Effacer les filtres
Effacer les filtres

Sum selected columns of a matrix

5 vues (au cours des 30 derniers jours)
federico nutarelli
federico nutarelli le 30 Août 2022
Hi all,
I have a matrix that looks as follow:
0.0486947932834768 0.590504451038576 0 0.0579865082127525 0.0490083905415712 0.672222222222222 0 0.0948620001376558
0.636363636363636 0 -0.0250828729281768 0.00478066895757889 0 -0.0254237288135594 -0.0299024510443391 0.662162162162162
-0.0480580939947781 0.0438218390804598 0.549848942598187 0 -0.00331564986737398 -0.00459016393442624 0.596590909090909 0
0 -0.0554096692111959 -0.0255076123713734 0.632241813602015 0.608591885441527 0 0.00929780525880808 0.0107530809183949
0 0 0 0 0 0 0 0
(see the picture for a better view of the matrix:)
What I would like to do is to average the columns where a 0 appears, a part from the last row. So for instance in the above case, since in the first column a zero, a part from the last row, appears in position 4, I would like to average this column with column number 6 (where another 0 appears in row 4). The second column should be averaged with the fifth as in both columns a 0 appears in second row and so on.
The expected result should look as follows:
1 0.1076.
2 0.1233
3 0.1463
4 0.1279
where 1,2,3,4 represent the row position of the 0 in the columns averaged.
How can I achieve this result?
Thank you

Réponse acceptée

Karim
Karim le 30 Août 2022
One approach is shown below, there are some comments to explain the steps.
A = [0.0486947932834768 0.590504451038576 0 0.0579865082127525 0.0490083905415712 0.672222222222222 0 0.0948620001376558;
0.636363636363636 0 -0.0250828729281768 0.00478066895757889 0 -0.0254237288135594 -0.0299024510443391 0.662162162162162;
-0.0480580939947781 0.0438218390804598 0.549848942598187 0 -0.00331564986737398 -0.00459016393442624 0.596590909090909 0;
0 -0.0554096692111959 -0.0255076123713734 0.632241813602015 0.608591885441527 0 0.00929780525880808 0.0107530809183949;
0 0 0 0 0 0 0 0];
% first find row (R) and column (C) indexes for the zeros, omit the last row
[R,C] = find( A(1:end-1,:) == 0 );
% determine the number of rows (omit the last one)
numrow = size(A,1) - 1;
% initialize a vector for the results
MyMean = zeros( numrow,1);
% loop over the number of rows
for i = 1:numrow
% first find the indexes for the columns with zeros on the i-th row
idx = R == i;
% extract the data from the matrix using the indexes
currData = A(:, unique( C(idx) ) );
% take the mean of the data
MyMean(i) = mean( currData , 'all');
end
% print the result
MyMean
MyMean = 4×1
0.1075 0.1233 0.1463 0.1279
  1 commentaire
Bruno Luong
Bruno Luong le 30 Août 2022
Déplacé(e) : Bruno Luong le 30 Août 2022
currData = A(:, unique( C(idx) ) )
I think unique statement is not needed.

Connectez-vous pour commenter.

Plus de réponses (1)

Bruno Luong
Bruno Luong le 30 Août 2022
Another approach
A = [0.0486947932834768 0.590504451038576 0 0.0579865082127525 0.0490083905415712 0.672222222222222 0 0.0948620001376558;
0.636363636363636 0 -0.0250828729281768 0.00478066895757889 0 -0.0254237288135594 -0.0299024510443391 0.662162162162162;
-0.0480580939947781 0.0438218390804598 0.549848942598187 0 -0.00331564986737398 -0.00459016393442624 0.596590909090909 0;
0 -0.0554096692111959 -0.0255076123713734 0.632241813602015 0.608591885441527 0 0.00929780525880808 0.0107530809183949;
0 0 0 0 0 0 0 0];
[row,col] = find(A(1:end-1,:)==0);
R = zeros(size(A));
R(:,col) = repmat(row',size(A,1),1);
meancol = accumarray(R(:),A(:),[],@mean)
meancol = 4×1
0.1075 0.1233 0.1463 0.1279

Catégories

En savoir plus sur Logical dans Help Center et File Exchange

Produits


Version

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by