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How could I find the start index of an "approximate" pattern in a binary vector

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Barry Troniqes
Barry Troniqes on 11 Jul 2017
Edited: Grzegorz Knor on 11 Jul 2017
I have a binary time series which represents the Hi/Low status of a sensors output over time. I'm looking for a method which will help me efficiently find the start and end indices of a specific pattern of on/off pulses. which looks something like this:
case1 = [0 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 0 0 0 0]
In this case there's a pattern (three high pulses, separated by low pulses of equivalent length) starting at case1(8) and ending at case1(22). I've been able to bodge a partial solution for finding this pattern out of an answer to a similar question which looks for zero islands (below) by looking for three 1 islands of length 3 with start indices quite close together in time.
https://stackoverflow.com/questions/3274043/finding-islands-of-zeros-in-a-sequence
However in my case the length of the 1 pulses and the number of zeros between them varies slightly due to mechanical issues with the sensor so we may get variations on the pattern, example:
case2 = [ 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 0 ]
Alternatively two patterns can occur very close to one another so that they merge
case3 = [0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0]
The rules for finding the pattern in case1 would therefore not work for case2 or case3. Could anyone therefore offer any advice on how to approach locating a pattern which 'approximately' matches a defined template with some degree of tolerance on the length of pulses within the pattern?

Answers (1)

Grzegorz Knor
Grzegorz Knor on 11 Jul 2017
Edited: Grzegorz Knor on 11 Jul 2017
I would try regular expressions :
case1 = [0 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 0 0 0 0]
case2 = [ 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 0 ]
case3 = [0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0]
% transform it to strings
case1str = strrep(num2str(case1), ' ', '')
case2str = strrep(num2str(case2), ' ', '')
case3str = strrep(num2str(case3), ' ', '')
% expression for regexp
expression = '11[1,0]?00[0,1]?11[0,1]?00[0,1]?11[0,1]?';
% find start indices
regexp(case1str,expression)
regexp(case2str,expression)
regexp(case3str,expression)
% expressions & indices
[expr, ind] = regexp(case3str,expression,'match')
Based on expr and ind you can calculate last index of the pattern.

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