20,30,40;];
nq=size(q,2);
ns=length(s);
nr=length(r);
ts=20;
tr=25;
% PREALLOCATION OF VARIABLES =========
s2=cell(T,ns);
tsd=s2;
r_opt =cell(big_num,ns);;
tsd_total=r_opt;
fmin_init=cell(big_num,1);
tsd_intrpln=cell(big_num,nq);
%THE MAIN PART OF PROGRAM WHICH WAS VECTORISED======
dr= bsxfun(@minus,r,ones(nq,nr).*tr).^2;
n=1;
while n<10000
fmin_init{n}=ones(ns,nq);
for t=T:-1:1
sum_qr= bsxfun(@minus,q(t,:)',r);
for ss=1:ns
tsd_total{n,ss}=ones(nq,nr)*888888 ;
r_opt{n,ss}=ones(nq,nr)*888888 ;
sum_qrs=s(ss)+ sum_qr;
s2_orgnl{t,ss}=sum_qrs;
ds=bsxfun(@minus,s(ss)*ones(nq,nr),ts).^2;
tsd_init=bsxfun(@plus, ds,dr);
index_out = sum_qrs>s_max | sum_qrs <s_min;
sum_qrs(index_out)=NaN;
tsd_init(index_out)= NaN;
s2{t,ss}= sum_qrs;
tsd{t,ss}=tsd_init;
if n==1 && t==T
tsd_total{n,ss}=tsd{t,ss};
else
sum=0;
for qq=1:nq
tsd_intrpln{n,qq}=ones(nq,nr) ;
tsd_intrpln{n,qq}=interp1(s(:),fmin_init{n-1}(:,qq), s2{t,ss},'nearest');
sum = sum+pr(t,qq) .*tsd_intrpln{n,qq};
end
tsd_total{n,ss}=sum + tsd{t,ss};
end
f_min{n,ss}= min(tsd_total{n,ss},[],2);
fmin_init_0{n,ss}= f_min{n,ss}';
end
fmin_init_1{n}= cell2mat(fmin_init_0(n,:));
fmin_init{n} = vec2mat(fmin_init_1{n},nq );
n=n+1;
end
%= below lines are criterias to stop program =
y_brk1=0;
if n>3*T
nT=fix(n/T);
for p=T:nT+1
if (n-1)==(p-1)*T;
c1 =(p-T-1)*T; c2=(p-T)*T; c3 =(p+1-T)*T;
f_1=cell2mat(tsd_total(c1+1:c2,:));
f_2=cell2mat(tsd_total(c2+1:c3,:));
f_3=cell2mat(tsd_total(c3+1:n-1,:));
dif_32=fix(f_3-f_2);
dif_21=fix(f_2-f_1);
if isequalwithequalnans(dif_32,dif_21)==1 ; y_brk1=1; end
end
if y_brk1==1; break; end
end
end
if y_brk1==1; break; end
end
toc
Any help would really be appropriated.

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采纳的回答

Jan
Jan 2012-5-3
You do not need BSXFUN for operations with scalars:
bsxfun(@minus, r,ones(nq, nr) .* tr) .^ 2
==> faster:
(r - tr) .^ 2
interp1 is very slow in Matlab. You can copy the code o INTERP1 and crop all unnecessary tests and branches.
The repeated multiplication "ones(nq,nr)*888888" wastes time. repmat(888888, nq, nr) would be faster. But "888888" looks like a magic number - an ugly anti-pattern of bad programming practice.
A faster breaking out of the loop:
y_brk1 = 0;
for p=T:nT+1
if (n-1)==(p-1)*T;
...
if isequalwithequalnans(dif_32,dif_21) % no "== 1"
y_brk1 = 1;
break; % Break out of "for p"
end
end
So before vectorizing, there is a lot of potential for accelerations in the loop methods already.
  1 个评论
som
som 2012-5-3
Thanks for your constructive comments.
As you are an expert in Matlab, Can I ask you to spend time for understanding whole of the program? I know its time consuming for you , but I 'm sure you can help me to resolve problems about my program's run time.
Thanks in advance

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