Arg max without using a loop

1 次查看(过去 30 天)
I am looking for a method to reproduce the following without using a loop.
The issues come from the fact that the matrices I am using are too big and I could save some space avoiding to create a third matrix C to do the job.
My aim is to find some function f that does something similar to B = f(B(policy)). Is it possible to do it?
A = rand(10,20,30);
B = rand(10,20,30);
[A_hat, policy] = max(A,[],3);
for i = 1:10
for ii = 1:20
C = B(i,ii,policy(i,ii));
end
end
Thanks for the attention.
Sincerely
Luca

采纳的回答

Adam
Adam 2017-8-8
编辑:Adam 2017-8-8
[ rowGrid, columnGrid ] = ndgrid( 1:size(A,1), 1:size(A,2) );
idx = sub2ind( size( B ), rowGrid, columnGrid, policy );
C = B( idx );
As far as I am aware you have to convert to linear indices to extract the data as above. I have done exactly this for something I am working on recently. There may be a more efficient way, but not that I am aware of.
I tested one or two values to check they were correct and also looking at
A( idx )
gives a good confidence level too since these numbers are obviously all towards the higher end of the 0-1 range, e.g.
>> stuff = A( idx );
>> mean( stuff(:) )
ans =
0.969286594713116
Note: This may be memory intensive though as you create two grids of size(A,1) * size(A,2) containing the row and column indices in order to be able to turn all your 3rd dimension indices into linear indices.

更多回答(1 个)

Luca Gagliardone
Luca Gagliardone 2017-8-8
Thanks Adam, I see your point.
It still seems to me impossible that Matlab does not provide a more direct way without recurring to linear indexing (which is also quite RAM-expensive with ndgrid). I will wait for a few days before accepting definitely your answer, hoping for an easier solution!
Thanks again for your help.
Luca
  2 个评论
Adam
Adam 2017-8-8
Well, everything is linear indexing under the hood anyway.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by