How to reshape a matrix

I need to reshape a Matrix from an [5 x n] matrix into a matrix of [n,0,0,0,0 ; 0,n,0,0,0 ; 0,0,n,0,0 ; 0,0,0,n,0 ; 0,0,0,0,n]

Example:

given the matrix: A = [1,2,3,4 ; 1,2,3,4 ; 1,2,3,4 ; 1,2,3,4 ; 1,2,3,4]

reshaped into:

1 2 3 4 0 0 0 0

0 1 2 3 4 0 0 0

0 0 1 2 3 4 0 0

0 0 0 1 2 3 4 0

0 0 0 0 1 2 3 4

How is this done in a simple way?

Answer 1

[John D'Errico]

This has nothing to do with what in MATLAB is describd as a reshape, since a reshape cannot change the number of elements in the matrix. The function reshape already exists, and is quite useful.

toeplitz([1;zeros(4,1)],[1:4,zeros(1,4)])
ans =
     1     2     3     4     0     0     0     0
     0     1     2     3     4     0     0     0
     0     0     1     2     3     4     0     0
     0     0     0     1     2     3     4     0
     0     0     0     0     1     2     3     4

toeplitz is a simple way to create that matrix. However, there are many alternatives. You could use diag, or perhaps spdiags. Or, you could create it as a circulant matrix. Lots of ways.

This alternative is a bit of a hack, but perhaps cute with the replicated cumsum:

tril(cumsum(cumsum(eye(5,8),2),2),3)
ans =
     1     2     3     4     0     0     0     0
     0     1     2     3     4     0     0     0
     0     0     1     2     3     4     0     0
     0     0     0     1     2     3     4     0
     0     0     0     0     1     2     3     4

[Joakim Mørk]

Thank you very much John!! you made my day!. i see what you mean with 'reshape' but it was more literally meant than of Matlab vocab :) but thanks ones more :)

Answer 2

[Adam Danz]

I would go with John D'Errico's approach (David Hill 's method is also good) but just to add another approach,

A = [1,2,3,4 ; 1,2,3,4 ; 1,2,3,4 ; 1,2,3,4 ; 1,2,3,4]; 
nPad = 4;   % number of 0s to pad to the end of the 1st row

m = zeros(size(A,1), size(A,2)+nPad); 
colIdx = bsxfun(@(x,y)x+y, 1:size(A,2), (0:size(A,2)).'); 
rowIdx = (1:size(A,1)).' .* ones(1,size(colIdx,2));  % Requires >= matlab r2016b
linIdx = sub2ind(size(m), rowIdx, colIdx); 
m(linIdx(:)) = A; 

m =
     1     2     3     4     0     0     0     0
     0     1     2     3     4     0     0     0
     0     0     1     2     3     4     0     0
     0     0     0     1     2     3     4     0
     0     0     0     0     1     2     3     4

Answer 3

[David Hill]

Lots of way to do it. Here is one:

a=[1 2 3 4 0 0 0 0];
y=a;
for k=1:4
y=[y;circshift(a,k)];
end