Finding unique slices of a multidimensional array

12 vues (au cours des 30 derniers jours)
Jake Martin
Jake Martin le 20 Nov 2020
Commenté : the cyclist le 20 Nov 2020
I'm trying to find unique slices of a multi-dimensional array. For example, if I have something like
A = repmat(magic(4),1,1,3);
A(:,:,2) = A(:,:,2) - 1
A(:,:,1) =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
A(:,:,2) =
15 1 2 12
4 10 9 7
8 6 5 11
3 13 14 0
A(:,:,3) =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
I would like to be able to call something (with the dimension along which I'm slicing specified) and get
[C, ia, ic] = highdim_unique(A, 3)
C(:,:,1) =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
C(:,:,2) =
15 1 2 12
4 10 9 7
8 6 5 11
3 13 14 0
ia =
1
2
ic =
1
2
1
Any chance that something like this exists or has a straightforward solution? Ideally I would want to do this for arbitrariliy many dimensions, but doing it in 3 would be fine.
Thanks,
Jake

Réponse acceptée

the cyclist
the cyclist le 20 Nov 2020
% Define a 3-d array of pretend data, that has a couple duplicate slices
% (Use your data instead)
rng default
A = rand(2,2,4) > 0.5;
% Find the size of A, for generality
[s1,s2,s3] = size(A);
% Reshape the slices into rows (by working down column-wise, and then across columns)
A_r = reshape(A,s1*s2,s3,1)';
% Find the unique rows (which corresponds to unique slices of original array)
A_ru = unique(A_r,'rows','stable');
% Reshape the rows back to 3-d
A_u = reshape(A_ru',s1,s2,[]);
  2 commentaires
Jake Martin
Jake Martin le 20 Nov 2020
Works great, thanks!
the cyclist
the cyclist le 20 Nov 2020
Excellent.
I didn't really think about how to do it for N-dimensional array for N>3, but I see no reason why you could not do the same "reshape into rows" method.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Creating and Concatenating Matrices dans Help Center et File Exchange

Produits


Version

R2020a

Community Treasure Hunt

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

Start Hunting!

Translated by