how can i get a horzenital vector?
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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 件のコメント
採用された回答
Andrei Bobrov
2011 年 9 月 14 日
X1 = reshape(X',1,[])
2 件のコメント
Jan
2011 年 9 月 14 日
@Fangjun: Without the transpose, you get X1 = [2,5,1,...], but Amal wants [2,1,5,...] - the data in row order.
その他の回答 (2 件)
Daniel Shub
2011 年 9 月 14 日
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 件のコメント
Sean de Wolski
2011 年 9 月 14 日
There is not. Though there was a bug with ctranspose in the original release of R2009b.
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