How can I repeat a 2-D array to create a 3-D array?

27 visualizzazioni (ultimi 30 giorni)
A is a 2x3 array. I want to create a 3-D 'stack' so that each layer of the 3-D stack is identical to A. I've found that the following code gives the desired result:
A =[1 2 3;4 5 6];
for j=1:5
B(j,:,:)=A;
end
disp (squeeze(B(3,:,:))) % an example showing that any layer of the 3-D array is the same as A
Is there a more elegant way to do this? I tried using repmat but couldn't get the same result.
  5 Commenti
Steve Francis
Steve Francis il 13 Lug 2022
Thanks again, James. You're right. Damn, I wish I'd thought about this earlier. I will spend some time seeing if I can re-write my scripts
Stephen23
Stephen23 il 13 Lug 2022
"That way the 2D slices are contiguous in memory..."
which also means that James Tursa's recommended approach will be more efficient (assuming that you mostly want to access those matrices).

Accedi per commentare.

Risposta accettata

Bruno Luong
Bruno Luong il 12 Lug 2022
A =[1 2 3;4 5 6];
B = repmat(reshape(A, [1 size(A)]),[5 1 1])
B =
B(:,:,1) = 1 4 1 4 1 4 1 4 1 4 B(:,:,2) = 2 5 2 5 2 5 2 5 2 5 B(:,:,3) = 3 6 3 6 3 6 3 6 3 6
  2 Commenti
Steve Francis
Steve Francis il 13 Lug 2022
Modificato: Steve Francis il 13 Lug 2022
Thanks very much!
Bruno Luong
Bruno Luong il 13 Lug 2022
Modificato: Bruno Luong il 13 Lug 2022
EDIT : my deleted comment moves here
If you do repeat 3rd dimension few other approaches
A =[1 2 3;4 5 6]; N = 2;
B = repmat(A,[1,1,N]), % James's comment
B =
B(:,:,1) = 1 2 3 4 5 6 B(:,:,2) = 1 2 3 4 5 6
B = repelem(A,1,1,N),
B =
B(:,:,1) = 1 2 3 4 5 6 B(:,:,2) = 1 2 3 4 5 6
B = A(:,:,ones(1,N)),
B =
B(:,:,1) = 1 2 3 4 5 6 B(:,:,2) = 1 2 3 4 5 6
B = A + zeros(1,1,N),
B =
B(:,:,1) = 1 2 3 4 5 6 B(:,:,2) = 1 2 3 4 5 6

Accedi per commentare.

Più risposte (1)

Voss
Voss il 12 Lug 2022
% what you have now:
A =[1 2 3;4 5 6];
for j=1:5
B(j,:,:)=A;
end
% another way, using repmat and permute:
B_new = repmat(permute(A,[3 1 2]),5,1);
% the result is the same:
isequal(B_new,B)
ans = logical
1

Categorie

Scopri di più su Structures in Help Center e File Exchange

Tag

Prodotti


Release

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by