# How to permute a row vector based on sepecfic method for permutation?

8 views (last 30 days)
Sarah A on 14 Sep 2018
Commented: Sarah A on 15 Sep 2018
Hello, suppose we have the matrix x,y,and z that is 1*28 dimension, I want to permute each one of them in this way:
x= [x(1,14),x(1,1), x(1,2), x(1,3), x(1,4), x(1,5), x(1,6),x(1,7), x(1,8),...
x(1,9), x(1,10), x(1,11), x(1,12), x(1,13), x(1,14),x(1,1), x(1,28), x(1,15),...
x(1,16), x(1,17), x(1,18), x(1,19), x(1,20), x(1,21), x(1,22), x(1,23), x(1,24), ...
x(1,25), x(1,26), x(1,27),x(1,28), x(1,15)];
So how can I do that also for y and z without repeating these lines for them also?
Regards,

David Goodmanson on 14 Sep 2018
Edited: David Goodmanson on 14 Sep 2018
Hi Sarah,
Since you have more than 28 elements in the final array I would not call it a permutation of elements, strictly speaking. Be that as it may, it makes sense work on an index.
ind = [14 1:14 1 28 15:28 15] % concatenate
xnew = x(1,ind);
ynew = y(1,ind);
znew = z(1,ind);
Since x,y,and z are row vectors and row 1 is the only possibility, you don't have to use the row index. In this case x(14) is the same as x(1,14). It's preferable to use
xnew = x(ind);
ynew = y(ind);
znew = z(ind);

#### 1 Comment

Sarah A on 15 Sep 2018
Great! Thank you T.T you are really great thank you.