On Mon, Feb 25, 2008 at 7:40 AM, Peter Dalgaard <[EMAIL PROTECTED]> wrote: > > Gabor Grothendieck wrote: > > On Mon, Feb 25, 2008 at 6:03 AM, Peter Dalgaard > > <[EMAIL PROTECTED]> wrote: > > > >> Gabor Grothendieck wrote: > >> > >>> In looking at this again here is a slight simplification. Its now > >>> only one line: > >>> > >>> > >>> > >>>> library(chron) > >>>> x <- c("01/00/05", "01/22/06") > >>>> as.chron(sub("/00/", "/15/", x)) + (regexpr("/00/", x) > 0) / 2 > >>>> > >>>> > >>> [1] (01/15/05 12:00:00) (01/22/06 00:00:00) > >>> > >>> > >> You don't really need chron here, do you? > >> > >> as.Date(sub("/00/", "/15/", x), format="%m/%d/%y") > >> > >> (The format spec seems to have been left out below. Also, beware the > >> system-dependence of %y.) > >> > > > > Yes, you need chron since entire point was to encode the missings as > > noon so one can reverse the procedure and Date does not support times. > > Also the format was omitted because its not required. m/d/y is the default > > for chron. > > > But the _original_ question involved as.Date and was missing the format > in its sample code: > > > > interesting.data$date > > > [1] "1/22/93" "1/22/93" "1/23/93" "1/00/93" "1/28/93" "1/31/93" "1/12/93" > > > > as.Date(interesting.data$date) > > > [1] "1993-01-22" "1993-01-22" "1993-01-23" NA "1993-01-28" "1993-01-31" > "1993-01-12" > > > Encoding missing values as a specific time of day was your own invention > and might as well be done otherwise, e.g. as > > missingDate <- (regexpr("/00/", x) > 0) > >
But the whole point of this seems to be to encode the data in a single vector of dates. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.