Counting subrows in each row of a matrix?

Hi all, I need an algorithm which counts how many adjacent and non-overlapping (1,1) I have in each row of a matrix A mx(n*2) without using loops. E.g.

A=[1 1 1 0 1 1 0 0 0 1; 1 0 1 1 1 1 0 0 1 1]  %m=2, n=5

Then I want

B=[2;3] %mx1

In fact A=[(1 1) (1 0) (1 1) (0 0) (0 1); (1 0) (1 1) (1 1) (0 0) (1 1)]. Then, according to this separation, I have 2 (1 1) in the first row and 3 (1 1) in the second row.

[Cedric]

And if one row was

 0 1 1 0 0 0

would you count this as

 (0 1) (1 0) ..

or still spot the 1 1 block?

Answer 1

[Cedric]

Assuming that n is defined previously as

 >> n = 5 ;

(for the case of your example) here is a one liner

 >> B = sum( reshape( all( reshape( A.', 2, [] )), n, [] )).'
 B =
     2
     3

[MRC]

What if instead (1,1) I want (0,1)?

[Cedric]

What is the purpose?

Answer 2

[Azzi Abdelmalek]

I think you need to use at least one loop

A=[1 1 1 0 1 1 0 0 0 1; 1 0 1 1 1 1 0 0 1 1]
for k=1:size(A,1)
  a=[0 A(k,:)  0];
  ii=strfind(a,[1 0])-strfind(a,[0 1]);
  jj=mod(ii,2)~=0;
  ii(jj)=ii(jj)-1;
  out(k)=sum(ii)/2;
end
disp(out)