Finding parfor baffling: Can anybody explain to me why this little bit of code works with for,but not with parfor?
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John Billingham
el 3 de Sept. de 2014
Comentada: John Billingham
el 4 de Sept. de 2014
function B = partest
A = 1:4; B = zeros(1,4);
parfor j=1:2
B([j j+2]) = A([j+2 j]);
end
The two bits of the loop access different bits of B, so there should be some way of doing this. My actual application involves large cell arrays, for which something similar holds for the function within the loop.
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José-Luis
el 3 de Sept. de 2014
Editada: José-Luis
el 3 de Sept. de 2014
Looks like the interpreter is not smart enough to detect that there is no race condition in the case you present. You could go around that using two loops:
A = 1:4; B = zeros(1,4);
parfor j=1:2
B(j) = A(j+2);
end
parfor j=1:2
B(j+2) = A(j);
end
I assume the operations you actually perform are more complicated than that. Otherwise Matt J's solution is the way to go.
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