Sum of previous elements in a matrix(dynamic way)

Hi, let's say I have the following vector matrix: A=[1 2 3; 4 5 6;7 8 9;10 11 12];

I'm looking for a dynamic way to sum the previous elements(Matrix B the results). I would like to have the variable L to do that.

Example 1: L=1 means I won't to sum all element with their previous, wich give the following result:

B=[5 7 9;11 13 15;17 19 21]

Example 2: L=2 means I won't to sum all element with their 2 previous elements, wich give the following result:

B=[12 15 18;21 24 27]

L need to be dynamic.

A concrete example would be appreciate a an awnser.

Thank you in adavance,

Answer 1

[Matt Fig]

Say your matrix is like:

A = [1 2 3; 
     4 5 6;
     7 8 9;
     10 11 12;
     13 14 15;
     16 17 18];  

% Now find the solution.
L = 3;  % This is dynamically set on 1<L<=M.
[M,N] = size(A); 
B = zeros(M-L+1,N);

for ii = 1:M-L+1
    B(ii,:) = sum(A(ii:ii+L-1,:));
end

.

.

.

.

EDIT

Note that this is exactly what the FILTER function can do:

   B = filter(ones(1,L),1,A);
   B = B(L:end,:)

[Matt Fig]

We could even make a little function to do this:

function B = spec_sum(A,L)
% Help 
% check A is m-by-n,L is in range, etc.
[M,N] = size(A); 
B = zeros(M-L+1,N);

for ii = 1:M-L+1
    B(ii,:) = sum(A(ii:ii+L-1,:));
end

[Matt Fig]

Or, a simpler function based on my edit above:

function B = spec_fun(A,L);
% Help 
% check A is m-by-n,L is in range, etc.
B = filter(ones(1,L),1,A);
B = B(L:end,:);

[Gimpy]

THANK YOU!

Answer 2

[Azzi Abdelmalek]

A=[1 2 3; 4 5 6;7 8 9;10 11 12];
out=A(1:end-1,:)+A(2:end,:)

Answer 3

[Walter Roberson]

T = cumsum(A);
T(L+1:end,:) - T(1:end-L)