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Find the consecutive positive and negative elements for the entire array

조회 수: 17 (최근 30일)
Hello all, I have a channel from which I take around 1 million samples now it contains both positive and negative values in it. My intention is to find the consecutive positive and negative integers(doesn't have to be same) and perform some operations on it. I have given my code below. chA is my channel from where i derive my inputs as values. The code is only giving me a value of 43.2600, which ideally should have given an array of numbers as there are lots of samples which are consecutive positive and negative.
for i = 1:1000000
if (chA(i)<0) && (chA((i+1) >0))
tan = ((chA(i+1))- chA(i));
deltaOfTime = tan/i;
end
thanks
  댓글 수: 5
Jayanta Deb
Jayanta Deb 2017년 3월 8일
Hi, your answer helped me, but sorry i cant thumbs up as i think i am not eligible to do so. its disabled. Anyways thanks :)
Jan
Jan 2017년 3월 8일
편집: Jan 2017년 3월 8일
@Jayanta Deb: You can accept and vote answers only, not your own question. When you are able to add a comment, you should have the power to vote also. Accepting an answer is useful for the forum, because the readers see, that the problem is solved.
If would be interesting to know and valuable for the forum, if these approaches are useful for your problem:
index = find(chA(1:end-1) < 0 & chA(2:end) > 0);
% Or: index = find(diff(chA < 0)); % For both directions
deltaOfTime = (chA(index + 1) - chA(index)) ./ index;

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채택된 답변

Jan
Jan 2017년 3월 7일
편집: Jan 2017년 3월 7일
If you want the start indices and the lengths of the runs: With FEX: RunLength:
x = randn(1, 1e6);
[B, N, Index] = RunLength(x < 0);
negStart = Index(B);
negLen = N(B);
posStart = Index(~B);
posLen = N(~B);
[EDITED]
Perhaps you want something like this:
deltaOfTime = zeros(1, 1000000); % Pre-allocate
c = 0;
for i = 1:1000000 - 1 % -1 to support chA(i+1)
if (chA(i) < 0) && (chA(i+1) >= 0)
% if (chA(i) < 0) == (chA(i+1) >= 0) % if both directions are wanted
c = c + 1;
deltaOfTime(c) = (chA(i+1) - chA(i)) /i;
end
end
deltaOfTime = deltaOfTime(1:c); % Crop unused elements
If this solves your problem, you can "vectorize" the code to increase the speed:
index = find(chA(1:end-1) < 0 & chA(2:end) > 0);
deltaOfTime = (chA(index + 1) - chA(index)) ./ index;
Or if you want to find both transicients even simpler:
index = find(diff(chA < 0));
deltaOfTime = (chA(index + 1) - chA(index)) ./ index;
NOTE: Using "tan" as name of a variable might be confusing, if the function tan() is used later on. Better avoid shadowing the names of built-in functions.
  댓글 수: 6
Star Strider
Star Strider 2017년 3월 8일
편집: John Kelly 2017년 3월 8일
If an Answer was unaccepted, only the OP has the ability to do it during the first 7 days after a Question is posted.
Neither Jan Simon nor anyone else with Editor privileges are able to accept or unaccept Answers during that time.
No one gets Reputation Points for accepting their own Answers.
DARSHAN N KANNUR
DARSHAN N KANNUR 2021년 3월 29일
I just have an array of 76140 data. What to do, if I want to know the starting index of consecutive negative elements and also thier count. Thank you in advance

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추가 답변 (1개)

John BG
John BG 2017년 3월 7일
편집: John BG 2017년 3월 7일
Jayanta
you are almost there, all left is is
1.
to accumulate the indices that your loop is already finding. You current loop only keeps the last zero crossing.
and
2.
include both transitions - to + and + to -
One way of doing both things would be
L=0
if ((chA(i)<0) && (chA((i+1) >0))) || ((chA(i)>0) && (chA((i+1) <0)))
L=[L i];
end
L=[]
if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG
  댓글 수: 2
Jayanta Deb
Jayanta Deb 2017년 3월 8일
yes that was helpful thanks John BG
DARSHAN N KANNUR
DARSHAN N KANNUR 2021년 3월 29일
I just have an array of 76140 data. What to do, if I want to know the starting index of consecutive negative elements and also thier count. Thank you in advance

댓글을 달려면 로그인하십시오.

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