rearrangement rows of matrices

1 vue (au cours des 30 derniers jours)
parham kianian
parham kianian le 17 Nov 2020
Commenté : Rik le 2 Fév 2022
Suppose:
A = repmat((1:15)',1,5);
Now I want to rearrange A such that it becomes matrix of size 5 by 15 in following form:
B = [A(1:5,:), A(6:10,:), A(11:15,:)];
The problem is that size of matrix A in my work is too big and its size may change in each iteration. So I can't use a fix command like above to evaluate B. On the other hand, using a for loop is too much time consuming. I tried function "reshape". But, it does not work well.
Is there any function to do this? or how should I call function reshape to get what I want?

Réponse acceptée

Bruno Luong
Bruno Luong le 17 Nov 2020
A = reshape(A,[5 3 5]);
A = permute(A,[1 3 2]);
A = reshape(A,[5 15]);

Plus de réponses (3)

Rik
Rik le 17 Nov 2020
The easiest way is probably to use a cell as an intermediary step:
A = repmat((1:15)',1,5);
B = [A(1:5,:), A(6:10,:), A(11:15,:)];
d=5;%or is this size(A,2)?
C=mat2cell(A,...
d*ones(1,size(A,1)/d),...
size(A,2));
C=C.';
C=cell2mat(C);
isequal(B,C)
ans = logical
1
This would all be much simpler if your example array is actually this repetitive, as you could use repelem.

KSSV
KSSV le 17 Nov 2020
A = repmat((1:15)',1,5);
B = [A(1:5,:), A(6:10,:), A(11:15,:)];
[r,c] = size(A) ;
n = 3 ;
C = permute(reshape(A',[c,r/n,n]),[2,1,3]);
D = reshape(C,5,[]) ;

Andrei Bobrov
Andrei Bobrov le 17 Nov 2020
A = reshape(1:75,15,[]);
a = 5;
m = size(A,1);
B = reshape(permute(reshape(A,a,m/a,[]),[1,3,2]),a,[]);
  2 commentaires
Antonio Carvalho
Antonio Carvalho le 2 Fév 2022
@Andrei Bobrov how are you. Can i reshape two long vectors? one has a lenght completely different to another and from a spefic point (time) compare them using a plot figure?
Rik
Rik le 2 Fév 2022
You can't. If you don't have as many x-values as you have y-values it is not possible to match them up and create a plot. You need to make sure you can make pairs of your x-y-values.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Logical dans Help Center et File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by