Using nested for loops inside parfor

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Louis
Louis il 2 Dic 2014
Commentato: Louis il 2 Dic 2014
Hi,
I have upper triangle portion of n by n matrix to fill with s sample values (consider undirected and fully connected network without repeating edges). Sample values are drawn from designated pools unique to each entry of the upper triangle. For the simplicity, I am just putting in random values in the sample code below. Since I am taking a very large number of samples, and this is the major bottleneck in running time, I have decided to use parfor. However, I keep getting this error message saying "The PARFOR loop cannot run due to the way variable 'a' is used". I have read parfor documentation several times and understood required structures, sliced variables, and etc, but I can't figure this out. Any help would be greatly appreciated! Thanks,
%Create the matrix and assign preallocate zeros on the upper triangle portion to fill with sample values
matrix = cell(n,n);
for i = 1:n
for j = i+1:n
a{i,j} = zeros(s,1);
end
end
parfor k = 1:s
for i = 1:n
for j = i+1:n
a{i,j}(k) = rand; %Instead of rand there will be some specific function
end
end
end

Risposta accettata

Guillaume
Guillaume il 2 Dic 2014
Modificato: Guillaume il 2 Dic 2014
In the statement
A{i, j}(k) = something
with k scalar, something has to be scalar for it to work. So, you could just as well do:
parfor k = 1:s
a{k} = zeros(n);
for i = 1:n
for j=i+1:n
a{k}(i,j) = rand; %or whatever scalar is returned by your function
end
end
end
a = cat(3, a{:}) %3d matrix rather than cell array
If you really want a cell array at the end, after the cat:
a = num2cell(a, 3);

Più risposte (1)

Matt J
Matt J il 2 Dic 2014
Modificato: Matt J il 2 Dic 2014
I agree the documentation doesn't make it obvious what the problem is. But I think the problem is you are trying to slice a variable in the 2nd level of indexing. One fix, at least for the example you've shown, is to make "a" into a 3D numeric array instead of a cell array,
a=zeros(s,n,n);
parfor k = 1:s
for i = 1:n
for j = i+1:n
a(k,i,j)=rand;
end
end
end
a=num2cell(a,1); %post-convert to cell if needed
  3 Commenti
Matt J
Matt J il 2 Dic 2014
This appears to fix it.
parfor k = 1:100
for i = 1:10
for j = 1:10
if j>=i+1
a(k,i,j) = rand;
end
end
end
end
Louis
Louis il 2 Dic 2014
Thank you!

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