More efficient alternative to the find function?
25 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Canberk Suat Gurel
le 8 Mar 2018
Commenté : Walter Roberson
le 9 Mar 2018
Hi all,
I am working on a path planning algorithm. I have generated a matrix A (consists of 1s and 0s.) which stores the neighbor node information. Using the following code I am finding the available neighbors and storing them in vector F.
F = find(A(current,:)==1);
However, unfortunately I found out that this line of code takes the longest time in my code (from 0.004018 to 0.023119s). (A matrix is a 37901 by 37901 matrix.)
Is there a more computationally efficient method to do this?
Thanks!
This is what the A matrix looks like:
0 commentaires
Réponse acceptée
James Tursa
le 9 Mar 2018
It is not clear from your post if A always has the pattern you show, but if it does then why not just this:
F = [current-1,current+1];
F(F<1|F>size(A,2)) = [];
2 commentaires
Walter Roberson
le 9 Mar 2018
locs = current+[-1, 1, -250, 250, -251, 251, -252, 252];
F = locs( logical(A(current, locs)) );
You can rearrange locs however is suitable.
The result in F is a list of indices, in the same order as you used to create locs.
You could improve performance by originally storing A as logical instead of as double.
Have you considered using diag() to extract the data for all of the rows at the same time? With a small bit of padding and some diag() you could build a 37901 x 8 matrix that focused on just the entries that might have useful information.
Plus de réponses (1)
Walter Roberson
le 8 Mar 2018
You can improve a little by using
F = find(A(current,:));
Your A values are only 0 and 1, so comparing to 1 is going to give you exactly the same as A since == returns 0 and 1.
1 commentaire
Voir également
Catégories
En savoir plus sur Creating and Concatenating Matrices 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!