Replicate vectors shifting them to the right - matlab

Question

In Matlab, I have two single row (1x249) vectors in a 2x249 matrix and I have to create a matrix A by replicating them many times, each time shifting the vectors of 2 positions to the right. I would like to fill the entries on the left with zeros. Is there a smart way to do this? Currently, I am using a for loop and circshift, and I add at each iteration I add the new row to A, but probably this is highly inefficient.
Code (myMat is the matrix I want to shift):
A = [];
myMat = [1 0 -1 zeros(1,246); 0 2 0 -2 zeros(1,245)];
N = 20;
for i=1:N-1
aux = circshift(myMat,[0,2*(i-1)]);
aux(:,1:2*(i-1)) = 0;
A =[A; aux];
end

Answer 1

As you are probably aware, loops in Matlab are not so efficient.
I know that the Mathworks keep saying this is no longer so with JIT
compilation, but I haven't experienced the fast loops yet.
I put your method for constructiong the matrix A in a function:
function A = replvector1(myMat,shift_right,width,N)
pre_alloc = true; % make implementation faster using pre-allocation yes/no
% Pad myMat with zeros to make it wide enough
myMat(1,width)=0;
% initialize A
if pre_alloc
A = zeros(size(myMat,1)*(N-1),width);
else
A = [];
end
% Fill A
for i=1:N-1
aux = circshift(myMat,[0,shift_right*(i-1)]);
aux(:,1:min(width,shift_right*(i-1))) = 0;
A(size(myMat,1)*(i-1)+1:size(myMat,1)*i,:) =aux;
end
Your matrix-operation looks a lot like a kronecker product, but the
block-matrixces have overlapping column ranges so a direct kronecker product
will not work. Instead, I constructed the following function:
function A = replvector2(myMat,shift_right,width,N)
[i,j,a] = find(myMat);
i = kron(ones(N-1,1),i) + kron([0:N-2]',ones(size(i))) * size(myMat,1);
j = kron(ones(N-1,1),j) + kron([0:N-2]',ones(size(j))) * shift_right;
a = kron(ones(N-1,1),a);
ok = j<=width;
A = full(sparse(i(ok),j(ok),a(ok),(N-1)*size(myMat,1),width));
You can follow the algorithm by removing semicolons and looking at intermediate
results.
The following main program runs your example, and can easily be modified to
run similar examples:
% inputs (you may vary them to see that it always works)
shift_right = 2;
width = 249;
myMat1 = [ 1 0 -1 0 ;
0 2 0 -2 ];
N = 20;
% Run your implementation
tic;
A = replvector1(myMat,shift_right,width,N);
disp(sprintf('\n original implementation took %e sec',toc))
% Run the new implementation
tic;
B = replvector2(myMat,shift_right,width,N);
disp(sprintf(' new implementation took %e sec',toc))
disp(sprintf('\n norm(B-A)=%e\n',norm(B-A)))

Answer 2

I've taken Nathan's code (see his answer to this question), and added another possible implementation (replvector3).
My idea here stems from you not really needing a circular shift. You need to right-shift and add zeros to the left. If you start with a pre-allocated array (this is really where the big wins in time are for you, the rest is peanuts), then you already have the zeros. Now you just need to copy over myMat to the right locations.
These are the times I see (MATLAB R2017a):
OP's, with pre-allocation: 1.1730e-04
Nathan's: 5.1992e-05
Mine: 3.5426e-05
^ shift by one on purpose, to make comparison of times easier
This is the full copy, copy-paste into an M-file and run:
function so
shift_right = 2;
width = 249;
myMat = [ 1 0 -1 0 ;
0 2 0 -2 ];
N = 20;
A = replvector1(myMat,shift_right,width,N);
B = replvector2(myMat,shift_right,width,N);
norm(B(:)-A(:))
C = replvector3(myMat,shift_right,width,N);
norm(C(:)-A(:))
timeit(#()replvector1(myMat,shift_right,width,N))
timeit(#()replvector2(myMat,shift_right,width,N))
timeit(#()replvector3(myMat,shift_right,width,N))
% Original version, modified to pre-allocate
function A = replvector1(myMat,shift_right,width,N)
% Assuming width > shift_right * (N-1) + size(myMat,2)
myMat(1,width) = 0;
M = size(myMat,1);
A = zeros(M*(N-1),width);
for i = 1:N-1
aux = circshift(myMat,[0,shift_right*(i-1)]);
aux(:,1:shift_right*(i-1)) = 0;
A(M*(i-1)+(1:M),:) = aux;
end
% Nathan's version
function A = replvector2(myMat,shift_right,width,N)
[i,j,a] = find(myMat);
i = kron(ones(N-1,1),i) + kron((0:N-2)',ones(size(i))) * size(myMat,1);
j = kron(ones(N-1,1),j) + kron((0:N-2)',ones(size(j))) * shift_right;
a = kron(ones(N-1,1),a);
ok = j<=width;
A = full(sparse(i(ok),j(ok),a(ok),(N-1)*size(myMat,1),width));
% My trivial version with loops
function A = replvector3(myMat,shift_right,width,N)
% Assuming width > shift_right * (N-1) + size(myMat,2)
[M,K] = size(myMat);
A = zeros(M*(N-1),width);
for i = 1:N-1
A(M*(i-1)+(1:M),shift_right*(i-1)+(1:K)) = myMat;
end