# Computing difference between maximum values in columns of matrix within predefined time interval

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Poulomi Ganguli le 15 Août 2017
Commenté : Poulomi Ganguli le 23 Oct 2017
Hello, I have a matrix of values of size 8760 x 6 (a small excerpt is shown below), in which 1st column is the year, 2nd month, 3rd is the day, 4th is the hour, 5th is value 1 and 6th is value 2. The nature of graph for val 1 and val 2 is sinusoidal, hence there is alternate peaks and troughs. I need to identify the peaks in val 1 and corresponding peaks in val 2 within +/- 3 hr window and then difference between the two values. For example, in the matrix below, the 1st peak value is 2.08 in val 1 and next value is +1 hr next to it, i.e., 1.75 in val 2. In the next positive cycle, the peak value is 2.02 and corresponding peak value in val2 is 1.82 and so on.
Year Mo Day hr val 1 val 2
2014 1 1 0 1.84 1.67
2014 1 1 1 1.74 1.56
2014 1 1 2 1.26 1.04
2014 1 1 3 0.47 0.23
2014 1 1 4 -0.4 -0.65
2014 1 1 5 -1.11 -1.40
2014 1 1 6 -1.43 -1.78
2014 1 1 7 -1.30 -1.68
2014 1 1 8 -0.76 -1.12
2014 1 1 9 0.078 -0.27
2014 1 1 10 0.89 0.58
2014 1 1 11 1.62 1.29
2014 1 1 12 2.08 1.74
2014 1 1 13 2.06 1.75
2014 1 1 14 1.61 1.33
2014 1 1 15 0.87 0.56
2014 1 1 16 -0.05 -0.37
2014 1 1 17 -0.87 -1.25
2014 1 1 18 -1.47 -1.86
2014 1 1 19 -1.59 -2.00
2014 1 1 20 -1.28 -1.63
2014 1 1 21 -0.56 -0.87
2014 1 1 22 0.327 0.022
2014 1 1 23 1.181 0.85
2014 1 2 0 1.85 1.521
2014 1 2 1 2.020 1.819
2014 1 2 2 1.736 1.637
2014 1 2 3 1.074 1.021
2014 1 2 4 0.142 0.115
2014 1 2 5 -0.81 -0.83
2014 1 2 6 -1.54 -1.58
2014 1 2 7 -1.82 -1.90
2014 1 2 8 -1.59 -1.68
2014 1 2 9 -0.92 -0.98
2014 1 2 10 0.002 -0.058
2014 1 2 11 0.892 0.825
2014 1 2 12 1.693 1.529
2014 1 2 13 2.138 1.914
2014 1 2 14 2.057 1.829
2014 1 2 15 1.547 1.287
2014 1 2 16 0.696 0.4018
Finally, the desired output is:
2014 1 1 2.08 1.76 0.32
2014 1 2 2.02 1.82 0.20
2014 1 2 2.14 1.91 0.22
Any help?
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### Réponse acceptée

Guillaume le 15 Août 2017
Modifié(e) : Guillaume le 15 Août 2017
The first step would be to identify the local maxima (and their row index) in val1 and val2 using findpeaks (requires signal processing toolbox) or the FileExchange peakfinder:
[val1max, rowval1] = findpeaks(yourtable.value1);
[val2max, rowval2] = findpeaks(yourtable.value2);
The next thing is to compute the time difference between the rows. Assuming that there's so many peaks that you'd run out of memory you could do it all in one go. But first, I'd convert your ymdh vectors into datetime objects:
dt = datetime([yourtable{:, 1:4}, repmat([0 0], height(yourtable), 1)]);
The time difference in hours is then:
hourdiff = hours(dt(rowval1) - dt(rowval2)') %require R2016b, in earlier versions use bsxfun for the subtraction
Then find where the time difference is within +/-3 hours:
[row, col] = find(abs(hourdiff) <= 3);
row is the index of val1max and rowval1 and col is the corresponding index of val2max and rowval2 where the absolute difference is within 3 hours.
You can build your final table with:
result = table(dt(rowvar1(row)), val1max(row), val2max(col), val2max(col)-val1max(row))
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Guillaume le 23 Oct 2017
It's a trade off between speed and memory. If you don't have enough memory you have to sacrify speed and implement the calculation as a for loop:
row = [];
col = [];
for c = 1:numel(rowval2)
hourdiff = dt(rowval1) - dt(rowval2(c);
r = find(abs(hourdiff) <= 3 | ...
(~any(abs(hourdiff) <= 3) & abs(hourdiff) <= 6));
row = [row; r(:)];
col = [col; repmat(c, numel(r), 1)];
end
Poulomi Ganguli le 23 Oct 2017
hmm now I am using a 12-core processor, now its working. But good to know about alternatives. Thanks a bunch!

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