清空筛选条件
清空筛选条件

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

1 次查看（过去 30 天）
John Billingham
John Billingham 2014-9-3
评论： John Billingham 2014-9-4
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.

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2014-9-3
编辑：Matt J 2014-9-3
A = reshape(1:4,2,2);
B = zeros(2);
parfor j=1:2
B(j,:) = A(j,end:-1:1);
end
  5 个评论
显示 3更早的评论
John Billingham
John Billingham 2014-9-3
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 2014-9-4
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.

请先登录，再进行评论。

更多回答（1 个）

José-Luis
José-Luis 2014-9-3
编辑：José-Luis 2014-9-3
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.

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by