Delete rows in a matrix that contain ONLY a negative number
4 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Hello
I've got a really large matrix (say 2000 x 3000) and I want to delete the rows that contain ONLY the negative number -999.
For example, I've got A = [5 3 -999 ; -999 -999 -999; 4 1 7 ; -999 -999 -999]
and I want my new matrix to be A = [5 3 -999 ; 4 1 7]
How can I do that?
Thanks!
0 comentarios
Respuesta aceptada
Más respuestas (3)
Kye Taylor
el 14 de En. de 2013
Editada: Kye Taylor
el 14 de En. de 2013
Try
isNegRow = all( A==-999, 2 );
A(isNegRow,:) = [];
0 comentarios
Shashank Prasanna
el 14 de En. de 2013
Editada: Shashank Prasanna
el 14 de En. de 2013
If you don't want to use loops and keep it simple in a single line, I suggest logical indexing as follows:
A(all(A==-999,2),:)=[]
This should do the trick.
0 comentarios
Amith Kamath
el 14 de En. de 2013
Editada: Amith Kamath
el 14 de En. de 2013
I'm quite positive that this does the trick. There would probably be more optimal ways to do it!
X = 100.*rand(12,5); %for example.
X([2 5],:) = -999; % artificially create rows to remove.
i = 1;
while i <= size(X,1)
if(sum(X(i,:) == -999) == size(X,2))
X(i,:) = [];
else
i = i+1;
end
end
3 comentarios
Amith Kamath
el 14 de En. de 2013
Edited: Thanks Walter! Didn't catch that case earlier! Your answer is much more elegant anyways.
James Tursa
el 15 de En. de 2013
Editada: James Tursa
el 15 de En. de 2013
Using a loop in this manner is very bad for performance. At each row that is deleted, potentially huge amounts of data will need to be copied. The same data can end up being copied many, many times. For a large matrix with many rows being deleted, this can easily result in a performance hit that is one or more orders of magnitude slower than the other answers shown above. (This is basically the same problem as increasing the size of an array in a loop). Best to get the indices of everything to be deleted first and then delete it all at once as shown above.
Ver también
Categorías
Más información sobre Loops and Conditional Statements 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!