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

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Jaya
Jaya 2021-9-3
编辑： Stephen23 2021-12-6
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

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采纳的回答

Stephen23
Stephen23 2021-12-4
编辑：Stephen23 2021-12-6
Ran in:
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
  1 个评论
Jaya
Jaya 2021-12-4
编辑：Jaya 2021-12-4
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

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更多回答（1 个）

Ive J
Ive J 2021-9-3
Ran in:
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
  2 个评论
Jaya
Jaya 2021-12-4
编辑：Jaya 2021-12-4
@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?

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