Repetition of rows of a matrix maintaining the number of columns

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Iván
Iván 2011 年 12 月 13 日
Hi all.
I also want to create a new matrix from a previous one (A) such all the elements in the first row are copied with N replications and the same should be done for the next row till I reach last row.
The order of new matrix is also N * [rows(A)*cols(A)]; but in my case its number of rows has to be N*rows(A) and the number of columns has to be the same as the previous (A) matrix.
For example, imagine my A matrix is A=[ 1 2
3 4
5 6]
If I replicate N=2 all the rows, my new A_ matrix should be A_=[1 2
1 2
3 4
3 4
5 6
5 6]
Can you help me, please?
Thanks and regards
  1 件のコメント
Sean de Wolski
Sean de Wolski 2011 年 12 月 13 日
+1 Well written question with sample data, operation, result and proof you searched the records (a rarity)!

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回答 (1 件)

Paulo Silva
Paulo Silva 2011 年 12 月 13 日
A=[ 1 2
3 4
5 6]
N=1; %number of repeats
kron(A,ones(N+1,1))
  1 件のコメント
Sean de Wolski
Sean de Wolski 2011 年 12 月 13 日
+1`But I'd recommend for readability purpose keeping N to be the number of repeats as a multiplicative number, i.e. 1 repeat = what you have, two repeats = Ivan's example etc.

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