How to find a trend in a sequence in MATLAB?

Question

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 (eg) 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 (ie 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

Answer 1

I'm not sure exactly what output you're after but you can find a group of 10 or more 1 s quite easily using convolution:

w = [0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1]
n = 10;
find(conv(w, ones(1,n)) == n) - n + 1

this gives you the indices of where groups of 10 start

Answer 2

Let the data be

w = [0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1];
step_index = 101;
step_duration = 10;

Then:

d = diff([0 w 0]);
starts = find(d==1); %// indices of starts of runs of 1's
ends = find(d==-1); %// indices of ends of runs of 1's (actually starts of 0's)
keep = find(ends-starts>=step_duration); %// detect runs that are long enough
starts = starts(keep)+step_index-1; %// keep only those runs and adjust with "-1"
ends = ends(keep)+step_index-2; %// "-2" because of how ends is defined

The result is in vectors starts and ends . In your example there is just one such window, so they contain just one element:

starts =
   109
ends =
   118

Answer 3

Solution based on strfind to find starting indices of N or N+ consecutive ones -

%// sample accessibility vector (different from the one in question) 
access_vec = [1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0]

N = 10; %// N or N+ consecutive ones are to be detected
indices = strfind([0 access_vec],[0 ones(1,N)])

Output -

indices =
     1    13

---

Another example -

access_vec = [ 0 1 zeros(1,10) 1]

Output -

indices =
     []

Answer 4

I think you are looking for a period with good weather, where this would be defined as having sufficient ones in a stretch of 10.

In that case you could use a filter to check how many ones occur in each stretch of 10. It could be something like this:

f= filter(ones(10,1),1,x)

Not sure about the syntax, but check it out. Filter should definitely get you there.

You could then follow up by finding good periods like so:

find(f==10)
find(f>=8)
find(f==max(f))