Filter löschen
Filter löschen

How to duplicate rows of a matrix if the number of copies for each row are different?

14 Ansichten (letzte 30 Tage)
Hi all,
I want to duplicate each rows of a matrix according to the numbers given in a vector.
For example, I have matrix A and vector , I want to duplicate the first row of A 2 times, second row by 4 times and third row by 5 times.
Is there any way to do this?
Thank you for your help.

Akzeptierte Antwort

Stephen23
Stephen23 am 9 Nov. 2021
Bearbeitet: Stephen23 am 9 Nov. 2021
out = repelem(A,b,1)
  3 Kommentare
Stephen23
Stephen23 am 9 Nov. 2021
Bearbeitet: Stephen23 am 9 Nov. 2021
Use REPELEM, not REPMAT. Assuming that your matrix A has as many rows as b has elements:
A = randi(9,3,4)
A = 3×4
1 5 2 6 5 9 8 6 2 2 7 4
b = [2,4,5];
out = repelem(A,b,1)
out = 11×4
1 5 2 6 1 5 2 6 5 9 8 6 5 9 8 6 5 9 8 6 5 9 8 6 2 2 7 4 2 2 7 4 2 2 7 4 2 2 7 4
This is much simpler and more efficient than using a WHILE or FOR loop.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

KSSV
KSSV am 9 Nov. 2021
Read about repmat.
b = [2 4 5] ;
A = repmat(b,2,1)
A = 2×3
2 4 5 2 4 5
B = repmat(b,3,1)
B = 3×3
2 4 5 2 4 5 2 4 5

Sulaymon Eshkabilov
Sulaymon Eshkabilov am 9 Nov. 2021
It can be done in a few different ways, e.g.:
A = magic(3)
A = 3×3
8 1 6 3 5 7 4 9 2
b = [2, 4, 5];
AA = [repmat(A(1,:), b(1),1);repmat(A(2,:), b(2),1); repmat(A(3,:), b(3),1)]
AA = 11×3
8 1 6 8 1 6 3 5 7 3 5 7 3 5 7 3 5 7 4 9 2 4 9 2 4 9 2 4 9 2
% Alternative way:
A = magic(3);
b = [2, 4, 5];
AA =[];
for ii = 1:numel(b)
AA = [AA; repmat(A(ii,:), b(ii),1)];
end
AA
AA = 11×3
8 1 6 8 1 6 3 5 7 3 5 7 3 5 7 3 5 7 4 9 2 4 9 2 4 9 2 4 9 2
  2 Kommentare
Hung Dao
Hung Dao am 9 Nov. 2021
Thank you very much.
I am wondering if there is a more efficient way than using a forloop?
Sulaymon Eshkabilov
Sulaymon Eshkabilov am 9 Nov. 2021
I don't believe there is an alternative more efficient way than using [for .. end] or [while .. end] loop.

Melden Sie sich an, um zu kommentieren.

Produkte


Version

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by