HOW TO RESHAPE 3D ARRAY WITH DIFFERENT DIRECTIONS?

1 vue (au cours des 30 derniers jours)
Triveni
Triveni le 21 Août 2021
Modifié(e) : DGM le 21 Août 2021
for i = 1:length(j)
NETPLP(:,:,i)= [BPavg1(1,1,i), SPavg1(1,1,i), NETPLP1(1,1,i); BPavg2(1,1,i), SPavg2(1,1,i), NETPLP2(1,1,i); BPavg3(1,1,i), SPavg3(1,1,i), NETPLP3(1,1,i); BPavg4(1,1,i), SPavg4(1,1,i), NETPLP4(1,1,i); BPavg5(1,1,i), SPavg5(1,1,i), NETPLP5(1,1,i); BPavg6(1,1,i), SPavg6(1,1,i), NETPLP6(1,1,i); BPavg7(1,1,i), SPavg7(1,1,i), NETPLP7(1,1,i); BPavg8(1,1,i), SPavg8(1,1,i), NETPLP8(1,1,i)];
end
PL1 = reshape(permute(NETPLP(1,:,:),[1,3,2]), size(NETPLP(1,:,:),1)*size(NETPLP(1,:,:),3), 3);
PL2 = reshape(permute(NETPLP(2,:,:),[1,3,2]), size(NETPLP(2,:,:),1)*size(NETPLP(2,:,:),3), 3);
PL3 = reshape(permute(NETPLP(3,:,:),[1,3,2]), size(NETPLP(3,:,:),1)*size(NETPLP(3,:,:),3), 3);
PL4 = reshape(permute(NETPLP(4,:,:),[1,3,2]), size(NETPLP(4,:,:),1)*size(NETPLP(4,:,:),3), 3);
PL5 = reshape(permute(NETPLP(5,:,:),[1,3,2]), size(NETPLP(5,:,:),1)*size(NETPLP(5,:,:),3), 3);
PL6 = reshape(permute(NETPLP(6,:,:),[1,3,2]), size(NETPLP(6,:,:),1)*size(NETPLP(6,:,:),3), 3);
PL7 = reshape(permute(NETPLP(7,:,:),[1,3,2]), size(NETPLP(7,:,:),1)*size(NETPLP(7,:,:),3), 3);
PL8 = reshape(permute(NETPLP(8,:,:),[1,3,2]), size(NETPLP(8,:,:),1)*size(NETPLP(8,:,:),3), 3);
PL= [PL1;PL2;PL3;PL4;PL5;PL6;PL7;PL8];
LENGTH(J) = 20. IN WHICH I GOT
I WANT VALUES OF PL WITHOUT ASSIGNING THE VARIABLE PL1, PL2, P3, PL4, PL5, PL6, PL7 AND PL8.
PLEASE HELP ME.
  3 commentaires
Triveni
Triveni le 21 Août 2021
Modifié(e) : darova le 21 Août 2021
I have tried this earlier but
PL=NETPLP(1:8,:,:)
PL(:,:,1) =
22 22.05 11.04
22 21.95 -13.96
20.95 21 11.06
21.05 21 -13.94
22 21.95 -13.96
22 20.95 -263.95
22.05 21 -263.95
20.95 21 11.06
I have to rearrange this array. I have to select 1st row from each array then 2nd row to each array and so on. then i have to merge all into single matrix.
DGM
DGM le 21 Août 2021
What are the sizes of BPavg#, SPavg#, NETPLP#?

Connectez-vous pour commenter.

Réponse acceptée

DGM
DGM le 21 Août 2021
Modifié(e) : DGM le 21 Août 2021
This is why using numbered variables instead of arrays is a bad idea.
No need for a loop. Depending on the size of the inputs, just address the arrays accordingly (pick one)
% assuming that BPavgN, SPavgN, NETPLPN are all mxnx20
NETPLP = [BPavg1(1,1,:), SPavg1(1,1,:), NETPLP1(1,1,:);
BPavg2(1,1,:), SPavg2(1,1,:), NETPLP2(1,1,:);
BPavg3(1,1,:), SPavg3(1,1,:), NETPLP3(1,1,:);
BPavg4(1,1,:), SPavg4(1,1,:), NETPLP4(1,1,:);
BPavg5(1,1,:), SPavg5(1,1,:), NETPLP5(1,1,:);
BPavg6(1,1,:), SPavg6(1,1,:), NETPLP6(1,1,:);
BPavg7(1,1,:), SPavg7(1,1,:), NETPLP7(1,1,:);
BPavg8(1,1,:), SPavg8(1,1,:), NETPLP8(1,1,:)];
% if they are 1x1x20
NETPLP = [BPavg1, SPavg1, NETPLP1;
BPavg2, SPavg2, NETPLP2;
BPavg3, SPavg3, NETPLP3;
BPavg4, SPavg4, NETPLP4;
BPavg5, SPavg5, NETPLP5;
BPavg6, SPavg6, NETPLP6;
BPavg7, SPavg7, NETPLP7;
BPavg8, SPavg8, NETPLP8];
The rest seems simple enough
% reshape the array
PL = reshape(permute(NETPLP,[3 1 2]),[],3);

Plus de réponses (0)

Catégories

En savoir plus sur Matrix Indexing dans Help Center et File Exchange

Produits

Community Treasure Hunt

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

Start Hunting!

Translated by