Single average value for same range of time each day throughout the year

Hello,

I am currently looking for a way to average my dataset for a period of 4 minutes each day. For my current analysis, I am only interested in the specific value at 5pm (17:00:00). Sometimes though, this data point is missing, so I had thought to take an average from 16:58 to 17:02. I had attempted to work through this, using my test Time_Vector (attached) as a reference. My current code unfortunately also returns 16:00:00 - 16:01:59 as well as 17:58:00 - 17:59:59, so at this time I am unable to accurately average my dataset. I would then use the index created to split up my Time_Vector against my y-data to find a specific value each day.

(i.e. Test_Y = Test_Y(mt_id) or Test_Y = Test_Y(Times_to_Keep_id))

H = hour(Time_Vector);

hr_id = H >= 16 & H <= 17;

Time_Vector2 = Time_Vector(hr_id);

mt_id = minute(Time_Vector2) > 58 | minute(Time_Vector2) < 02;

Time_Vector3 = Time_Vector2(mt_id);

% Times_to_Keep_id = hr_id & mt_id;

Thank you for your help.

Answer 1

[Cris LaPierre]

You need to combine your conditions for hours and minutes. Consider using the timeofday function instead.

load Time_Vector
Tind = timeofday(Time_Vector) >= duration(16,58,0) & timeofday(Time_Vector) <= duration(17,02,0);
Time_Vector(Tind,1)

ans = 24×1 datetime array
   19-Jul-2017 16:58:03
   19-Jul-2017 16:58:18
   19-Jul-2017 16:58:33
   19-Jul-2017 16:58:48
   19-Jul-2017 16:59:03
   19-Jul-2017 16:59:18
   19-Jul-2017 16:59:33
   19-Jul-2017 16:59:48
   19-Jul-2017 17:00:03
   19-Jul-2017 17:00:18
   19-Jul-2017 17:00:33
   19-Jul-2017 17:00:48
   19-Jul-2017 17:01:03
   19-Jul-2017 17:01:18
   19-Jul-2017 17:01:33
   19-Jul-2017 17:01:48
   20-Jul-2017 16:58:03
   20-Jul-2017 16:58:18
   20-Jul-2017 16:58:33
   20-Jul-2017 16:58:48
   20-Jul-2017 16:59:03
   20-Jul-2017 16:59:18
   20-Jul-2017 16:59:33
   20-Jul-2017 16:59:48

[Eric Beamesderfer]

You make it look easy. Thank you, Cris!