How can I shift Right which is an array of numbers?

I need is a new shifted array and eventually I want to add the arrays.(New Shifted array + Original Array). Can I do something like e.g . aa =[11 22 33 44]; bb = aa(2:4); % this is shifting left It gives bb = [22 33 44], but one value is dropped out. If possible can someone provide method to shift right and how to pad the dropped out value appropriately so that the size of the new array remains same as the original array?

Answer 1

[Azzi Abdelmalek]

use circshift function

   circshift(aa,[0 -1])

or

   circshift(aa,[0 1])

[Micky]

Thanks Azzi for the response, but there is no change in the new array

[Azzi Abdelmalek]

aa =[11 22 33 44]
out=circshift(aa,[0 -1])

out and aa are the same?

If it's not what you need then what should be the result for this case?

[Micky]

I am trying something like this:

x= [11;22;33;44;55;66]

x =

    11
    22
    33
    44
    55
    66

>> k=2 % desired number of shifts

k =

     2
>> s = x([end-k+1:end 1:end-k])
s =

    55
    66
    11
    22
    33
    44

Here, it is working fine but I do not want the shifted out values to come circularly back to front of the array, I want to pad the number of shifts by my own desired values like zeros etc. Can you suggest some change?

[Azzi Abdelmalek]

 x= [11;22;33;44;55;66]
circshift(x,[2 0])

[Micky]

This is what I got

x= [11;22;33;44;55;66]

circshift(x,[2 0])

x =

    11
    22
    33
    44
    55
    66

The shifted values are circulated back to the front, and that is what I dont want. ans =

    55
    66
    11
    22
    33
    44

[Azzi Abdelmalek]

n=2
x= [11;22;33;44;55;66]
x=circshift(x,[n 0])
x(1:n)=0  % or what you want

[Micky]

Thanks for the help Azzi.

Answer 2

[Image Analyst]

Not quire sure what you're describing about not dropping off any values and having bb be the same size as aa, but how about this:

aa =[11 22 33 44]; 
bb = zeros(size(aa))
bb(2:4) = aa(2:4)
% or
bb1 = zeros(size(aa))
bb1(1:3) = aa(2:4)

In the command window:

bb =
     0    22    33    44
bb1 =
  22    33    44     0

You can see there is a zero and no value is dropped off or lost and the size of bb is the same as aa. If one of those methods is not what you want, explain in more detail.

[Micky]

Thanks for the response. I am trying something like this:

x= [11;22;33;44;55;66]

x =

    11
    22
    33
    44
    55
    66
>> k=2 % desired number of shifts

k =

     2
>> s = x([end-k+1:end 1:end-k])
s =
    55
    66
    11
    22
    33
    44
Here, it is working fine but I do not want the shifted out values to come circularly back to front of the array, I want to pad the number of shifts by my own desired values like zeros etc. Can you suggest some change?

I am trying out your suggestion as well. May be the above description helps you understand the problem more in detail.

[Image Analyst]

My solution will work. You just have to generalize it to "k" instead of fixed numbers of 2:4 like your example.

x = [11;22;33;44;55;66]
% k = -2;
k = +2;
s = zeros(size(x));
if k >= 1
  s(k+1:end) = x(1:end-k)
elseif k <= -1
  s(1:end+k) = x(-k+1:end)
end

[Micky]

Thanks for the help.

Answer 3

[komal]

Ques= x =[11;22;33;44]

i want this results

x = [0;11;22;33;44]

How can i do it:

[vishwajit jadhav]

x = [0; x]