delete certain elements in a matrix

Is there any way to delete certain rows(with a pattern) in a matrix

like to delete rows 1,6,11,21,26,..... using a loop

Thanks.

Answer 1

[Azzi Abdelmalek]

A=rand(40,4)  
v=[1 6 11 21 26]
w=sort(v);
for k=0:numel(v)-1
  A(w(k+1)-k,:)=[]
end

[dav]

thanks I want to delete all the rows. not just the above 4.

But all the rows in the pattern

1,6,11,16,21,26,31,..............101

[Azzi Abdelmalek]

A=rand(110,4)  
w=1:5:101
for k=1:numel(v)
  A(w(k)-k+1,:)=[]
end

[dav]

could you please explain this ?

[dav]

I made a mistake in my question.

I want to delete all the OTHER ROWS EXCEPT THE ROWS 1,6,11,16,......

[Azzi Abdelmalek]

% for k=1, w(1)=1; A(w(k)-k+1,:)=A(1+1-1)=A(1,:)
% for k=2, w(2)=6;  A(w(k)-k+1,:)=A(6+2-1)=A(6,:)
% and so on, all those rows will be deleted by A(w(k)-k+1,:)=[]

[Jan]

@dav: Simply set a breakpoint in the code an let it run. In the first iteration the first row is deleted. In the 2nd iteration, you should not delete the 6th row, because in the first iteration the 6th row became the 5th one. Therefore (k-1) is subtracted from the index in the k.th iteration.

Nevertheless, a loop is an inefficient approach and suffers from the old Schlemiel the painter problem.

[dav]

Thanks a lot

Answer 2

[Sven]

In a loop? Not really.

All at once? Absolutely.

DATA = rand(50,5);
rowsToDelete = [1,6,11,21,26];
DATA(rowsToDelete,:) = [];

If you mean that your rows to delete has the pattern of rows 1,6,11,16, etc, then just do this:

numRows = size(DATA,1);
rowsToDelete = [1:10:numRows 6:10:numRows];
DATA(rowsToDelete,:) = [];

[dav]

thanks, but I made a mistake in my question.

I want to delete all the OTHER ROWS EXCEPT THE ROWS 1,6,11,16,......

Could you please adjust your code to do this?

Thanks

[Jan]

This is much faster than a loop for large arrays. Shrinking arrays in a loop has the same drawback as letting them grow: In each iteration the new array must be allocated and the contents of the old one is copied. This is very expensive.

[dav]

Thanks a lot

Answer 3

[Jan]

Alternative to Sven's approach:

Data            = rand(50,5);
index           = false(1, 50);
index(1:10:end) = true;
index(6:10:end) = true;
Data(index, :)  = [];

[dav]

Thanks.