Circshift the columns of an array with different shiftsize withou using for loop

As the tittle suggests I am wondering if it is possible to circshift the columns of an array with a different shiftsize in each column without using a for loop.

Example:

a=randi(10,5,4);

I want to do this

a(:,4)=circshift(a(:,4),[-1 0]);

a(:,3)=circshift(a(:,3),[-2 0]);

without a loop. Is it possible?

[Azzi Abdelmalek]

a=randi(10,5,4);  % the size of a is [5 4] , which means a(:,5) does not exist

[Giorgos Papakonstantinou]

oops.. I am sorry

a(:,3)=circshift(a(:,3),[-2 0]);

not..5

[Azzi Abdelmalek]

But it still not clear, when it starts and when it ends

[Giorgos Papakonstantinou]

I want to avoid this loop:

 a=randi(10,5,4)
  shiftindex=[0 1 2 3];

for uu=1:length(shiftindex);
    a(:,uu)=circshift(a(:,uu),[shiftindex(uu) 0]);
end

[Giorgos Papakonstantinou]

Thank you

[Matt J]

In your example, you modify 'a' in-place. If that's really what you're after, for-loops are going to be pretty competitive. Notice that no iteration of your loop ever allocates any additional memory larger than 1 column a(:,uu). I think there are tools on the FEX that can get rid of even that.

Answer 1

[Matt J]

    [m,n]=size(a);
    S=full(sparse(mod(shiftindex,m)+1,1:n,1,m,n));

    a_new=ifft(fft(a).*fft(S),'symmetric')

[Giorgos Papakonstantinou]

Matt thank you for your answer above. Actually it does its work. However I find very difficult to understand its function when you implement the fft and the ifft. But indeed I find it awesome.

p.s. you were missing +1 :-). but ok you made it better!

[Giorgos Papakonstantinou]

Can you explain me why a_new matrix has trailing zeros (even in format short) but a matrix does not? I don't get it.

[Matt J]

a_new does not consist of exact integers. There are floating point residuals due to the fft/ifft operations used to transform a.

[Giorgos Papakonstantinou]

Thank you!

Answer 2

[Andrei Bobrov]

[m,n] = size(a);
b = rem(shiftindex-1,m)+1;
c = rem(bsxfun(@plus,m + 1 - b - m*(b == 0),(0:m-1)')-1,m)+1;
out = a(bsxfun(@plus,c,m*(0:n-1)));

[Matt J]

This solution doesn't use n at all and therefore only seems to make shifted copies of the first column of a:

    a =

         1     6    11    16
         2     7    12    17
         3     8    13    18
         4     9    14    19
         5    10    15    20

    out =

         5     4     3     1
         1     5     4     2
         2     1     5     3
         3     2     1     4
         4     3     2     5

[Matt J]

I think maybe this was the idea

    [m,n] = size(a);

    b=mod(bsxfun(@plus,(0:m-1).',-shiftindex(:).' ),m)+1;
    b=bsxfun(@plus,b,(0:n-1)*m);

    out = a(b)

[Giorgos Papakonstantinou]

Indeed Matt thee solution was making shifted copies of the first column. What you proposed last does the shifting correctly. I have question in your second solution.

What is the purpose of the dots in the second line? When you want to square the elements of a matrix you have to put the dot in order this to do it elementwise. But bsxfun is doing the same operation, I think.

[Matt J]

What is the purpose of the dots in the second line?

Not sure which "dots" you mean. I think you know what the dots in (0:m-1) does. The colon operator produces a vector 0,1,2,..m-1.

Answer 3

[Giorgos Papakonstantinou]

Where is Anrei's answer??? It was here before 2 minutes ago.

Is an answer deleted after accepting a previous one?

I had some question on it that why?

He proposed this:

    [m,n] = size(a);
  out = a(bsxfun(@plus,toeplitz(1:m,[1,m-(0:n-2)]),(0:n-1)*m));

Apart from the fact that its hardcore, my question would be what would be the case if the shiftindex was not linear..

ex.

shiftindex=[2 2 10 5];