Duration in double between two datenum

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.

[Stephen23]

"The answer is 24"

The answer is actually -24

[Luis Ruiz]

I edited my question to match the answers.

Answer 1

[Stephen23]

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

[Luis Ruiz]

This one seems to be the right answer, but then, does it mean that operations between two datenum values are always in days?

[Stephen23]

@Luis Ruiz: yes, datenum always returns days. But the conversion to seconds is trivial, as my answer shows.

Answer 2

[Peter Perkins]

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

[Stephen23]

@Peter Perkins: which of these will give a numeric value 24 for further processing? From the displayed representation it seems that none of these are numeric types, so it would be of interest to know which one could be used in an algorithm as a numeric value. (e.g. I wrote a hash function to encode data based on the seconds, which requires arithmetic operations... mod, times, etc).

[James Tursa]

To turn it into a double, e.g.

seconds(dur)

[Peter Perkins]

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.

Answer 3

[Image Analyst]

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.