Find the consecutive positive and negative elements for the entire array

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

[Jan]

What is the wanted output for e.g.:

x = [-1, -1, 2, 3, 0, -1]

Is the 0 treated as positive?

[Jayanta Deb]

Hi, The output should be all the values from the variable deltaOfTime. It should contain all the value after each iteration which fulfills the conditions it can be as an array/vector. I just need to have the values of deltaOfTime. yes you can consider zero as positive

[John BG]

Jayanta

you are almost there, all left is is

1. to accumulate the indices that your loop is already finding.

2.

include both transistions - 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(1)=[]

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

<mailto:[e-mail] [e-mail]>

[Jayanta Deb]

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]

@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;

Answer 1

[Jan]

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.

[Jayanta Deb]

sorry i dont find it relevant.

consider the array as [0,1,-3,4,5,6,7,8,9,-19]
for i = 1:1000000 % loops strats from 1 and ends at 1000000
if (chA(i)<0) && (chA((i+1) >0)) % if i = 1, i+1 = -3 <it satisfy the condition>
tan = ((chA(i+1))- chA(i)); %calculate it
deltaOfTime = tan/i; %store the value here in the vector deltaOfTime
end
now in the next iteration it should be able to find out the next consecutive positive and negative value which is 9,-19

[Jan]

Then I did not understand the expression "consecutive positive and negative integers". Can you create the wanted output for [0,1,-3,4,5,6,7,8,9,-19] manually?

See [EDITED] in my answer.

[Jayanta Deb]

well that works exactly how i wanted. thanks Jan Simon

[Jan]

You are welcome.

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[DARSHAN N KANNUR]

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

Answer 2

[John BG]

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

<mailto:[e-mail] [e-mail]>

[Jayanta Deb]

yes that was helpful thanks John BG

[DARSHAN N KANNUR]

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