# Find the consecutive positive and negative elements for the entire array

조회 수: 10 (최근 30일)
Jayanta Deb . 2017년 3월 7일
댓글: DARSHAN N KANNUR . 2021년 3월 29일
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: 5표시 이전 댓글 수: 4숨기기 이전 댓글 수: 4
Jan 2017년 3월 8일
편집: Jan 님. 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 2017년 3월 7일
편집: Jan 님. 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: 6표시 이전 댓글 수: 5숨기기 이전 댓글 수: 5
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 2017년 3월 7일
편집: John BG 님. 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=[]
John BG
##### 댓글 수: 2표시 이전 댓글 수: 1숨기기 이전 댓글 수: 1
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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