How to count time steps

Hi, I have X has several numbers inside say: X = [1 4 5 3 2 7 8 4 6 9 3 2 8 5 3 10] I want to first to select the values where X falls below 4, then need to count how many time steps does it take each one to go above 8.

Hope I am clear Thanks

[Image Analyst]

So it first goes above 4 at element 3 (where it's 5). Then to go from 5 to 9 (the first time it goes above 8) is 7 steps. Then it goes above 4 at element 14 (where it's 5) then it's 3 steps to get to 10. So is the answer [7, 3]? Or are your rules different than that?

What is the reason why you want this quirky thing?

[Karoline Qasem]

It is slightly different, I am looking at values of dissolved oxygen in water, at some time of the year, it goes BELOW 4 mg/L and it takes couple of days to go to 8 mg/L. My question is how many days does it need to go back from a threshold value (any value less than 4) to reach 8 mg/L

So the only thing not correct in your explanation is that I am looking at values less than 4 not above it. So, looking at the example: the first value below 4 is the first element which is 1, it needs 6 time steps to get to 8.

now moving after that number 4 (the one after 8) it needs two time steps to get to the 9.

etc

Thanks for the comment

Answer 1

[Image Analyst]

I imagine you've already modified by solution for your updated/clarified question. But anyway, here is the code:

% Define vector.
x = [1 4 5 3 2 7 8 4 6 9 3 2 8 5 3 10]

% Plot it.
plot(x, 'b*-', 'LineWidth', 2, 'MarkerSize', 13)
grid on;
hold on;
line(xlim, [4,4], 'Color', 'r', 'LineWidth', 2);

% Do the scan.
startingIndex = 1;
loopcounter = 1;
while startingIndex < length(x)
  subvector = x(startingIndex:end);
  % Find where it goes below 4 next.  Note ABOVE, not above or equal to.
  nextBelow4Index = find(subvector < 4, 1, 'first') + startingIndex - 1;
  % Find where it goes above 8 next.  Note ABOVE, not above or equal to.
  subvector2 = x((nextBelow4Index+1):end);
  % Count the number of steps for that to happen.
  numSteps(loopcounter) = find(subvector2 >= 8, 1, 'first');
  % Move to the next index past the one that exceeded 8.
  startingIndex = nextBelow4Index + numSteps(loopcounter) + 1;
  loopcounter = loopcounter + 1;
end
numSteps % Report results to command window.

Answer is [6,2,1]

[Karoline Qasem]

Thanks for the reply, when I put my original data I got this error Error in SelectRecoveryPeriod (line 54) numSteps(loopcounter) = find(subvector2 >= 8, 1, 'first'); Thanks

[Image Analyst]

Well you didn't post your original data. You posted this: X = [1 4 5 3 2 7 8 4 6 9 3 2 8 5 3 10] and it works for that. If you can't fix it for your other "original" data, post it, but I'll bet you can figure out what's going wrong since you're a smart engineer.

[Karoline Qasem]

Thanks a lot for the comment, I will try

[Karoline Qasem]

Attached is the sample data, the data am interested in is column 6 the DO. I tried to work it still got the same error, apologize for bothering you but, if you have time to look at I will appreciate. Thanks

[Image Analyst]

That is a file with lots of columns, many of which have gaps in them. How did you read it in (please give your code)? What column or part of a column was the "X" vector?

[Karoline Qasem]

its the whole column 6, I am attaching it again with it only

thanks

Answer 2

[Image Analyst]

Try this:

% Define vector.
x = [1 4 5 3 2 7 8 4 6 9 3 2 8 5 3 10]

% Plot it.
plot(x, 'b*-', 'LineWidth', 2, 'MarkerSize', 13)
grid on;
hold on;
line(xlim, [4,4], 'Color', 'r', 'LineWidth', 2);

% Do the scan.
startingIndex = 1;
loopcounter = 1;
while startingIndex < length(x)
  subvector = x(startingIndex:end);
  % Find where it goes above 4 next.  Note ABOVE, not above or equal to.
  nextAbove4Index = find(subvector > 4, 1, 'first') + startingIndex - 1;
  % Find where it goes above 8 next.  Note ABOVE, not above or equal to.
  nextAbove8Index = find(subvector > 8, 1, 'first') + startingIndex - 1;
  % Count the number of steps for that to happen.
  numSteps(loopcounter) = nextAbove8Index - nextAbove4Index;
  % Move to the next index past the one that exceeded 8.
  startingIndex = nextAbove8Index + 1;
  loopcounter = loopcounter + 1;
end
numSteps % Report results to command window.

The result is [7, 3] as expected.