Selecting random cells in a matrix without repetition

14 vues (au cours des 30 derniers jours)
Lorson Blair
Lorson Blair le 1 Oct 2021
Commenté : Lorson Blair le 1 Oct 2021
Good day guys. This may be a simple question, but I cannot seem to find an answer on here for it. I am trying to select random cells in a matrix without repeating the cells. For example. I have a 10x10 matrix, and I'm trying to select 20 cells randomly without repeating the cells. Any help will be appreciated.
Thanks.

Réponse acceptée

dpb
dpb le 1 Oct 2021
Modifié(e) : dpb le 1 Oct 2021
N2Choose=20;
R=M(randperm(numel(M),N2Choose));
ERRATUM
Stupid auto-complete put the silly parens in instead of letting me when/where wanted...my story and I'm stickin' to it!!! :)
ADDENDUM
As far as the comment re: positions, just wrap the indices when generate them if those are the desired return values instead of the values.
[r,c]=ind2sub(randperm(numel(M),N2Choose),size(M));
In some ways this is less useful than the linear indices; you'll need arrayfun or similar construct to pull the elements that way as MATLAB will expand the two vectors to all combinations of each instead of just accessing the two vector elements pairwise.
  4 commentaires
Kelly Kearney
Kelly Kearney le 1 Oct 2021
One parentheses was out of place on dpb's example:
M = rand(10);
N2Choose=20;
idx = randperm(numel(M), N2Choose) % the indices
idx = 1×20
31 46 11 16 79 30 40 90 3 36 47 6 81 94 96 66 2 89 83 42
[r,c] = ind2sub(size(M), idx) % if you need row/column
r = 1×20
1 6 1 6 9 10 10 10 3 6 7 6 1 4 6 6 2 9 3 2
c = 1×20
4 5 2 2 8 3 4 9 1 4 5 1 9 10 10 7 1 9 9 5
R=M(idx) % the values
R = 1×20
0.0584 0.3064 0.4200 0.4720 0.8760 0.4465 0.7343 0.7369 0.6275 0.2472 0.9182 0.7630 0.1024 0.7336 0.3009 0.4893 0.5159 0.3713 0.5391 0.0316
Lorson Blair
Lorson Blair le 1 Oct 2021
Thanks for this guys. Really helpful.

Connectez-vous pour commenter.

Plus de réponses (0)

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!

Translated by