Any efficient way to identify a set of 1s in a big array?

I have an array called link_slots of 800 elements made of 1, 0 and -1.

E.g. 1 1 1 -1 0 0 0 0 1 1 -1 0 0 1 1 1 1 -1 0 0 .... So, 1 denotes occupied, 0 for unoccupied and -1 just to mark the end of a set of 1s.

I want to know the start and end indices of each set of 1s. E.g.,here, start as [1,9,14] and end as [3,10,17]. My code works but found out through Profiler that it takes a lot of time. Is there any efficient way to solve this? Considering that I have to do this thing for multiple arrays for size 800 elements.

                        i=1;
                        while(i<numel(link_slots(1,:)) ) %to cycle through whole array
                            j=i
                            if(link_slots(1,i)==1)  %i.e. if occupied 
                                startt(i)=i %store it in the start array
                                j=i
                                while(link_slots(index,j+1)~=-1) 
                                    j=j+1
                                end
                                endd(i)=j  %store the end index upon encountering -1
                            end
                            i=j+1
                        end

Answer 1

[Stephen23]

A simple, efficient, robust solution:

a = [1,1,1,-1,0,0,0,0,1,1,-1,0,0,1,1,1,1,-1,0,0];
d = diff([false,a==1,false]);
s = find(d>0) % start

s = 1×3

1 9 14

e = find(d<0)-1 % end

e = 1×3

3 10 17

And tested on your new data set:

a = [1,1,-1,1,1,1,-1,0,0];
d = diff([false,a==1,false]);
s = find(d>0) % start

s = 1×2

1 4

e = find(d<0)-1 % end

e = 1×2

2 6

[Jaya]

Thank you so much. I want to accept your answer as well. Thinking that my comments on a old question won't be visible to wider audience, I asked a fresh question linking to this question here. You may please paste your answer there so that I can accept it there.

https://www.mathworks.com/matlabcentral/answers/1602735-efficiently-identifying-a-set-of-1s-follow-up-question-after-months-later

Answer 2

[Ive J]

Try this

a = [1 1 1 -1 0 0 0 0 1 1 -1 0 0 1 1 1 1 -1 0 0];
astart = [1, 9, 14];
astop = [3, 10, 17];

dda = diff([0, 0, diff(a)]);

start = find(dda == 1);
if a(1) > 0
    start = [1, start];
end
stop = find(circshift(a < 0, -1));

all(astart == start)

ans = logical
   1

all(astop == stop)

ans = logical
   1

[Jaya]

This really helped to speed up the implementation. Thanks.

[Jaya]

@Ive J Hello, I am the same person who asked the original question. Your solution worked perfectly. But now I find that when my array is like below then it isn't able to identify the start properly. But stop is ok.

a=[1     1    -1     1     1     1    -1     0    0];
%start is only coming as -1 It should instead come as [1 4]
%stop is working fine. I.e. [2 6]

The difference w.r.t the original question array and this array is only that now, the set of 1s are next to another set of 1s separated by the marker, -1. Whereas in the original question, there were some 0's in between. Can you please help me on this?