convert Julian Date to yyyy-mm-dd

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Qiming Sun
Qiming Sun on 19 Oct 2020
Edited: Stephen23 on 17 Jul 2023
I am analyzing the vertical profiles of the profiling robotic float ( ). 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!
  1 Comment
James Tursa
James Tursa on 19 Oct 2020
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?

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Accepted Answer

meghannmarie on 19 Oct 2020
Edited: meghannmarie on 19 Oct 2020
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);
  1 Comment
Stephen23 on 17 Jul 2023
Edited: Stephen23 on 17 Jul 2023
+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

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More Answers (1)

Adam Danz
Adam Danz on 19 Oct 2020
Edited: Adam Danz on 23 Oct 2020
Are you looking for this?
ans = datetime
which makes sense given,
datetime(0,'convertfrom','juliandate','Format','yyy-MMM-dd') + days(25856)
ans = datetime
Inspired by this answer.


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