Hi there
I have a matrix that i need to make twice as long. i.e i need to double the length of columns. eg i need:
[0 1 0 0]
[0 2 0 0]
to become:
[0 1 0 0]
[0 1 0 0]
[0 2 0 0]
[0 2 0 0]
Thanks

1 Kommentar

Stephen23
Stephen23 am 3 Jan. 2022
Bearbeitet: Stephen23 am 5 Jan. 2022
A = [0,1,0,0;0,2,0,0]
A = 2×4
0 1 0 0 0 2 0 0
B = repelem(A,2,1)
B = 4×4
0 1 0 0 0 1 0 0 0 2 0 0 0 2 0 0

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Star Strider
Star Strider am 23 Jan. 2019
Bearbeitet: Star Strider am 23 Jan. 2019

0 Stimmen

Try this:
M = [0 1 0 0
0 2 0 0];
M2 = repmat(M, 1, 2);
M2 = reshape(M2', [], 4)'
producing:
M2 =
0 1 0 0
0 1 0 0
0 2 0 0
0 2 0 0
You could combine both calculations into a one-line calcualtion if you want to.
EDIT —
Generally, for ‘N’ repitions, the code becomes:
N = 4; % Number Of Repititions
Mn = repmat(M, 1, N)
Mn = reshape(Mn', [], 2*N)'
producing:
Mn =
0 1 0 0
0 1 0 0
0 1 0 0
0 1 0 0
0 2 0 0
0 2 0 0
0 2 0 0
0 2 0 0

3 Kommentare

Matthew Allison
Matthew Allison am 3 Jan. 2022
elegant solution.
to make this a truly generic solution, the 2*N should be replaced by m*N, where m is the number of rows being duplicated. The 2 assumes that the original array has only 2 rows.
Stephen23
Stephen23 am 3 Jan. 2022
Bearbeitet: Stephen23 am 3 Jan. 2022
@Matthew Allison if you think that is elegant, just wait until you see REPELEM in action:
M = [0,1,0,0;0,2,0,0]
M = 2×4
0 1 0 0 0 2 0 0
Mn = repelem(M,2,1)
Mn = 4×4
0 1 0 0 0 1 0 0 0 2 0 0 0 2 0 0
REPELEM was introduced in R2015a, it is not clear why StarStrider avoided using it here.
Star Strider
Star Strider am 3 Jan. 2022
Bearbeitet: Star Strider am 4 Jan. 2022
@Matthew Allison — Thank you!
@Stephen — Because I wanted to make it general enough to work with earlier versions as well.

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