find non-overlapping sequences of zeros in matlab arrays - matlab

Question

This is related to:
Finding islands of zeros in a sequence.
However, the problem is not exactly the same:
Let's take the same vector with the above postfor the purpose of comparison:
sig = [1 1 0 0 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0];
What I am trying to find are the starting indices of islands of n consecutive zeros; however, overlapping is not allowed. For example for n=2, I want the result:
v=[3, 5, 14, 25];
I found the solution of Amro brilliant as a starting point (especially with regards to strfind), but the second part of his answer does not give me the result that I expect. This is a non-vectorized solution that I have so far:
function v=findIslands(sig, n)
% Finds indices of unique islands
% sig --> target vector
% n --> This is the length of the island
% This will find the starting indices for all "islands" of ones
% but it marks long strings multiple times
startIndex = strfind(sig, zeros(1,n));
L=length(startIndex);
% ongoing gap counter
spc=0;
if L>0 % Check if empty
v=startIndex(1);
for i=2:L
% Count the distance
spc=spc+(startIndex(i)-startIndex(i-1));
if spc>=n
v=[v,startIndex(i)];
% Reset odometer
spc=0;
end
end
else
v=[];
display('No Islands Found!')
end
I was wondering if someone has a faster vectorized solution to the above problem.

Answer 1

You can convert everything into strings and use regular expressions:
regexp(sprintf('%d', sig(:)), sprintf('%d', zeros(n, 1)))
Example
>> sig = [1 1 0 0 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0];
>> n = 2;
>> regexp(sprintf('%d', sig(:)), sprintf('%d', zeros(n, 1)))
ans =
3 5 14 25

Answer 2

Do this:
As an example let's look at the case where the run length you want is 2.
Convert vector to binary number
Set index = size-1, set starting = []
Loop until n < 4:
Is n divisible by 4?
Yes? Append index to starting. Set n = n / 4
No? Set n = n / 2
Goto 3
For any other run length replace 4 with 2**run.

Answer 3

Use gnovice's answer from the same linked question. It's vectorized, and the runs where duration == n are the ones you want.
https://stackoverflow.com/a/3274416/105904
Take the runs with duration >= n, and then divide duration by n, and that'll tell you how many consecutive runs you have at each position and how to expand the index list. This could end up faster than the regexp version, if your island density isn't too high.