How can I find the time difference between each consecutive rows?
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Ammy
le 3 Juil 2018
Réponse apportée : M.Devaki Mohanarangam
le 8 Sep 2023
I have a column in which date and time is mentioned, I want to find the time difference between every consective rows . Also how can I find the time difference in seconds , minutes, hours , days , weeks , months and years.
TimeStamp ; TimeDiff;
2014-09-02 07:48:09.567; NULL;
2014-09-02 07:51:02.810 ; 00:03:00
2014-09-02 08:06:13.387 ; 00:15:00
2014-09-02 08:37:09.647; 00:31:00
2014-09-02 14:32:00.113 ; 05:55:00
2014-09-02 16:16:42.593; 01:44:00
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Réponse acceptée
Stephen23
le 3 Juil 2018
Modifié(e) : Stephen23
le 3 Juil 2018
8 commentaires
Stephen23
le 3 Juil 2018
Modifié(e) : Stephen23
le 3 Juil 2018
@Noor Bano: yes, you can easily calculate this yourself. Serial date numbers (e.g. the output from datenum) are given in days, so...
- Multiply the days by 24 to get hours (there are 24 hours in 1 day).
- Multiply again by 60 to get minutes (there are 60 minutes in 1 hour).
- Multiply again by 60 to get seconds (there are 60 seconds in 1 minute).
For example, to get the differences in minutes:
>> C = {'2014-09-02 07:48:09.567';'2014-09-02 07:51:02.810';'2014-09-02 08:06:13.387';'2014-09-02 08:37:09.647';'2014-09-02 14:32:00.113';'2014-09-02 16:16:42.593'};
>> N = datenum(C,'yyyy-mm-dd HH:MM:SS.FFF');
>> D = diff(N);
>> D*24*60
ans =
2.8874
15.1763
30.9377
354.8411
104.7080
Plus de réponses (1)
M.Devaki Mohanarangam
le 8 Sep 2023
Time = [16/04/2023 14:44:22
16/04/2023 14:44:26
16/04/2023 14:44:31
16/04/2023 14:44:36
16/04/2023 14:44:42
16/04/2023 14:44:46
16/04/2023 14:44:51
16/04/2023 14:44:56
16/04/2023 14:45:02
16/04/2023 14:45:06 ]
how to find time difference in loop condtion? plz help
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