How to replace elements in part of a matrix that meet a certain criteria?

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Sobhan 2024-3-21

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评论： Sobhan 2024-3-21

I want to replace elements in certain rows and columns with 0, if they are greater or less than 0. As I am using a sparse matrix (that's 10 000 x 10 000) I don't think I can just set those rows equal to zero, so I am specifically looking for all nonzero elements and replacing them. I can do this row by row in a for loop over i: A( rows(i) , abs(A(rows(i),:))>0 )=0, if rows is a 1D array with row indices. I was wondering if there is a way to do this for the whole matrix without looping? I tried A( rows , abs(A(rows,:))>0 )=0, but that doesn't work as my logical array is 2D.

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Answer 1

Hassaan 2024-3-21

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https://ww2.mathworks.cn/matlabcentral/answers/2097416-how-to-replace-elements-in-part-of-a-matrix-that-meet-a-certain-criteria#answer_1429171

编辑：Hassaan 2024-3-21

在 MATLAB Online 中打开

% Assuming A is your sparse matrix and 'rows' is the 1D array of row indices you're interested in
[rowIdx, colIdx, values] = find(A);
% Initialize a logical vector to mark elements to change
toChange = false(size(values));
% Loop through the specific rows you're interested in
for i = 1:length(rows)
    % Find indices in rowIdx that match the current row of interest
    inRow = rowIdx == rows(i);
    % Mark these for change (you can adjust this condition as needed)
    toChange(inRow) = abs(values(inRow)) > 0;
end
% Now, toChange contains true for elements to be set to zero
% Replace values to be changed with 0
values(toChange) = 0;
% Create the new sparse matrix
A_new = sparse(rowIdx, colIdx, values, size(A,1), size(A,2));

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Sobhan 2024-3-21

This is similar to what I did, but my question was more specificly if there is a way to do it without looping.

My original solution was:

bottomD = [3,5,7]; %example

K = rand(10,10); %example (although it's sparse)

for i = 1:length(bottomD)

K(bottomD(i),abs(K(bottomD(i),:))>0)=0;

end

I figured out that K(bottomD,:)=sparse(length(bottomD),10) works as I don't care about the values in those rows, so I can just replace the entire rows.

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