Replicate vector and shift each copy by 1 row down

Question

I would like replicate a vector N times to create a matrix with each copy shifted 1 row down. See image (first column is the vector 1 to 5). It would be great if this can be achieved without using for loop.

So far was able to to do this repmat(my_vector, 1, 5) to create an N x 5 matrix.

Answer 1

Solution Code

This seems to be a fast approach based on repmat and bsxfun as the benchmarks listed in the next section might convince us -

%// Concatenate one zero at the end of a column vector version of the input vector.
%// Then, replicate the whole vector along columns to have a 2D matrix. 
%// Then "progressively" set elements from each column as zeros corresponding 
%// to the starting zeros of the desired output.
val = repmat([A(:);0],1,N).*bsxfun(@le,[1:N+1]',N:-1:1);  %//'

%// Chop-off at N x N length and reshape to have the final output
out = reshape(val(1:N*N),N,N);

Benchmarking

In this section we will cover runtime benchmarking for the various approaches listed on this page for the stated problem.

Benchmarking Code -

%datasizes = [10 20 50 70 100 200 500 700 1000]; %// Set -1
datasizes = [1000 2000 5000 7000 10000];         %// Set -2
fcns = {'repvecshiftdown_flipud_hankel','repvecshiftdown_toeplitz',...
'repvecshiftdown_repmat_bsxfun','repvecshiftdown_tril_toeplitz'};%//approaches
tsec = zeros(numel(fcns),numel(datasizes));

for k1 = 1:numel(datasizes),
    A = randi(9,1,datasizes(k1)); %// Creare random input vector
    for k2 = 1:numel(fcns), %// Time approaches  
        tsec(k2,k1) = timeit(@() feval(fcns{k2}, A), 1);
        fprintf('\tFunction: %s (%3.2f sec)\n',fcns{k2},tsec(k2,k1));
    end
end

figure;  %% Plot Runtimes
plot(datasizes,tsec(1,:),'-rx'), hold on
plot(datasizes,tsec(2,:),'-bo')
plot(datasizes,tsec(3,:),'-k+')
plot(datasizes,tsec(4,:),'-g.')
set(gca,'xgrid','on'),set(gca,'ygrid','on'),
xlabel('Datasize (# elements)'), ylabel('Runtime (s)')
legend(upper(strrep(fcns,'_',' '))),title('Runtime')

Associated function codes (all approaches) -

function out = repvecshiftdown_repmat_bsxfun(A)
N = numel(A);
val = repmat([A(:);0],1,N).*bsxfun(@le,[1:N+1]',[N:-1:1]); %//'
out = reshape(val(1:N*N),N,N);
return;

function out = repvecshiftdown_tril_toeplitz(A)
out = tril( toeplitz(A) );
return;

function out = repvecshiftdown_toeplitz(A)
out = toeplitz(A, zeros(size(A)));
return;

function out = repvecshiftdown_flipud_hankel(A)
out = flipud( hankel( flipud(A(:)) ) );
return;

Runtime plots -

Set #1 [From 10 till 1000 datasizes]:

Set #2 [From 1000 till 10000 datasizes]:

Answer 2

You can do it with toeplitz and tril;

a = [1 2 3 4 5]
out = tril( toeplitz(a) )

or

out = toeplitz(a, a*0)
%// out = toeplitz(a, zeros(size(a)) )  %// for large arrays

or if you don't mind some happy flipping:

out = flipud( hankel( flipud(a(:)) ) )