Forgot you were asking for the end date, so just subtract a day: > seq(x, by='month', length=2)[2] - 24*3600 [1] "2005-12-31 EST"
On 12/19/05, jim holtman <[EMAIL PROTECTED]> wrote: > > Here is one way using POSIX: (you can create a function to do this) > > > x <- as.POSIXlt('2005-12-16') # a date > > x > [1] "2005-12-16" > > dput(x) #structure of the date > structure(list(sec = 0, min = 0, hour = 0, mday = 16, mon = 11, > year = 105, wday = 5, yday = 349, isdst = 0), .Names = c("sec", > "min", "hour", "mday", "mon", "year", "wday", "yday", "isdst" > ), class = c("POSIXt", "POSIXlt")) > > x$mday <- 1 # reset to first of the month > > seq(x, by='month', length=2)[2] # select 2nd number in the sequence > [1] "2006-01-01 EST" > > > > > > On 12/19/05, t c <[EMAIL PROTECTED]> wrote: > > > > I have a vector of dates. > > > > I wish to find the month end date for each. > > > > Any suggestions? > > > > e.g. > > > > For 12/15/05, I want 12/31/05, > > > > For 10/15/1995, I want 10/31/1995, etc > > > > > > __________________________________________________ > > > > > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help@stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide! > > http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html> > > > > > > -- > Jim Holtman > Cincinnati, OH > +1 513 247 0281 > > What the problem you are trying to solve? -- Jim Holtman Cincinnati, OH +1 513 247 0281 What the problem you are trying to solve? [[alternative HTML version deleted]] ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html