Borrar filtros
Borrar filtros

Finding parfor baffling: Can anybody explain to me why this little bit of code works with for,but not with parfor?

1 visualización (últimos 30 días)
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.

Respuesta aceptada

Matt J
Matt J el 3 de Sept. de 2014
Editada: Matt J el 3 de Sept. de 2014
A = reshape(1:4,2,2);
B = zeros(2);
parfor j=1:2
B(j,:) = A(j,end:-1:1);
end
  5 comentarios
John Billingham
John Billingham el 3 de Sept. de 2014
OK. Thanks. I'll have a look at that tomorrow and see what I can do. I expect I'll have more questions. Really struggling to get my head around this. Sorry.
John Billingham
John Billingham el 4 de Sept. de 2014
Having given this some thought, I realize that your answer is great, but that I'm asking the wrong question, so I'm going to ask the right question instead.

Iniciar sesión para comentar.

Más respuestas (1)

José-Luis
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.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Etiquetas

Productos

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by