Delete rows in a matrix that contain ONLY a negative number

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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!

Respuesta aceptada

Walter Roberson
Walter Roberson el 14 de En. de 2013
idx = all(A == -999, 2);
A(idx, :) = [];

Más respuestas (3)

Kye Taylor
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,:) = [];

Shashank Prasanna
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.

Amith Kamath
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
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
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.

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