Reversing every second row in a matrix

Hello!

I have a question regarding matrix transformation. The simplified version of what I have is:

1 2 3

4 5 6

7 8 9

10 11 12

etc

What I want is that every second row is reversed, so resulting in:

1 2 3

6 5 4

7 8 9

12 11 10

I have tried the flip-function, but as far as I found out that reverses every row. Any idea how to do this?

Thanks!

Answer 1

[the cyclist]

M = [ 1  2  3;
      4  5  6;
      7  8  9;
     10 11 12];

M(2:2:end,:) = fliplr(M(2:2:end,:))

[Melissa2305]

This was exactly what I was looking for :) Thanks!

Answer 2

[Star Strider]

This works:

M = [1 2 3
    4 5 6
    7 8 9
    10 11 12];

M(2:2:end,:) = M(2:2:end, end:-1:1);

Answer 3

[harjeet singh]

dear melissa use this code

clear all
close all
clc

a=[1 2 3;4 5 6;7 8 9;10 11 12];
for i=1:size(a,1)
    if(mod(i,2)==0)
        b(i,:)=(flipud(a(i,:)'))';
    else
        b(i,:)=a(i,:);
    end
end
b