Duration in double between two datenum

16 vues (au cours des 30 derniers jours)
Luis Ruiz
Luis Ruiz le 7 Sep 2018
Commenté : Peter Perkins le 12 Sep 2018
I have two dates given in text format, I want to have the real duration in seconds between the two values.
The answer is -24, and I can do it parsing the strings. But does MATLAB have a function to do it nice and quick?
If I do the following the answer is not a -24 that I can use as a double:
datenum('2018-09-07 18:36:05.079')-datenum('2018-09-07 18:36:29.079')
I need this time for a Simulink simulation. For example, I might need the duration in seconds between two days.
  2 commentaires
Stephen23
Stephen23 le 7 Sep 2018
"The answer is 24"
The answer is actually -24
Luis Ruiz
Luis Ruiz le 10 Sep 2018
I edited my question to match the answers.

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Réponse acceptée

Stephen23
Stephen23 le 7 Sep 2018
Modifié(e) : Stephen23 le 7 Sep 2018
To get seconds simply multiply the days by 60*60*24:
>> F = 'yyyy-mm-dd HH:MM:SS.FFF';
>> D = datenum('2018-09-07 18:36:05.079',F)-datenum('2018-09-07 18:36:29.079',F);
>> D*60*60*24
ans = -24.000
  2 commentaires
Luis Ruiz
Luis Ruiz le 10 Sep 2018
Modifié(e) : Luis Ruiz le 10 Sep 2018
This one seems to be the right answer, but then, does it mean that operations between two datenum values are always in days?
Stephen23
Stephen23 le 10 Sep 2018
@Luis Ruiz: yes, datenum always returns days. But the conversion to seconds is trivial, as my answer shows.

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Plus de réponses (2)

Peter Perkins
Peter Perkins le 7 Sep 2018
If possible, don't use datenum. Use datetimes:
>> fmt = 'yyyy-MM-dd HH:mm:ss.SSS';
>> dur = datetime('2018-09-07 18:36:05.079','Format',fmt) - datetime('2018-09-07 18:36:29.079','Format',fmt)
dur =
duration
-00:00:24
>> dur.Format = 's'
dur =
duration
-24 sec
  3 commentaires
James Tursa
James Tursa le 7 Sep 2018
To turn it into a double, e.g.
seconds(dur)
Peter Perkins
Peter Perkins le 12 Sep 2018
As James says, you can convert, but the point of duration is that you may not need a number. duration supports all kinds of time arithmetic. Hard to know if that's possible in your case.

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Image Analyst
Image Analyst le 7 Sep 2018
Modifié(e) : Image Analyst le 10 Sep 2018
Try the etime() function.
t1 = datevec('2018-09-08 18:36:05.079','yyyy-mm-dd HH:MM:SS.FFF')
t2 = datevec('2018-09-07 18:36:29.079','yyyy-mm-dd HH:MM:SS.FFF')
elapsedTime = etime(t1, t2) % Results in seconds.

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