how to get following matrix using bsxfun
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I have a matrix x=[0 0 0;5 8 0; 7 6 0].
I want a matrix m=[0 0 0; 5 8 0 ; 7 6 0 ; 0 0 8; 5 8 8; 7 6 8; 0 0 16; 5 8 16; 7 6 16] I want that the third column of matrix x gets multiplied each time by 8 while other two columns remain same. I want this to continue until the value in third column reaches 72. how can i do it with bsxfun??
0 commentaires
Réponse acceptée
Geoff Hayes
le 7 Fév 2015
Saurabh - if x is your defined as
x=[0 0 0;5 8 0; 7 6 0];
then you could replicate it 10 times as
m = repmat(x,10,1);
then just replace the third column with
m(:,3) = cell2mat(arrayfun(@(u)repmat(u*8,3,1),0:9,'UniformOutput',false)');
to produce the desired result. Try the above and see what happens!
3 commentaires
Geoff Hayes
le 8 Fév 2015
Saurabh - try the following
x=[0 0 0;5 8 0; 7 6 0];
cell2mat(arrayfun(@(u)x + repmat([u*5 u*9 u*8],3,1),0:9, 'UniformOutput',false)')
5, 9, and 8 are added to columns 1, 2, and 3 respectively of each new matrix. We use repmat to create the matrix to add to x at each iteration u. We apply this add using arrayfun for iterations 0 through 9. The result is a cell array, so we transpose the output from arrayfun and convert to a matrix using cell2mat.
Plus de réponses (1)
Andrei Bobrov
le 7 Fév 2015
Modifié(e) : Andrei Bobrov
le 7 Fév 2015
k = 0:72/8;
M = repmat(x,numel(k),1);
add = 8*ones(size(x,1),1)*k;
M(:,3) = add(:);
with bsxfun
m = reshape(bsxfun(@plus,permute(x,[1 3 2]),...
bsxfun(@times,reshape([0 0 1],1,1,[]),0:72/8)),[],3);
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!