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Help. I'm trying to created a code that...........

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Javier
Javier am 16 Jul. 2013
Geschlossen: MATLAB Answer Bot am 20 Aug. 2021
I'm trying to created a code that doing something like this, when v (time serie) changes only from negative to positive (zero crossings),this was separated by at least 4 elements from other zero crossings(left and right direction), and if I take 8 elements (from zero crossings)to the right, at least the 80% of the elements were positive,and if i take 6 elements to the left (from zero crossings), at least the 60% of the elements were negative. (I have a huge data serie , and I'm trying to generate a code that I can rearrange the parameters.
Regards)
v=[-1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 1 -1 -1 1 1 1 1 -1 -1
1 -1 -1 1 1 1 1 1 1 1 1 -1 1]
regards,
  4 Kommentare
Jan
Jan am 17 Jul. 2013
How large is "huge"? It matters if you want to process 1GB of data or only some MB.
Javier
Javier am 17 Jul. 2013
Bearbeitet: Javier am 17 Jul. 2013
Hello Jan, are only a few mb...
>> size(v)
ans =
1 180407
regards

Antworten (2)

Andrei Bobrov
Andrei Bobrov am 16 Jul. 2013
a = v == 1;
ii = [true;diff(a(:)) ~= 0];
i0 = find(ii);
iii = reshape(diff([i0;numel(a)+1]),2,[])';
i1 = reshape(i0,2,[])';
indexout = i1(all(bsxfun(@le,[.6 .8],bsxfun(@rdivide,iii,[6 8])),2),2);
  1 Kommentar
Javier
Javier am 17 Jul. 2013
Bearbeitet: Javier am 17 Jul. 2013
Thank you so much Andrei, but for example, if I wanna (rearrange the parameters) change this part of the code("when v (time serie) changes only from negative to positive (zero crossings),this was separated by at least 4 elements from other zero crossings(left and right direction)"), change 4 by 10, or other number
regards

Jan
Jan am 17 Jul. 2013
Bearbeitet: Jan am 17 Jul. 2013
Try a simple FOR loop at least as a proof of concept: (See FEX: RunLength)
[value, rep, index] = RunLength(v);
n = numel(value);
match = zeros(1, n); % Pre-allocate
iMatch = 0;
first = find(index > 6, 1, 'first'); % Or <= ?
last = find(index < numel(v) - 8, 1, 'last'); % Or <= ?
for k = first:last
if n(k-1) < 4 && n(k) < 4 % Or <= ?
continue;
end
q = index(k);
if sum(v(q:q+7) == 1) / 8 < 0.8 % Or q+1:q+8 ?
continue;
end
if sum(v(q-6:q-1) == 1) / 6 < 0.6 % Or q-5:q ?
continue;
end
% All conditions match:
iMatch = iMatch + 1;
match(iMatch) = k;
end
match = match(1:iMatch);
Some points are not clear yet, e.g. if "8 elements to the right" is inclusive or exclusive the current element. But if this is cleared, e.g. the conditiosn could be vectorized easily. But it is not sure if this is faster and perhaps the speed of the loop is sufficient already.

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