How to find strings of ones in a vector

Question

Maddie 2011-8-27

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https://ww2.mathworks.cn/matlabcentral/answers/14651-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).

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Answer 1

Jan 2011-8-27

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https://ww2.mathworks.cn/matlabcentral/answers/14651-how-to-find-strings-of-ones-in-a-vector#answer_19905

在 MATLAB Online 中打开

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.

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Fangjun Jiang 2011-8-28

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

Andrei Bobrov 2011-8-28

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

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Answer 2

Fangjun Jiang 2011-8-28

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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(:)'

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Fangjun Jiang 2011-8-28

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 2011-8-28

+1

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Answer 3

the cyclist 2011-8-27

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在 MATLAB Online 中打开

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)))

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Fangjun Jiang 2011-8-28

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]

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Answer 4

the cyclist 2011-8-28

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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)));