Sum selected columns of a matrix

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

Answer 1

[Karim]

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

[Bruno Luong]

currData = A(:, unique( C(idx) ) )

I think unique statement is not needed.

Answer 2

[Bruno Luong]

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