Simple reformat data question

I have

a=[1:16]

and I need to get

b=    1    2   4   3
      5    6   8   7
     13   14   16  15
      9   10   12  11

 have a number of such vectors that I want to convert into this format..

Answer 1

[Thomas]

There might be a shorter way of doing this but this is a start..

a=[1:16]
b=reshape(a,4,[])'
c=b(:,3)
b(:,3)=[]
b=[b c]
d=b(3,:)
b(3,:)=[]
b=[b;d]

[Rahul]

trying your solution.. thanks..

Answer 2

[Sean de Wolski]

If you can explain what's going on we might be able to help you with a more general solution. Otherwise, there doesn't appear to be a pattern and the solution is thus:

b = a([1 2 4 3; 5 6 8 7; 13 14 16 15;9 10 12 11])

More per comments

A=reshape(1:16,4,4)'
B = A;
B([end-1 end],:) = B([end end-1],:);
B(:,[end-1 end]) = B(:,[end end-1]);

[Rahul]

I have to exchange the last and second last row and last and second last column. This is ok for the example matrix, but I have a number of such vectors that I need to reformat.. I'm trying the above solution by Thomas..

[Sean de Wolski]

see update

Answer 3

[Matt Tearle]

How much does this need to generalize? If a might be very long, using a transpose is inefficient. Will the number of elements of a be fixed? If not, will the number of elements of a always be a square (so it can be reshaped into an n-by-n matrix)?

If numel(a) can be large, but will always be square:

A = 1:49; % or whatever

n2 = length(A);
n = sqrt(n2);
idx = mod(0:n:(n*n2-1),n2-1)+1;
B = reshape(A(idx),n,n);
B(n,n) = n2;

B([end-1 end],:) = B([end end-1],:);
B(:,[end-1 end]) = B(:,[end end-1]);