How to find strings of ones in a vector

I have... [3 1 1 5 7 2 1 1 1 9 10]; and I would like to find the indices for my "runs" of ones. I would like my output to be: [2 3 7 9]. (The start and end of the runs of ones).

Answer 1

[Jan]

Another solution:

x = [3 1 1 5 7 2 1 1 1 9 10];
isOne = [false, (x==1), false];
index = [strfind(isOne, [false, true]); ...
         strfind(isOne, [true, false]) - 1];
index = reshape(index, 1, []);

This takes about the half time of Oleg's method - if x has 10'000 elements.

[Oleg Komarov]

You mean findseq?

[Fangjun Jiang]

Jan, I found one solution without using strfind(). It's a little faster.

[Andrei Bobrov]

+1. Hi Jan! This is my favorite method, Matt Fig - first use for such tasks.

Answer 2

[Fangjun Jiang]

Can use find() directly. And it's faster than strfind().

x=[3 1 1 5 7 2 1 1 1 9 10];
a=diff([0 x 0]==1);
b=[find(a==1);find(a==-1)-1];
c=b(:)'

[Fangjun Jiang]

Speed test results:

%%

x=rand(1,1000);y=x>0.5;

x(y)=1; x(~y)=0;

t1=0;t2=0;

for k=1:1000;

%Jan's method

tic;

isOne = [false, (x==1), false];

index = [strfind(isOne, [false, true]); ...

strfind(isOne, [true, false]) - 1];

index = reshape(index, 1, []);

t1=t1+toc;

%Fangjun's method

tic;

a=diff([0 x 0]==1);

b=[find(a==1);find(a==-1)-1];

c=b(:)';

t2=t2+toc;

end

fprintf('Jan''s time: %f\n',t1);

fprintf('Fangjun''s time: %f\n',t2);

Jan's time: 0.082744

Fangjun's time: 0.080910

[Andrei Bobrov]

+1

Answer 3

[the cyclist]

There are probably less pedantic ways, but I think this does what you want.

x = [3 1 1 5 7 2 1 1 1 9 10];

isOne = (x==1);

isSameAsPrev = diff([NaN x])==0;
isSameAsNext = diff([x NaN])==0;

indexToFirstOrLastOne = find(isOne & (not(isSameAsPrev) | not(isSameAsNext)))

[Fangjun Jiang]

Need to improve to handle single 1 case. For example, if x=[1 2 1 1 1]; result should be [1 1 3 5]. Otherwise, if result is [1 3 5], won't be able to tell if x=[1 1 1 2 1]

Answer 4

[the cyclist]

Slightly more efficient version of my other answer:

isOne = (x==1);

isSameAsPrevOrNext = diff([NaN x NaN])==0;

isFirstOrLastOne = find(isOne & not(isSameAsPrevOrNext(1:end-1) & isSameAsPrevOrNext(2:end)));