parallel computation for a for loop

2019 4 12
1 个回答
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更新时间：2019 4 12
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0 个投票

Below is the part of the code I have written, in which I am trying to use parallel computation. But it does give me an error as below:
"Error: The variable k_1 in a parfor cannot be classified."
In each parfor iteration some specific rows of "k_1" matrix will be updated irrelavant to the rest, so I cannot see why I get this message. Any help in this regard will be highly appreciated. Please find the parallel portion of my code below:
k_1 = k;
M_1 = M;
% pi = 0;
parfor ppi=1:NP
for pii=1:kkk
% pi= pi + 1;
pi = (ppi-1)*kkk+pii;
Tempo1 = zeros(1,1);
Tempo1_M = zeros(1,1);
Tempo2 = zeros(1,1);
Tempo2_M = zeros(1,1);
for irow = 1:p(pi)
[xx,inside_angle] = find(irow==[p;q]);
Tempo1(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[k(p(inside_angle),p(pi))+lambda(pi)*k(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[k(q(inside_angle),p(pi))+lambda(pi)*k(q(inside_angle),q(pi))];
Tempo1_M(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[M(p(inside_angle),p(pi))+lambda(pi)*M(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[M(q(inside_angle),p(pi))+lambda(pi)*M(q(inside_angle),q(pi))];
end
for irow = 1:q(pi)
[xx,inside_angle] = find(irow==[p;q]);
Tempo2(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[alpha(pi)*k(p(inside_angle),p(pi))+k(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[alpha(pi)*k(q(inside_angle),p(pi))+k(q(inside_angle),q(pi))];
Tempo2_M(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[alpha(pi)*M(p(inside_angle),p(pi))+M(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[alpha(pi)*M(q(inside_angle),p(pi))+M(q(inside_angle),q(pi))];
end
%Assign tempos to k
for irow = 1:p(pi)
k_1(irow,p(pi)) = Tempo1(irow);
k_1(p(pi),irow) = Tempo1(irow);
M_1(irow,p(pi)) = Tempo1_M(irow);
M_1(p(pi),irow) = Tempo1_M(irow);
end
for irow = 1:q(pi)
k_1(irow,q(pi)) = Tempo2(irow);
k_1(q(pi),irow) = Tempo2(irow);
M_1(irow,q(pi)) = Tempo2_M(irow);
M_1(q(pi),irow) = Tempo2_M(irow);
end
end
end
k=k_1;
M=M_1;
Please note that k and M matrices are defined by me at the start of the code!

5 个评论
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Matt J
Matt J 2019-4-12
编辑：Matt J 2019-4-12
The computation does not look parallel. All of your parfor workers are trying to write into the same matrices k and M.
Maryam
Maryam 2019-4-12
Thank you so much for your answer Matt. Would you please tell me how can I change it such that the computation will become parallel?
Maryam
Maryam 2019-4-12
In other words, I don't understand why it is not parallel, and what changes should I make to convert it to parallel!
Matt J
Matt J 2019-4-12
编辑：Matt J 2019-4-12
No, you need to explain to us why you think it is parallelizable. Parallelizable means the operations done by your loop iterations (over ppi) are independent of one another and could just as well be done on separate computers. Since all loop iterations in the code you've shown appear to be writing into the same matrix, it is hard to see what kind of independence from each other you think the loop iterations can have.
Maryam
Maryam 2019-4-12
Well, the reason I think it's parallel is that in each iteration (over ppi) I produce 4 arrays, which are independant. Then I assign the arrays to the columns of k_1 and M_1, which the array numbers are not the same. For example for a 6x6 matrix, if "NP" will be equal 3 (which means I want to use 3 processors to do the job independantly from each other), then processor one will override 2 specific columns of k_1 (lets say 2 and 3), processor 2 override another 2 columns (1 and 4 for example), and the last processor should override the last two columns (5 and 6). These pairs have computed at the start of my code (which is not mention here), and it is prooved that there aren't any repeated number in pairs!
So by the explanation I provided, would you please tel me in what part I am making the mistake? In another words, what part of my explanation is wrong based on the parallel computation concept?
Again thank you so much for your time and considerations.

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 采纳的回答

Catalytic
Catalytic 2019-4-12

0 个投票

parfor pi=1:NP*kkk
Tempo1 = zeros( p(pi) ,1 );
Tempo1_M = zeros( p(pi) ,1);
Tempo2 = zeros(q(pi) ,1);
Tempo2_M = zeros(q(pi) ,1);
pSubs = zeros( p(pi) ,2 ); %new
qSubs = zeros(q(pi) ,2);
for irow = 1:p(pi)
[xx,inside_angle] = find(irow==[p;q]);
Tempo1(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[k(p(inside_angle),p(pi))+lambda(pi)*k(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[k(q(inside_angle),p(pi))+lambda(pi)*k(q(inside_angle),q(pi))];
Tempo1_M(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[M(p(inside_angle),p(pi))+lambda(pi)*M(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[M(q(inside_angle),p(pi))+lambda(pi)*M(q(inside_angle),q(pi))];
pSubs(irow,:)=[irow,p(pi)]; %new
end
for irow = 1:q(pi)
[xx,inside_angle] = find(irow==[p;q]);
Tempo2(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[alpha(pi)*k(p(inside_angle),p(pi))+k(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[alpha(pi)*k(q(inside_angle),p(pi))+k(q(inside_angle),q(pi))];
Tempo2_M(irow) = [(xx-1)*alpha(inside_angle)+(2-xx)]* ...
[alpha(pi)*M(p(inside_angle),p(pi))+M(p(inside_angle),q(pi))] + ...
[(2-xx)*lambda(inside_angle)+(xx-1)]* ...
[alpha(pi)*M(q(inside_angle),p(pi))+M(q(inside_angle),q(pi))];
qSubs(irow,:)=[irow,q(pi)]; %new
end
subsCell{pi}=[pSubs;qSubs]; %new
kValCell{pi}=[Tempo1;Tempo2];
MValCell{pi}=[Tempo1_M;Tempo2_M];
end
subs=cell2mat(subsCell);
kVal=cell2mat(kValCell);
MVal=cell2mat(MValCell);
k_1=accumarray(subs,kVal,size(k));
M_1=accumarray(subs,MVal,size(M));
k=k_1 + tril(k_1.',-1); %make symmetric
M=M_1 + tril(M_1.',-1);

6 个评论
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Maryam
Maryam 2019-4-12
Thank you so much Catalytic for answering. I have tried the code and I get the error as below:
"
Error using cat
Dimensions of arrays being concatenated are not consistent.
Error in cell2mat (line 75)
m{n} = cat(2,c{n,:});
Error in real_parallel_modified_MATLAB (line 191)
subs=cell2mat(subsCell);
"
Catalytic
Catalytic 2019-4-12
Try instead,
subsCell{pi,1}=[pSubs;qSubs]; %new
kValCell{pi,1}=[Tempo1;Tempo2];
MValCell{pi,1}=[Tempo1_M;Tempo2_M];
Matt J
Matt J 2019-4-12
编辑：Matt J 2019-4-12
Or better yet, pre-llocate
[subsCell,kValCell,MValCell]=deal(cell(NP*kkk,1));
parfor pi=1:NP*kkk
...
end
Maryam
Maryam 2019-4-12
Also another think is because the number of parallel processors are limited (lets say 5 for instance), this form of formulation requires "NP*KKK", which might be so large, about 1000 for instance. That was why I tried to do block parallel and divide the jobs between certain processors. This is the reason instead of using "parfor pi=1:NP*kkk" I used:
"parfor ppi=1:NP
for pii=1:kkk"
Matt J
Matt J 2019-4-12
编辑：Matt J 2019-4-12
You are doing manually what the parfor machinery already does for you. Parfor is smart enough, without your intervention, to break the loop into chunks and distribute them to the workers.
Maryam
Maryam 2019-4-12
I understand! Well I wasn't aware of that. Thank you so much for clarification and your help.

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