find '1' in an array
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hi ı wanna ask question about find a 1's number in array. for example the array should be like that 1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 and ı only want to find a group of 1's. In array 0 1 0 or 1 0 1 or 0 0 0 1 0 something like that is an error in array. How can we do that. thanks
2 comentarios
Jan
el 1 de Oct. de 2015
What is the desired output for the given data? How should these "errors" be treated?
Respuestas (5)
Guillaume
el 1 de Oct. de 2015
If you mean you want to find the start of the sequences of at least two consecutive 1, then it's simple using diff. You'll find the start of a sequence when the diff with the previous value is 1 (transition from 0 to 1), and you'll know it's followed by a 1 when the diff with the next value is 0, so:
A = [1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1]
startseq1 = find(A & diff([0 A]) == 1 & diff([A 0]) == 0)
4 comentarios
Guillaume
el 1 de Oct. de 2015
I have no idea what you mean. The code I've written work:
>>A = [1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1];
>>startseq1 = find(A & diff([0 A]) == 1 & diff([A 0]) == 0)
startseq1 =
1 7 13
>>endseq1 = find(A & diff([A 0]) == -1 & diff([0 A]) == 0)
endseq1 =
3 9 16
>>sequences = arrayfun(@(s,e) A(s:e), startseq1, endseq1, 'UniformOutput', false);
>>celldisp(sequences)
sequences{1} =
1 1 1
sequences{2} =
1 1 1
sequences{3} =
1 1 1 1
Thorsten
el 1 de Oct. de 2015
It is not entirely clear what you want to achieve; if you just want to get rid of the 0's, use
x = [1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1];
y = x(x==1);
Jan
el 1 de Oct. de 2015
data = [1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1];
[b, n] = RunLength(data);
result = sum(b == 1 & n > 1);
If the data are small (e.g. just some kB), or you do not have a compiler installed already, use RunLength_M instead from the same submission.
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