interesting matrix indexing question without for loops

1 vue (au cours des 30 derniers jours)
Chris Hooper
Chris Hooper le 2 Oct 2018
Commenté : Chris Hooper le 3 Oct 2018
size(matrixA)=[b,n,m]
size(matrixB)=[n,m]
how can I create
matrixC(i,j)=matrixA(matrixB(i,j),i,j)
without using for-loops? What is this kind of indexing called? Thanks!

Réponse acceptée

Stephen23
Stephen23 le 3 Oct 2018
Modifié(e) : Stephen23 le 3 Oct 2018
Use sub2ind to get the linear indices:
A = reshape(1:6*3*2,6,3,2);
B = [6,1;4,3,;5,2];
% Solution with loops:
for ii = 1:size(B,1)
for jj = 1:size(B,2)
C(ii,jj) = A(B(ii,jj),ii,jj);
end
end
% Solution with SUB2IND:
S = size(B);
[I,J] = ndgrid(1:S(1),1:S(2));
X = sub2ind(size(A),B,I,J);
D = A(X)
% Compare:
isequal(C,D)
  1 commentaire
Chris Hooper
Chris Hooper le 3 Oct 2018
Stephen you're awesome thanks! I knew there was something like that... it's been a while!

Connectez-vous pour commenter.

Plus de réponses (1)

Bruno Luong
Bruno Luong le 3 Oct 2018
Modifié(e) : Bruno Luong le 3 Oct 2018
C = A(B+reshape(0:numel(B)-1,size(B))*size(A,1))
  1 commentaire
Chris Hooper
Chris Hooper le 3 Oct 2018
Awesome I'll check this one, cannot even wrap my head around it!!

Connectez-vous pour commenter.

Catégories

En savoir plus sur Matrix Indexing dans Help Center et File Exchange

Tags

Produits

Community Treasure Hunt

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

Start Hunting!

Translated by