Summing of non-zero elements of columns using indexing

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D_coder
D_coder el 16 de Sept. de 2018
Comentada: D_coder el 16 de Sept. de 2018
I have idx, is there a way of using idx to make the sum(matrix,1) in the below code cheaper. In another words, is it possible to use idx for calculating sum of 'matrix'(see below code) along the columns to get a row vector.
matrix(matrix<thresh) = 0 ;
idx = find(matrix~=0);
while condition is true
matrix(idx) = matrix(idx).*another matrix(idx) %matrix is sparse and has lot of zeros ;numel(matrix) = 1000000
new_matrix(a,:) = b*sum(matrix,1) %here b is an integer, make this line efficient and faster so it takes only non zero elements for summing
end

Respuestas (2)

Walter Roberson
Walter Roberson el 16 de Sept. de 2018
Yes, it is possible to use idx to reduce the number of additions.
matrix(matrix<thresh) = 0 ;
idx = find(matrix~=0);
[idxr, idxc] = ind2sub(size(matrix), idx);
numcol = size(matrix,2);
while condition is true
matrix(idx) = matrix(idx).*another matrix(idx) %matrix is sparse and has lot of zeros ;numel(matrix) = 1000000
for col = 1 : num2col
matched_rows = idxr(idxc==col);
new_matrix(a,col) = b * sum(matrix(matched_rows, col));
end
end
I would be surprised if this is more efficient, but it is possible . It would probably be more efficient to loop calculating the column-wise indices into a cell array and indexing into that.
  1 comentario
D_coder
D_coder el 16 de Sept. de 2018
Editada: D_coder el 16 de Sept. de 2018
do you have any alternative code without use of for loop coz the while loop runs 9000 times so thats gonna consume a lot of time.accumarray()?

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Akira Agata
Akira Agata el 16 de Sept. de 2018
If your matrix contains many zeros, I would recommend converting it to sparse matrix and applying sum function, like the following. It runs faster.
matrix = sparse(matrix); % Convert to sparse matrix
summedValue = sum(matrix,1);
  3 comentarios
Walter Roberson
Walter Roberson el 16 de Sept. de 2018
But you would convert to sparse outside the loop.
D_coder
D_coder el 16 de Sept. de 2018
No then my element by element multiplication would suffer

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