I am working with a 1627x11 array, labelled A, and I want to delete the entries that have 0 in the 9th column (i.e delete the whole row).
V = A(:,9);
vals=find(V==0);
A(vals)=[ ];
after running that code, the array converts to an 1x16617 double - which is not wat I anticipated nor want.
To reiterate, the 9th column which has a sort of counter is used to keep track of which point I am working with, and I do not want to include the rows which are in the 0th point.
Rows 0 to 1280 contains 0 in the 9th column (this was done by observing the array), so I should have a 347x11 array after running the code.
What am I doing wrong

 Réponse acceptée

Brian Hart
Brian Hart le 29 Jan 2019

1 vote

Hi Ramitha,
You almost got it. For your last line of code, try
A(vals,:)=[];
The way you tried it at first uses linear lindexing (see sub2ind)

3 commentaires

Hans123
Hans123 le 29 Jan 2019
Thanks Brian, ti worked perfectly!
Stephen23
Stephen23 le 30 Jan 2019
Simpler and more efficient to use logical indexing (without find):
A(A(:,9)==0,:) = [];
Hans123
Hans123 le 30 Jan 2019
that's great! Thanks Stephen

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Plus de réponses (1)

madhan ravi
madhan ravi le 29 Jan 2019
Modifié(e) : madhan ravi le 30 Jan 2019

0 votes

A(~any(A(:,9),2),:)=[]

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