Counting subrows in each row of a matrix?

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MRC
MRC 2014-5-7
评论： Cedric 2014-5-7
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.
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Cedric
Cedric 2014-5-7
编辑：Cedric 2014-5-7
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?

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Cedric
Cedric 2014-5-7
编辑：Cedric 2014-5-7
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

更多回答（1 个）

Azzi Abdelmalek
Azzi Abdelmalek 2014-5-7
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)

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