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Effacer les filtres

Select dominated columns in a large matrix

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valentino dardanoni
valentino dardanoni le 18 Déc 2023
Commenté : Image Analyst le 19 Déc 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 commentaires
Matt J
Matt J le 18 Déc 2023
Modifié(e) : Matt J le 18 Déc 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?
valentino dardanoni
valentino dardanoni le 18 Déc 2023
Unfortunately it is not sparse, but I have 128 GB of memory

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Image Analyst
Image Analyst le 18 Déc 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 commentaires
valentino dardanoni
valentino dardanoni le 19 Déc 2023
Thank you very much Image Analiyst... Yes, it works perfectly, I run it and timed it, now I am timing my naive method, I am acceptingbyour answer and I will let you know later the psedd inv=crease
Image Analyst
Image Analyst le 19 Déc 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.

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