Re: [R] given a mid-month date, get the month-end date

From: jim holtman <jholtman_at_gmail.com>
Date: Tue 20 Dec 2005 - 03:12:37 EST

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 <jholtman@gmail.com> 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 <quantpm@yahoo.com> 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]]
> >
> > ______________________________________________
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> >
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> >
>
>
>
> --
> 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?

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Received on Tue Dec 20 03:22:42 2005

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