Select dominated columns in a large matrix
2 views (last 30 days)
Show older comments
valentino dardanoni
on 18 Dec 2023
Commented: Image Analyst
on 19 Dec 2023
Consider a MxN real valued matrix F with nonnegative elements. I say that a column Fn is dominated if there is another column which has all elements greater than Fn.
A simple way to find the set of dominated columns is
z=zeros(1,size(F,2));
for j=1:size(F,2);
z(j)=max(min((F-F(:,j)))>0);
end
However, I need to do it for very large F (say 10,000 x 500,000). What is a more efficient way to do it?
2 Comments
Matt J
on 18 Dec 2023
Edited: Matt J
on 18 Dec 2023
(say 10,000 x 500,000)
If so, then this would be a sparse matrix?
If not, then you have 37 GB to hold such a matrix in double floats?
And if it is sparse, what is the sparsity? And are the zero-elements to be included in the determination of whether a column is dominated?
Accepted Answer
Image Analyst
on 18 Dec 2023
How about (untested)
[rows, columns] = size(F)
itsDominated = false(1, columns); % Keep track of which columns are dominated.
for col = 1 : columns
% Get one columns to check.
thisColumn = F(:, col);
% Check it against all other columns.
for col2 = 1 : columns
if col2 == col
continue; % Don't check column against itself.
end
fprintf('Checking column %d against column %d.\n', col, col2);
% See if all the values of column2 are greater than the column
% we're checking on.
thisColumn2 = F(:, col2);
isDom = all(thisColumn2 > thisColumn);
if isDom
% It's dominated. Log this fact and then skip on to the next
% reference column.
itsDominated(col) = true;
fprintf(' Column %d dominates column %d.\n', col2, col);
% Break out of the col2 loop.
break; % Don't bother checking any other columns against this one.
end
end
end
4 Comments
Image Analyst
on 19 Dec 2023
OK, thanks. I tought my method was "naive". It's not particularly clever. It's basically just a brute force comparison method. About the only clever things about it might be bailing out after the first dominant row is found, and the use of all().
You could speed it up quite a bit by getting rid of the fprintf() statements.
More Answers (0)
See Also
Categories
Find more on Matrix Indexing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!