Using Date objects, calculates given
dates as fractional years.
Usage
get.yrs(dates, format = "%Y-%m-%d", year.length = "approx")
Arguments
dates
a vector or column of Date objects or right kind of character strings, see Details
format
a character string; if dates is a character vector,
specifies the format; see as.Date
year.length
character string, either 'actual' or
'approx'; can be abbreviated; see Details
Details
dates should preferably be a date, Date or IDate
object,
although they can also be character strings in a format
specified by format (passed to as.Date).
When year.length = 'actual', fractional years are calculated as
year + day_in_year/365 for non-leap-years
and as year + day_in_year/366 for leap years.
If year.length = 'approx', fractional years are always
calculated as in year + day_in_year/365.242199.
There is a slight difference, then, between the two methods
when calculating durations between fractional years. For
meticulous accuracy one might instead want to calculate durations using
dates (days) and convert the results to fractional years.
Note that dates are effectively converted to fractional years at
00:00:01 o'clock:
get.yrs("2000-01-01") = 2000, and
get.yrs("2000-01-02") = 2000 + 1/365.242199.