Version of IDL used to prepare this article: IDL 8.2. JulTime = ncdfTime / (24*60*60.0D) + JulDay(1,1,1970,0,0,0)ĬalDat, julTime, month, day, year, hour, min, secĪlternatively, you can use the "calendar" format from above to do the conversion for you. The process and use the CalDat function to do the conversion for you. To convert numbers in CF-compliant seconds back to a date, you simply reverse In both cases, the value printed is 1384331482.000.Ĭonverting CF-Compliant Seconds to a Date Note that this can also be accomplished using the “calendar format” specification in IDL. This will work just as well for a string array, too). Would do something like this (using a scalar string as the example, but Would read the two columns of data as a single string array. I need to convert this informaiton to the "time" variable according to NetCDF CF convention, which requires the time to be in seconds since 00:00:00 (in UTC). ![]() QUESTION: I have the date and the UTC time in two columns of my data file as, where YYYY is Year, MM is Month, DD is day, HR is Hours, MI is Minutes, and SE is Seconds. IDL> print, places->map(lambda(x: x + ‘, CO’)) Boulder, CO Denver, CO Fort Collins, CO. ![]() Coyote's Guide to Traditional IDL Graphics.
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