matrix of circulating vectors without loop

hi all , I want to built a matrix of circulating vectors without using loops

 for example

input:

     1     2     3     4
     4     5     6     7
     3     8     4     6

output :

     1     2     3     4
     4     1     2     3
     3     4     1     2
     2     3     4     1
     4     5     6     7
     7     4     5     6
     6     7     4     5
     5     6     7     4
     3     8     4     6
     6     3     8     4
     4     6     3     8
     8     4     6     3

I use the code

v =[1,2,3,4;4,5,6,7;3,8,4,6]

%notice the length of each vector in v is equal 

len_rows=length(v(:,1)); 

   len_cols=length(v(1,:)); 

   mat_shifted=zeros(len_rows*len_cols,len_cols);

   j=1;

   for i=1:len_rows

       vec=v(i,:);

mat=toeplitz([vec(1) fliplr(vec(2:end))], vec)

mat_shifted(j:j+len_cols-1,:)=mat

j=j+len_cols;

   end

is there a smart way to do it without using loops , thanks in advance

Answer 1

[Honglei Chen]

Not necessarily better in performance, but if you really want to avoid explicitly calling a loop, try

y = arrayfun(@(r)gallery('circul',x(r,:)),1:size(x,1),'UniformOutput',false);
cat(1,y{:})

HTH

[fatema hamodi]

hi , thanks :) , I want to avoid using loops since I want to work with huge matrix so I want matlab to use all the possible processors of the computer , so for my purpose do you think that the way you suggested would be better ?

[Stephen23]

"do you think that the way you suggested would be better ?"

Using arrayfun will be slower than using a for loop. The main advantage of arrayfun / cellfun is more compact code. For a large array you probably would not want to swap a fast loop for slow arrayfun.

[Honglei Chen]

Here is another way which should be more efficient and without explicit loop usage. But overall you want to find a tradeoff between time and memory when you deal with large data set.

x = [1 2 3 4;4 5 6 7;3 8 4 6];
[M,N] = size(x);
r = kron((1:M)',ones(N));
c = repmat(gallery('circul',1:N),M,1);
x(sub2ind([M,N],r,c))

HTH

[fatema hamodi]

thanks a lot !!

Answer 2

[Stephen23]

I think your concept is quite reasonable You could simplify the code a bit by using a cell array for the output, rather than keeping track of the rows:

M = [1,2,3,4;4,5,6,7;3,8,4,6];
N = size(M,1);
C = cell(N,1);
for k = 1:N
    V = M(k,:);
    C{k} = toeplitz([V(1),V(end:-1:2)],V);
end
Z = vertcat(C{:})

[fatema hamodi]

thanks!

Answer 3

[Carlos Sanchis]

Still with a for loop, but somewhat more readable...

input = [1 2 3 4; ...
         4 5 6 7; ...
         3 8 4 6];

[m, n] = size(input);
output = zeros(m * n, n);  
for s = 0:n-1
  output((0:m-1) * n + s + 1, :) = circshift(input, s, 2);
end
   
expected = [1 2 3 4; 4 1 2 3; 3 4 1 2; 2 3 4 1; ...
            4 5 6 7; 7 4 5 6; 6 7 4 5; 5 6 7 4; ...
            3 8 4 6; 6 3 8 4; 4 6 3 8; 8 4 6 3];
assert(isequal(output, expected));

[fatema hamodi]

thanks for your suggestion