How to reshape any matrix using while loop or any other method?
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Sushil Pokharel
am 29 Jun. 2022
Bearbeitet: Sushil Pokharel
am 30 Jun. 2022
Hello there, I have a matrix B of size 432000x120 and I want another matrix A of same size in such a way that:
A(: , 1) = B(: , 1)
A(: , 2) = B(: , 7)
A(: , 3) = B(: , 13) and so on
I have done by using two for loops but I wanted to solve this problem by other method (may be by using while loop or any other efficient method). You help will be greatly appreciated.
0 Kommentare
Akzeptierte Antwort
Voss
am 29 Jun. 2022
Here's one way, based on the assumption that it goes
A(:,[1,2,3,...,20,21,22,...]) = B(:,[1,7,13,...,115,2,8,...])
% 5x120 matrix B:
B = reshape(1:5*120,120,[]).'
n = 6;
% construct A by reordering the columns of B:
A = B(:,(1:n)+(0:n:size(B,2)-1).')
% another way to do the same reordering:
A = B(:,reshape(1:size(B,2),n,[]).')
% check first and last column of each 20-column sequence:
isequal( ...
A(:,[1 20 21 40 41 60 61 80 81 100 101 120]), ...
B(:,[1 115 2 116 3 117 4 118 5 119 6 120]))
Weitere Antworten (1)
Walter Roberson
am 29 Jun. 2022
A = B(:, 1:6:end) ;
5 Kommentare
Voss
am 29 Jun. 2022
Bearbeitet: Voss
am 29 Jun. 2022
I don't think that works correctly when the number of columns is not 6^2:
B = (1:120) + (1:5).'*100;
A = reshape(permute(reshape(B, size(B,1), [], 6), [1 3 2]), size(B,1), [])
I guess it should be:
A = reshape(permute(reshape(B, size(B,1), [], size(B,2)/6), [1 3 2]), size(B,1), [])
Siehe auch
Kategorien
Mehr zu Matrices and Arrays finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!