How to find a trend in a sequence?

I have an accessibility vector with 1's and 0's. 1 means the weather is good; and 0 meaning the weather is not good and the place is not accessible.

I have a step_duration of (e.g.) 10 hrs. Considering the step_index (start of the step) as 101,I need to find a window of straight 10 hrs of good weather.

Expected solution: With 10 hours of expected weather, say the accessibility vector is [0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1] . So, the indexes that we find the window are from 109-118. And the corresponding weather delay (considering we could not find straight hours) is from index 101-108 (i.e. 9 hours.) I need to write a code for such an algorithm.

Some sample code that I could think of is as follows( though this is not exactly what I want):

window = 0;     % Initialize the counter for finding the weather window.

        for tt = step_index

            if Accessibility(tt) == 0
                % bad weather, move to next index
                % reset window.
                window = 0;
                k = k + 1;
                possible_window(k) = window;
                continue
            else
                % Increase window
                window = window + 1;

                % sote window history
                k = k + 1;
                possible_window (k) = window;
            end

        end
        tt = tt + 1;
end
  if true
    % code
  end

[Andy L]

Does each element correspond to one hour?

[Ashish]

Yes, each element corresponds to 1 hour

Answer 1

[Star Strider]

You can use the filter function (here, ‘av’ is the accessibility vector):

av = randi(2, 1, 100)-1;        % Create Data
av(20:29) = ones(1,10);         % Insert Series of 10 1’s

y  = filter(ones(1,10), 10, av);
yf = find(y > 0.99);

The ‘yf’ variable will find the end index of the sequences, and the size of ‘yf’ will be the number of sequences in a given accessibility vector.

[Andy L]

This is a nice solution!

[Star Strider]

Thank you!

[Ashish]

Thanks for the answer and introducing me to the new function. But, I have more than one steps (more than 100 with different step_duration). And the accessibility vector is of length of 87600 indices. For the given step, I have to limit the search from the point of "step_index" till the time I get the window for 10 hours (e.g.)

[Star Strider]

My code finds the end-indices of your ‘step_duration’ as I understand it. To find the beginnings, count back 10. (Also, if there are 11 consecutive 1s, it will return 2 indices, if 12, 3 indices, and so on.) I don’t understand how ‘step_index’ enters into it, since it wasn’t part of the ‘accessibility vector’ you posted.

If you have fewer than 10 consecutive 1s, it will generate values <1. If you have other criteria for the step_duration — fewer or more than 10 consecutive 1s — change the filter coefficients appropriately, with the ‘b’ vector a ones vector of appropriate length, and changing the ‘a’ vector (the 10 in my code) to the length of the ones vector.

If you want to find the occurrences of ‘[1 0 1]’, use the same filter idea with a different pattern:

av = randi(2, 1, 100)-1;
si = filter([1 0 1], 2, av);
sidx = find(si > 0.99);

The ‘si’ vector is 1 where the the filter detects the end of a ‘[1 0 1]’ sequence. The ‘b’ value has to be the sum of the elements of the ‘a’ vector.

I don’t understand the algorithm you need to write, but my code gets you very close. It can detect the ‘step_index’ with the ‘sidx’ variable, and detect the ‘step_duration’ in the ‘yf’ variable as you defined it in your Question, and can give you the indices for both. The filter calls will take your entire 87600-element vector and produce the indices for the required patterns. It seems to me all you then need to do is program the logic relating to the ‘step_index’ and ‘step_duration’ indices to do what you want.

Answer 2

[Andy L]

Try this thread for a similar problem

https://www.mathworks.co.uk/matlabcentral/newsreader/view_thread/244401