Logical indexing a 2D array into a 2D array
I am having trouble finding a way to remove all rows and columns that are all zeros from a 2D matrix to get a resulting 2D matrix using logical indexing.
Z = [56,0,0,0,0,55;
0,0,0,0,0,0;
27,0,0,0,0,0;
0,0,0,0,0,0;
0,0,0,0,0,0;
100,0,0,0,0,25];
zidx = Z ~= 0;
Z2 = Z(zidx);
results in column vector of:
Z2 = [56;27;100;55;25]
However I really need it to be:
Z2 = [56,55; 27,0; 100,25];
Any help is greatly appreciated.
Accepted Answer
More Answers (2)
2 votes
So, to clarify, you want to have your columns turned into rows?
Because you are looking to roughly keep the row and column locations I would suggest using a for loop to keep the non-zero values from each row. Before that though, because you have different numbers of values you're going to need to either initialize the output array as the max size in zeroes, or to pad as you add new rows.
Z2 = []; for k = 1:size(Z,1); % Easier to do row wise because of needing to pad rows, rather than columns. row = Z(k,Z(k,:)~=0); if isempty(row); elseif isempty(Z2); Z2 = row; elseif size(Z2,2)<length(row); % Check if previous row was smaller than new row Z2(:,size(Z2,2)+1:length(row)) = 0; Z2 = vertcat(Z2,row); elseif size(Z2,2)>length(row); % Check if previous row was bigger than new row row(length(row)+1:size(Z2,2)) = 0; Z2 = vertcat(Z2,row); else Z2 = vertcat(Z2,row); end end
Z2 = Z2';
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