convert Julian Date to yyyy-mm-dd

I am analyzing the vertical profiles of the profiling robotic float ( ftp://ftp.ifremer.fr/ifremer/argo/dac/coriolis/6902905/ ). The dates are there stored as Julian Dates, field JULD. As I was converting them into format yyyy-mm-dd on Matlab, difficulties occurred. The Julian date used here is not as usual (first 2 digits year and last 3 digits days w.r.t. Jan 1st, nor the number of days w.r.t. 4713 BCE ). For example here 25856 represents 17-Oct-2020, which is confusing to me. Could someone please provide some instruction on this? Thank you very much!

[James Tursa]

Can you explain how 25856 represents 17-Oct-2020? If the "first 2 digits year" is 25, how does that and 856 days result in a date in 2020?

Answer 1

[meghannmarie]

Your number represents "days since 1950-01-01 00:00:00 UTC". Most netcdf times are stored like this, you need to look at attribute on the time field in the netcdf file to so see whether it is hours since or days since and the date of origin. To convert do this:

epoch = datetime(1950,01,01);
time = epoch + days(jd);

[Stephen23]

+1 Possibly CALDAYS is required:

D = datetime(1950,01,01, 'Format','u-MM-dd HH:mm:ss') + caldays(25856)

D = datetime
   2020-10-16 00:00:00

Answer 2

[Adam Danz]

Are you looking for this?

datetime(25856,'convertfrom','juliandate','Format','yyy-MM-dd')

ans = datetime
   4643-09-09

which makes sense given,

datetime(0,'convertfrom','juliandate','Format','yyy-MMM-dd') + days(25856)

ans = datetime
   4643-Sep-09

Inspired by this answer.