how can i get a horzenital vector?
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
X = [2 1 5 1
5 4 8 1
1 4 5 7];
if i use X1 = X(:);
the output is X1=[2
5
1
1
4
4
5
8
5
1
1
7]
but i wante the out put is X1 =[2 1 5 1 5 4 8 1 1 4 5 7];
thank you very much
0 Commenti
Risposta accettata
Andrei Bobrov
il 14 Set 2011
X1 = reshape(X',1,[])
2 Commenti
Jan
il 14 Set 2011
@Fangjun: Without the transpose, you get X1 = [2,5,1,...], but Amal wants [2,1,5,...] - the data in row order.
Più risposte (2)
Daniel Shub
il 14 Set 2011
If there is any chance that you matrix will have complex numbers you need to be careful with the difference between ' (ctranspose) and .' (transpose). To be clear I might go with:
X1 = transpose(X(:));
I do not know if there is a performance difference between ctranspose and transpose for real matrices.
1 Commento
Sean de Wolski
il 14 Set 2011
There is not. Though there was a bug with ctranspose in the original release of R2009b.
Vedere anche
Categorie
Scopri di più su Matrix Indexing in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!