Get duration of timerange

I have a timetable. I also have a timerange corresponding to data in that timetable. I am having a hard time thinking of how to get the length, or duration, of that timerange.

I feel it should be simple but I cannot think of a compact way to do it.

Answer 1

[Walter Roberson]

R1 = timerange('2018-01-01','2018-05-01','quarters')

R1 = 
	timetable timerange subscript:

		Select timetable rows with times in: QUARTERS
		  Starting at, including:   01-Jan-2018 00:00:00
		  Ending at, but excluding: 01-Jul-2018 00:00:00

	See Select Times in Timetable.

R1S = struct(R1);

Warning: Calling STRUCT on an object prevents the object from hiding its implementation details and should thus be avoided. Use DISP or DISPLAY to see the visible public details of an object. See 'help struct' for more information.

R1S.first

ans = datetime
   01-Jan-2018

R1S.last

ans = datetime
   01-Jul-2018

R1S.unitOfTime

ans = 'quarters'

R1S.last - R1S.first

ans = duration
   4344:00:00

R2 = timerange(seconds(2),seconds(4),'closed')

R2 = 
	timetable timerange subscript:

		Select timetable rows with times in the closed interval:
		[2 sec, 4 sec]

	See Select Times in Timetable.

R2S = struct(R2);

Warning: Calling STRUCT on an object prevents the object from hiding its implementation details and should thus be avoided. Use DISP or DISPLAY to see the visible public details of an object. See 'help struct' for more information.

R2S.first

ans = duration
   2 sec

R2S.last

ans = duration
   4 sec

R2S.unitOfTime

ans =

  0×0 empty char array

R2S.last - R2S.first

ans = duration
   2 sec

But what does the "length" or "duration" mean of a time range? In the case of a time range constructed from seconds it is fairly clear, but if the time range was constructed from quarter then do you need the "length" to be the number of quarters? If it was constructed from months do you need the length to be the number of months?

You can see in the quarters example that the difference between last and first is a duration, which by default gets expressed by hours minutes seconds. But because the range was constructed in quarters (a calendar duration) then looking at the duration in hours in misleading -- if you were to add that number of hours to a quarter at a different point in the year the end would probably not be the end of a quarter (quarters are not exactly equal length, and cannot be because 365 is not exactly divisible by 4)

You might have noticed the Warning message in the struct() call. timerange() does not expose any public methods or properties that allow you to get at the endpoint information, so you have to hack it.

[Poison Idea fan]

This is helpful. I needed to get the time in minutes of an average period from a time range I select from data stored in a timetable. I had only previously implemented the default parameters on timerange, so I did not think to use 'quarters', or return a structure.

Thanks for the help, Walter.

[Walter Roberson]

Note that I am not suggesting using 'quarters' or 'months' or whatever... I had no idea how you had expressed your time ranges, and my Answer was pointing out that the meaning of the question was not clear and could depend upon which interval you had used.

Converting an object to a struct() is not recommended, but it can be useful sometimes. By the way you can use

warning('off', 'MATLAB:structOnObject')

but it is recommended that you turn the warning back on afterwards.

[Walter Roberson]

You can call hours on a duration that you get by subtracting last-first

[Poison Idea fan]

Yeah I just realized I don't need to use quarters. Thanks for the tip on turning the warning off.