Convert from Coordinated UTC to datetime?

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Shayma Al Ali
Shayma Al Ali on 5 Nov 2021
Commented: Dave B on 5 Nov 2021
I have a list of times formatted in coordinated UTC time. For example, the first data entry point I have is
1.511092230713565e+05
where the first six digits are the year, month and day (so 11/09/2015) and the rest of the digits are the fraction of the day. How do I extract the the first six digits and then convert the rest of the digits so I can convert the time to a datenum array?

Answers (2)

Dave B
Dave B on 5 Nov 2021
Edited: Dave B on 5 Nov 2021
You can do this easily with datetime:
x=1.511092230713565e+05;
% The date part is the part before the decimal, so use floor
d = datetime(string(floor(x)),'InputFormat','yyMMdd')
d = datetime
09-Nov-2015
% The time part is fractions of a day, remove the date part and call days
% to get a duration
t = days(x-floor(x))
t = duration
0.22307 days
% The result is the date part (which will be midnight) plus the fraction of
% a day
dt = d+t
dt = datetime
09-Nov-2015 05:21:13
% If you really must use datenum (though this is not recommended, datetime is so much better!)
datenum(dt)
ans = 7.3628e+05

Star Strider
Star Strider on 5 Nov 2021
A different approach —
format long g
nr = 1.511092230713565e+05;
dt = datetime(num2str(fix(nr)),'InputFormat','yyMMdd') + timeofday(datetime(datevec(rem(nr,1))))
dt = datetime
09-Nov-2015 05:21:13
dn = datenum(dt)
dn =
736277.223071357
Also, Specify Time Zones might be of interest.
.
  1 Comment
Dave B
Dave B on 5 Nov 2021
I like the rem(nr,1) better than my x-floor(x), and I think fix is technically the accurate choice if there might be some BC dates?
Less sure about timeofday(datetime(datevec())) vs. days()

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