Extract sub array from d-dimensional array given indices for each dimension
13 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Michael Van de Graaff
le 20 Mai 2022
Modifié(e) : Michael Van de Graaff
le 21 Mai 2022
Suppose I have a d-dimensional array A which is size d_1 by d_2 by....by d_d and for each dimension i have a index vector vd_i such that vd_i(1)>=1 and vd_i(end)<= size(A, i). How do I extract the sub array A(vd_1,vd_2,...,vd_3) when d is arbitrary?
Edit: as mentioned by the wonderful Matt J, I meant that 1<= vd_i(j)<= size(A,i) for all i,j.
2 commentaires
Matt J
le 20 Mai 2022
Modifié(e) : Matt J
le 21 Mai 2022
and for each dimension i have a index vector vd_i such that vd_i(1)>=1 and vd_i(end)<= size(A, i).
I think what you really mean is 1<= vd_i(j)<= size(A,i) for all i,j. If that's not what you meant then, well, you should check that it's true, because it's a requirement in any subscript indexing operation.
Réponse acceptée
Plus de réponses (2)
Voss
le 20 Mai 2022
% random 4-D (4x8x5x10) array A:
d_siz = [4 8 5 10];
A = randn(d_siz);
size(A)
% make the index vectors for each dimension
% into a cell array:
vd = {[2 3] [1 3 5 7] [2 3 4] 4:8};
% index into A in each dimension:
A_subs = A(vd{:});
size(A_subs)
2 commentaires
dpb
le 20 Mai 2022
Did you try the obvious???
Smallish example, but works in general...
>> d=3;A=reshape(1:4^d,4,4,[]) % make up a sample array to play with that can read
>> A
A(:,:,1) =
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16
A(:,:,2) =
17 21 25 29
18 22 26 30
19 23 27 31
20 24 28 32
A(:,:,3) =
33 37 41 45
34 38 42 46
35 39 43 47
36 40 44 48
A(:,:,4) =
49 53 57 61
50 54 58 62
51 55 59 63
52 56 60 64
>>
>> v1=2:3;v2=v1;v3=v1; % pick middle two of each dimension
>> A(v1,v2,v3)
ans(:,:,1) =
22 26
23 27
ans(:,:,2) =
38 42
39 43
>>
is, by inspection, the array elements of 2nd, 3rd plane and the central four elements of each, respectively.
0 commentaires
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!