# How to find a chunk of a certain number of zeros inside a vector

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Pedro Cavaco on 21 Jun 2011
Hi all,
I have a vector of ones and zeros randomly distributed.
i.e: A = [0;1;1;0;0;0;0;1;1;1;1;0;1;]
What I want is to find the location of the first zero of the first chunk with 4 OR MORE zeros appearing in the vector.
In this example the result would be:
pos = 4;
The size of the group of zeros doesn't have to be necessarily 4, this was just an example.
I cannot find a simple way to do this but most probably there's a command for for this kind of operations that I cannot recall.
Many thanks in advance,
Pedro Cavaco

David Young on 21 Jun 2011
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;]
n = 4;
To find the first group of 4 or more zeros:
p = regexp(char(A.'), char(zeros(1, n)), 'once')
To find the first group of exactly 4 zeros:
zz = char(zeros(1,n));
p = regexp(char(A.'), ['(?<=^|' char(1) ')' zz '(' char(1) '|\$)'], 'once')
David Young on 21 Jun 2011
By the way, in the case of n or more zeros, it's not obvious whether to use my first answer, with regexp, or Andrei's answer, with strfind. For very long strings, it may be faster to use regexp because of its 'once' option; however, strfind is simpler and will have a lower overhead.

### More Answers (3)

Andrei Bobrov on 21 Jun 2011
EDIT
A1 = A(:)';
out = strfind([1 A1],[1 0])-1; % all groups zeros
strfind([A1 1],[0 0 1]); % all groups two zeros
...
strfind([A1 1],[zeros(1,4) 1]); % all groups 4 zeros
Andrei Bobrov on 21 Jun 2011
it's idea of Matt Fig

Gerd on 21 Jun 2011
Hi Pedro,
just programming straigforward I would use
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;];
cons = 4;
indices = find(A==0);
for ii=1:numel(indices)-cons
if (indices(ii+1)-indices(ii) == 1) && (indices(ii+2)-indices(ii+1)==1) && indices(ii+3)-indices(ii+2)==1
disp(indices(ii));
end
end
Result is 4
Gerd
Gerd on 21 Jun 2011
Hi Pedro,
I tried both solution in a .m-file(David's and mine)
Please have a look at the result.
tic;
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;];
cons = 4;
indices = find(A==0);
for ii=1:numel(indices)-cons
if (indices(ii+1)-indices(ii) == 1) && (indices(ii+2)-indices(ii+1)==1) && indices(ii+3)-indices(ii+2)==1
disp(indices(ii));
end
end
t1 = toc;
tic;
A = [0;1;1;0;0;0;0;1;1;1;1;0;1;];
n = 4;
p = regexp(char(A.'), char(zeros(1, n)), 'once');
disp(p);
t2 = toc;
With your testvector the result is really fast.

David Young on 21 Jun 2011
Another approach to finding the first group of 4 or more zeros:
A = [0;1;1;1;0;0;0;0;1;1;1;1;0;1;0;0;0;1];
n = 4;
c = cumsum(A);
pad = zeros(n, 1)-1;
ppp = find([c; pad] == [pad; c]) - (n-1);
p = ppp(1)
EDIT Code corrected - n replaced by (n-1) to give correct offset.
David Young on 21 Jun 2011
Sorry, you are right!