Creating a daily date vector

4 vues (au cours des 30 derniers jours)
Lejla Latifovic
Lejla Latifovic le 29 Avr 2022
Hello,
I hope this is a fairly easy question for the community. I have a half hourly datetime vector which I would like to convert to a daily datetime vector. I don't seem to be having any luck. There must be a simple code to do this? Any help would be much appreciated.
Here is an example of my data: date2017 (17520x1 datetime)
date2017 =
'01/01/2017 00:30:01'
'01/01/2017 01:00:01'
'01/01/2017 01:30:01'
'01/01/2017 02:00:01'
'01/01/2017 02:30:01'
'01/01/2017 03:00:01'
'01/01/2017 03:30:01'
'01/01/2017 04:00:01'
'01/01/2017 04:30:01'
'01/01/2017 05:00:01'
'01/01/2017 05:30:01'
'01/01/2017 06:00:01'
'01/01/2017 06:30:01'
'01/01/2017 07:00:01'
'01/01/2017 07:30:01'
'01/01/2017 08:00:01'
'01/01/2017 08:30:01'
'01/01/2017 09:00:01'
'01/01/2017 09:30:01'
'01/01/2017 10:00:01'
'01/01/2017 10:30:01'
'01/01/2017 11:00:01'
'01/01/2017 11:30:01'
'01/01/2017 12:00:01'
'01/01/2017 12:30:01'
'01/01/2017 13:00:01'
'01/01/2017 13:30:01'
'01/01/2017 14:00:01'
'01/01/2017 14:30:01'
I'd like to do this so that I can match it to daily mean values I have for temperature.
I tried this but something went wrong. I ended up with a DateDaily (367x0 timetable) vector, so no date values.
T = table(date2017, 'VariableNames' , {'Date'});
D = table2timetable(T);
DateDaily = retime(D, 'daily');
Thank you!

Réponse acceptée

Steven Lord
Steven Lord le 29 Avr 2022
Based on your stated goal, I'd probably store the data in a timetable array and then use retime to change it to a 'daily' timetable.
n = 15;
randomHourVector = randi(6, n, 1); % n random integers between 1 and 6
d = datetime('now') + hours(cumsum(randomHourVector));
x = (1:n).';
tt = timetable(d, x)
tt = 15×1 timetable
d x ____________________ __ 29-Apr-2022 21:32:15 1 30-Apr-2022 00:32:15 2 30-Apr-2022 06:32:15 3 30-Apr-2022 08:32:15 4 30-Apr-2022 11:32:15 5 30-Apr-2022 12:32:15 6 30-Apr-2022 17:32:15 7 30-Apr-2022 20:32:15 8 01-May-2022 02:32:15 9 01-May-2022 04:32:15 10 01-May-2022 08:32:15 11 01-May-2022 09:32:15 12 01-May-2022 13:32:15 13 01-May-2022 15:32:15 14 01-May-2022 16:32:15 15
tt2 = retime(tt, 'daily', @mean)
tt2 = 3×1 timetable
d x ___________ __ 29-Apr-2022 1 30-Apr-2022 5 01-May-2022 12
  7 commentaires
Steven Lord
Steven Lord le 29 Avr 2022
The output of cumsum is not going to be suitable for use as an aggregation method in retime. It doesn't reduce a vector of data down to a scalar value like sum and mean do. But if I understand what you want, you can do that using cumsum after the retime call.
n = 15;
randomHourVector = randi(6, n, 1); % n random integers between 1 and 6
d = datetime('now') + hours(cumsum(randomHourVector));
x = (1:n).';
tt = timetable(d, x)
tt = 15×1 timetable
d x ____________________ __ 29-Apr-2022 22:57:36 1 30-Apr-2022 00:57:36 2 30-Apr-2022 03:57:36 3 30-Apr-2022 06:57:36 4 30-Apr-2022 08:57:36 5 30-Apr-2022 11:57:36 6 30-Apr-2022 13:57:36 7 30-Apr-2022 16:57:36 8 30-Apr-2022 19:57:36 9 01-May-2022 00:57:36 10 01-May-2022 03:57:36 11 01-May-2022 05:57:36 12 01-May-2022 09:57:36 13 01-May-2022 11:57:36 14 01-May-2022 15:57:36 15
tt2 = retime(tt, 'daily', @mean)
tt2 = 3×1 timetable
d x ___________ ____ 29-Apr-2022 1 30-Apr-2022 5.5 01-May-2022 12.5
tt2.CumulativeSum = cumsum(tt2.x)
tt2 = 3×2 timetable
d x CumulativeSum ___________ ____ _____________ 29-Apr-2022 1 1 30-Apr-2022 5.5 6.5 01-May-2022 12.5 19
Lejla Latifovic
Lejla Latifovic le 29 Avr 2022
Yes, that is exactly it. Wonderful! Thank you for all of your help, much appreciated.

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