convert Julian Date to yyyy-mm-dd
56 次查看(过去 30 天)
显示 更早的评论
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!
1 个评论
James Tursa
2020-10-19
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?
采纳的回答
meghannmarie
2020-10-19
编辑:meghannmarie
2020-10-19
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);
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Dates and Time 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!