Select dominated columns in a large matrix

2 visualizaciones (últimos 30 días)
valentino dardanoni
valentino dardanoni el 18 de Dic. de 2023
Comentada: Image Analyst el 19 de Dic. de 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 comentarios
Matt J
Matt J el 18 de Dic. de 2023
Editada: Matt J el 18 de Dic. de 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 el 18 de Dic. de 2023
Unfortunately it is not sparse, but I have 128 GB of memory

Iniciar sesión para comentar.

Respuesta aceptada

Image Analyst
Image Analyst el 18 de Dic. de 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 comentarios
valentino dardanoni
valentino dardanoni el 19 de Dic. de 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 el 19 de Dic. de 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.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by