Sum N many of the same array, which are each offset by an integer M

16 ビュー (過去 30 日間)
Patrick Bevington
Patrick Bevington 2019 年 11 月 27 日
編集済み: Turlough Hughes 2019 年 11 月 27 日
I have an array A = [1; 2; 3; 2; 1] and I would like to make an new array B, which is equal to the sum of N many of A, where array N+1 is off set by M many interfers from array N.
Say N = 3 and M = 1, the new array would be B = [1+0+0; 2+1+0; 3+2+1, 2+3+2; 1+2+3; 0+1+2; 0+0+1].
I need A to be an array of any size and I would like to have control over the size of N and M. A will always be 1D in this case
I have attched a picutre of the problem excel. In one case M = 1 and the other M = 2. In both cases N = 3.
Offset arrays and sum together.png
Thanks for any help you can offer

採用された回答

David Hill
David Hill 2019 年 11 月 27 日
function B = movSum(A,n,m)
B=A;
for k=2:n
B=[B;zeros(m,1)]+[zeros((k-1)*m,1);A];
end
end

その他の回答 (1 件)

Turlough Hughes
Turlough Hughes 2019 年 11 月 27 日
編集済み: Turlough Hughes 2019 年 11 月 27 日
The following should do the job:
N = 3; M = 2; % Inputs
A_t = [1:N N-1:-1:1]'; %single vector 1,2,3,...,N,...,3,2,1
A = zeros((N-1)*M+length(A_t),N); % Preallocate memory for A with zeros
for c = 1:N
A(c*M-1:c*M-2+length(A_t),c) = A_t; % Insert A_t and offset by additional M rows each time
end
The results are then:
A
B=sum(A,2)

カテゴリ

Help Center および File ExchangeMatrix Indexing についてさらに検索

タグ

製品


リリース

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by