Find indices where runs of zeros and ones ends

Assuming that I have a vector

H = [0 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 0];

I want to know the indices where long runs of 1's and 0's ends. Let's say if a bit repeats 4 times or greater, then it is a run of length greater than 4. According to this, I want to know the indices where all runs ends. First run is of zeros which ends at index 4. Second run is of ones which ends at 8. Third run is of zeros again and it ends at 17.

So far, I am able to count the run length only as following

i = find(diff(H));
run_length = [i numel(H)] - [0 i];

run_length =

     4     4     1     1     1     1     5

However, the needed answer is [4 8 17]. cumsum would not work because it will give cummulative sum of whole vector as

cumsum(run_length)

ans =

     4     8     9    10    11    12    17

Any suggestions will be helpful.

Answer 1

[Jonas]

i would use

H = [0 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 0];
Hstr=erase(num2str(H),' ');
[~,onesEnds]=regexp(Hstr,{'0000+','1111+'});
sort(cell2mat(onesEnds))

[Awais Saeed]

It worked like a charm. Thank you.

[Jonas]

my pleasure. you can write the regexp also as follwed if you want to be flexible with minimum length (just change the 4 in this case according to your wishes)

[~,onesEnds]=regexp(Hstr,{'0{4,}','1{4,}'});

[Awais Saeed]

Yes, this one is more flexible. Thank you.

[Stephen23]

Simpler and more efficient:

H = [0,0,0,0,1,1,1,1,0,1,0,1,0,0,0,0,0];
S = sprintf('%u',H);
X = regexp(S,'(0{4,}|1{4,})','end') % only zeros and ones

X = 1×3

4 8 17

X = regexp(S,'(.)(??$1{3,})','end') % any character

X = 1×3

4 8 17