converting UTC time formatted as 'ddd:HH:mm:ss.SSSSSSSS' from a string to a date time.

I have an array to strings that I need converted to a date time, but they are in an unusual format. The timezone is UTC, but the format of the date doesnt include the year. I do have the option of having the user input the the year or parsing it from another place. However the format of the time is 'ddd:HH:mm:ss.SSSSSSSSS', and I am struggling to figure out the right set of datetime comands in to convert it out of a string.

example:

User Input: Current year = 2022

UTC String: 036:19:45:30.123581351

Date: 02-05-2022 07:45:30.123581351 pm

Help would be appriciated! Thanks!

[Stephen23]

"02-05-2022"

This is an international community, that date format is ambiguous and should be avoided.

https://en.wikipedia.org/wiki/Date_format_by_country

To be unambiguous use ISO 8601 date formats:

https://en.wikipedia.org/wiki/ISO_8601

https://xkcd.com/1179/

Answer 1

[Stephen23]

A = '2022';
B = '036:19:45:30.123581351';
C = sprintf('%s:%s',A,B);
D = datetime(C,'InputFormat','u:D:H:m:s.SSSSSSSSS')

D = datetime
   05-Feb-2022 19:45:30

You can adjust the display format by specifying the 'Format' property. Note that changing the 'Format' makes absolutely no difference to the date/time stored in memory, only to how it looks when displayed.

[Peter Perkins]

It turns out that datetime will assume the current year, so this can be even simpler:

>> t = datetime('036:19:45:30.123581351','InputFormat','D:H:m:s.SSSSSSSSS')
t = 
  datetime
   05-Feb-2022 19:45:30

If the data are not from the current year, it's easy to change them en masse:

>> t.Year = 2020 % works on a vector of datetimes too
t = 
  datetime
   05-Feb-2020 19:45:30

[Stephen23]

"If the data are not from the current year, it's easy to change them en masse:"

@Peter Perkins: does that simply change the unit value, or does it keep a fixed duration from the start of the year?

For example, what happens if we change the unit to/from a leap year?

[Peter Perkins]

I'm not sure I fully understand the question. Setting the year property is "as if" you got all six y/mo/d/h/mi/s components, changed the year, and recreated. The clockface stays the same except for the year:

>> d = datetime(2022,2,27:31)
d = 
  1×5 datetime array
   27-Feb-2022   28-Feb-2022   01-Mar-2022   02-Mar-2022   03-Mar-2022
>> d.Year = 2023
d = 
  1×5 datetime array
   27-Feb-2023   28-Feb-2023   01-Mar-2023   02-Mar-2023   03-Mar-2023

And if you did not start out with a 29 Feb, you won't end up with one if you set to a leap year:

>> d.Year = 2020
d = 
  1×5 datetime array
   27-Feb-2020   28-Feb-2020   01-Mar-2020   02-Mar-2020   03-Mar-2020
>> caldiff(d,"days")
ans = 
  1×4 calendarDuration array
   1d   2d   1d   1d

I think you are asking, "yeah, but what happens to 29 Feb if I set a leap year to a non-leap year?" There are several ways to specify a non-existent date or time, and that has to be handled. In this case "29-Feb-2022" gets rolled into 1-Mar-2022:

>> d = datetime(2020,2,27:31)
d = 
  1×5 datetime array
   27-Feb-2020   28-Feb-2020   29-Feb-2020   01-Mar-2020   02-Mar-2020
>> d.Year = 2022
d = 
  1×5 datetime array
   27-Feb-2022   28-Feb-2022   01-Mar-2022   01-Mar-2022   02-Mar-2022
>> caldiff(d,"days")
ans = 
  1×4 calendarDuration array
   1d   1d   0d   1d

There's no right answer here, calendar arithmetic is messed up and there's nothing to be done about that.

[Tim Starbuck]

Both of these were very helpful, and gave me the solution I needed! Thank you!