Delete rows in a matrix that contain ONLY a negative number

조회 수: 5 (최근 30일)
Christina
Christina 2013년 1월 14일
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!

채택된 답변

Walter Roberson
Walter Roberson 2013년 1월 14일
idx = all(A == -999, 2);
A(idx, :) = [];

추가 답변 (3개)

Kye Taylor
Kye Taylor 2013년 1월 14일
편집: Kye Taylor 2013년 1월 14일
Try
isNegRow = all( A==-999, 2 );
A(isNegRow,:) = [];

Shashank Prasanna
Shashank Prasanna 2013년 1월 14일
편집: Shashank Prasanna 2013년 1월 14일
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 2013년 1월 14일
편집: Amith Kamath 2013년 1월 14일
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
  댓글 수: 3
Amith Kamath
Amith Kamath 2013년 1월 14일
Edited: Thanks Walter! Didn't catch that case earlier! Your answer is much more elegant anyways.
James Tursa
James Tursa 2013년 1월 15일
편집: James Tursa 2013년 1월 15일
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.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Matrix Indexing에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by