Vectorize a parfor loop to save time

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Novince
Novince il 15 Ago 2018
Commentato: Novince il 17 Ago 2018
parfor ii=1:(subnumX-1)*(subnumY-1)
[h1,p1] = subsolverCB(S1Group{ii}, Hhat1Group{ii}, tau, alpha, kappa, gamma, nIn,H1{ii},P1{ii},1); %Omega1%
H1{ii}=h1;
p1x=size(p1,1);
p1y=size(p1,2);
p1y=p1y/2;
P1{ii}(1:p1x, 1:p1y) =p1(:,1:p1y);
P1{ii}(1:p1x, subsizeY+1:(subsizeY+p1y)) =p1(:,(p1y+1):end);
end
As is shown above, I used a parfor loop within a while loop, but elapsed time increased several times compared to a for loop. Maybe it's better to vectorize this part in order to save time,but I don't know how to realize it. Thanks in advance!
Here is the body of my own function subsolverCB,all 'if' is true since I deleted false cases:
if true
function [hSub,pSub] = subsolverCB(s1, h1hat, tau, alpha, kappa, gamma,
nIn,HGroup,pInit,flag)
[m,n] =size(h1hat);
hSub = HGroup;
hcheck = HGroup;
c=size(pInit,2)/2;
Es =zeros(size(h1hat));
errSub=zeros(nIn,1);
if flag==1
px=pInit(1:m-1,1:n-1);
py=pInit(1:m-1,c+1:c+n-1);
Es(1:end-1,1:end-1)=s1;
end
iterationSub = 1;
while iterationSub <= nIn
hold = hSub;
[hx, hy] = grad2(hcheck);
if flag==1
Rhx=hx(1:end-1,1:end-1);
Rhy=hy(1:end-1,1:end-1);
end
ptildex = px + kappa*Rhx;
ptildey = py + kappa*Rhy;
Denom = sqrt(ptildex.^2+ptildey.^2);
px = ptildex ./ max(Denom, 1); py = ptildey ./ max(Denom, 1);
if flag==1
PPtmp=[px,zeros(m-1,1),py,zeros(m-1,1)];
PP=[PPtmp;zeros(1,size(PPtmp,2))];
end
Edivp = div2(PP);
htilde = hSub + gamma*(Edivp);
tmp= (tau*htilde + gamma*h1hat-tau*gamma*alpha*Es)/(tau+gamma);
hSub=max(min(tmp,1),0);
hcheck = 2*hSub - hold;
errSub(iterationSub)=norm(hSub-hold,'fro');
iterationSub = iterationSub+1;
end
pSub=[px py];
end
  3 Commenti
OCDER
OCDER il 15 Ago 2018
Modificato: OCDER il 15 Ago 2018
We need to know subsolverCB function - maybe you can vectorize this instead. While we wait for that information, here is a simple code cleanup. Note that using parfor will not always guaranteed improved speeds, as overhead for distributing and collecting jobs can exceed the speed for just using a regular for loop.
parfor ii = 1:(subnumX-1)*(subnumY-1)
[H1{ii}, p1] = subsolverCB(S1Group{ii}, Hhat1Group{ii}, tau, alpha, kappa, gamma, nIn,H1{ii},P1{ii},1); %Omega1%
[p1x, ply] = size(p1);
m = floor(ply/2); %what if ply is odd? this prevents error.
P1{ii}(1:p1x, [1:m (subsizeY+1):(subsizeY+(p1y-m))]) = p1;
end
Novince
Novince il 16 Ago 2018
Sorry guys,I'm not quite familiar with this forum yet. The subsolverCB is one of my own functions which intends to compute part of the picture.What I am trying to do is replacing a loop with a vectorized function.I am not quite sure if it's feasible.I was inspired by the assignment of a matrix.

Accedi per commentare.

Risposta accettata

OCDER
OCDER il 16 Ago 2018
I've looked through the code and it is mostly vectorized already. See comments though for minor improvements. At this point, you'll have to test out different matlab operations to see if you can find a shortcuts, prevent unnecessary matrix copies, etc. You are essentially doing micro-optimization, which is often time-consuming but could be worth it, depending on speed requirements.
As for parfor, see comment above. It seems difficult to vectorize the parallel computing of subsolverCB.
%function [hSub,pSub] = subsolverCB(s1, h1hat, tau, alpha, kappa, gamma, nIn, HGroup, pInit, flag)
function [hSub,pSub] = subsolverCB(s1, h1hat, tau, alpha, kappa, gamma, nIn, hSub, pInit, flag)
[m,n] = size(h1hat);
%hSub = HGroup; %Why do you need 2 copies of HGroup? How about just replacing input HGroup with hSub?
hcheck = hSub;
c = size(pInit,2)/2;
Es = zeros(size(h1hat));
errSub = zeros(nIn,1);
if flag %== 1 , no need to check flag==1 if flag is either true or false
px = pInit(1:m-1,1:n-1);
py = pInit(1:m-1,c+1:c+n-1);
Es(1:end-1,1:end-1) = s1;
end
% iterationSub = 1;
% while iterationSub <= nIn
tmpC1 = gamma*h1hat - tau*gamma*alpha*Es; %To prevent repeated calculations in for loop
for iterationSub = 1:nIn
hld = hSub; %hold = hSub; %don't override matlab function "hold", used for holding plots.
[hx, hy] = grad2(hcheck);
if flag %== 1
Rhx = hx(1:end-1,1:end-1);
Rhy = hy(1:end-1,1:end-1);
end
ptildex = px + kappa*Rhx;
ptildey = py + kappa*Rhy;
Denom2 = ptildex.^2+ptildey.^2; %Denom = sqrt(ptildex.^2+ptildey.^2); %don't sqrt early to save computing power
MaxDenom = sqrt(max(Denom2, 1)); %sqrt for a smaller set saves time
px = ptildex ./ MaxDenom; %max(Denom, 1);
py = ptildey ./ MaxDenom; %max(Denom, 1);
if flag %== 1
%PPtmp = [px,zeros(m-1,1),py,zeros(m-1,1)];
%PP = [PPtmp;zeros(1,size(PPtmp,2))]; %Why not just initialize in one shot?
PP = vercat([px,zeros(m-1,1),py,zeros(m-1,1)], zeros(1, size(PPtmp, 2)));
end
%Edivp = div2(PP);
%htilde = hSub + gamma*(Edivp);
htilde = hSub + gamma*(div2(PP)); %condensing into 1 line
%tmp = (tau*htilde + gamma*h1hat - tau*gamma*alpha*Es)/(tau+gamma); %Prevent repeated calculation( "gamma*h1hat, tau*gamma*alpha*Es, etc.
tmp = (tau*htilde + tmpC1)/(tau+gamma);
hSub = max(min(tmp,1),0);
hcheck = 2*hSub - hld;
errSub(iterationSub) = norm(hSub-hld,'fro');
end
pSub = [px py];
See if you can avoid creating temporary variable px and py, which is just a subset of pInit.

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