Sum of previous elements in a matrix(dynamic way)

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Gimpy
Gimpy 2012-11-27
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,

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采纳的回答

Matt Fig
Matt Fig 2012-11-27
编辑：Matt Fig 2012-11-27
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,:)
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Matt Fig
Matt Fig 2012-11-27
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,:);

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更多回答（2 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-11-27
编辑：Azzi Abdelmalek 2012-11-27
A=[1 2 3; 4 5 6;7 8 9;10 11 12];
out=A(1:end-1,:)+A(2:end,:)
Walter Roberson
Walter Roberson 2012-11-27
编辑：Walter Roberson 2012-11-27
T = cumsum(A);
T(L+1:end,:) - T(1:end-L)

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