MATLAB Answers


swap columns of a matrix

Asked by Rasmus
on 25 Feb 2014
Latest activity Commented on by Helen Joan on 30 Apr 2019
Hello guys/girls
How do i swap columns??
I have this
k_minus =
-46 -43 -26 -14 7 19 11 32 39 45 45
0 -4 -7 -7 -44 -44 -7 -7 -15 -15 0
and I want the columns to be in opposite order - How do I do this? and can do this in one go?


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3 Answers

Answer by Thomas
on 25 Feb 2014
Edited by Thomas
on 25 Feb 2014
 Accepted Answer

You could also use
out= fliplr(k_minus)

  1 Comment

on 25 Feb 2014
That is actually easier! :)

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Mischa Kim
Answer by Mischa Kim
on 25 Feb 2014
Edited by Mischa Kim
on 25 Feb 2014

k_minus_rev = k_minus(:, [length(k_minus(1,:)):-1:1])


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Mischa Kim
on 25 Feb 2014
Basically, when you do
you extract the sub-matrix consisting of all rows in the first column. With
you extract all rows of the second, and the first columns, in that order. You can now generalize the last bit
to extract all columns ( length(k_minus(1,:)) ), placing the last column first, and the first column last, in descending order ( -1 ).
Jos (10584)
on 25 Feb 2014
You do not need the length statement by using the keyword END:
MyMatrix_withReversedColumns = MyMatrix(:,end:-1:1)
but I do suggest you stick to FLIPLR. It is the same, but much easier to read!
on 2 Apr 2019
Thank you for this answer! It has helped me "merge" two matrices, i.e. creating a matrix consisting of the first column of matrix A, first column of matrix B, second column of matrix A, second column of matrix B, etc.

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Answer by rishabh gupta on 12 Jan 2018

you can also use: k_minus_rev = k_minus(:, [length(k_minus):-1:1])


The square brackets are totally superfluous, and using end is simpler than using length (of course length should be avoided generally because its output changes depending on the input array size: better to use numel or size with a specific dimension).
So a better (simpler, clearer, less buggy) is exactly as Jos already gave four years ago:
Helen Joan on 30 Apr 2019
is it possible to do this with a loop?

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