フィルターのクリア

Info

この質問は閉じられています。 編集または回答するには再度開いてください。

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

1 回表示 (過去 30 日間)
Javier
Javier 2013 年 7 月 16 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
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 件のコメント
Jan
Jan 2013 年 7 月 17 日
How large is "huge"? It matters if you want to process 1GB of data or only some MB.
Javier
Javier 2013 年 7 月 17 日
編集済み: Javier 2013 年 7 月 17 日
Hello Jan, are only a few mb...
>> size(v)
ans =
1 180407
regards

回答 (2 件)

Andrei Bobrov
Andrei Bobrov 2013 年 7 月 16 日
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 件のコメント
Javier
Javier 2013 年 7 月 17 日
編集済み: Javier 2013 年 7 月 17 日
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 2013 年 7 月 17 日
編集済み: Jan 2013 年 7 月 17 日
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.

製品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by