Info
La pregunta está cerrada. Vuélvala a abrir para editarla o responderla.
Help. I'm trying to created a code that...........
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
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 comentarios
Jan
el 17 de Jul. de 2013
How large is "huge"? It matters if you want to process 1GB of data or only some MB.
Respuestas (2)
Andrei Bobrov
el 16 de Jul. de 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);
Jan
el 17 de Jul. de 2013
Editada: Jan
el 17 de Jul. de 2013
[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.
0 comentarios
Ver también
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!