circshift using index values

Question

I was looking for a way to circshift using index values.

I know I can shift all values using the circshift command see below

a=[1:9]
b=circshift(a,[0,1])

But how do I shift every 3rd value over 1? example: Note: variable a can be any length

a=[1,2,30,4,5,60,7,8,90] %note variable `a` could be any length

I'm trying to get b to be

b=[1,30,2,4,60,5,7,90,8]  % so the values in the index 3,6 and 9 are shifted over 1.

Answer 1

You're not going to be able to do this with the standard usage of circshift. There are a couple of other ways that you could approach this. Here are just a few.

Using mod to create index values

You could use mod to subtract 1 from the index values at locations 3:3:end and add 1 to the index values at locations 2:3:end.

b = a((1:numel(a)) + mod(1:numel(a), 3) - 1);

Explanation

Calling mod 3 on 1:numel(a) yields the following sequence

mod(1:numel(a), 3)
%  1     2     0     1     2     0     1     2     0

If we subtract 1 from this sequence we get the "shift" for a given index

mod(1:numel(a), 3) - 1
%   0     1    -1     0     1    -1     0     1    -1

Then we can add this shift to the original index

(1:numel(a)) + mod(1:numel(a), 3) - 1
%   1     3     2     4     6     5     7     9     8

And then assign the values in a to these positions in b.

b = a((1:numel(a)) + mod(1:numel(a), 3) - 1);
%   1    30     2     4    60     5     7    90     8

Using reshape.

Another option is to reshape your data into a 3 x N array and flip the 2nd and 3rd rows, then reshape back to the original size. This option will only work if numel(a) is divisible by 3.

tmp = reshape(a, 3, []);

% Grab the 2nd and 3rd rows in reverse order to flip them
b = reshape(tmp([1 3 2],:), size(a));