I have a column of dates in the format: 16Jun10 and I would like to extract the Julian day. I have various years.
I have tried the functions julian and mdy.date an
Similarly:
require(lubridate) x = as.Date('2010-06-10') yday(x) [1] 161
Also note, using lubridate:
> dmy('16Jun10') [1] "2010-06-16 UTC"