[R] Month end calculations

[Gabor Grothendieck]

The last line is wrong (see below for correction):

On 8/30/07, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:
> The zoo package includes the "yearmon" class to facilitate such
> manipulations.  Here are a few solutions assuming you store
> you series in a zoo variable:
>
> # test data
> library(zoo)
> z <- zoo(1001:1100, as.Date(101:200))[-(45:55)]
>
> # Solution 1.  tapply produces indexes of last of month
> tt <- time(z)
> z[ c(tapply(seq_along(tt), as.yearmon(tt), tail, 1)) ]
>
> # If we want to create a last variable which corresponds
> # to last in sas then do it this slightly longer way:
>
> # Solution 2
> tt <- time(z)
> last <- seq_along(tt) %in% tapply(seq_along(tt), as.yearmon(tt), tail, 1)
> z[last]
>
> # Solution 3. another solution with a last variable.  f(x) is
> # vector same length as x with all 0's except last element is 1.
> tt <- time(z)
> f <- function(x) replace(0*x, length(x), 1)
> last <- ave(seq_along(tt), as.yearmon(tt), FUN = f)
> z[last]

This last line should be:

z[last == 1]


>
> In all these solutions the last point in the series is always
> included.
>
> We have not assumed that every day is necessarily included in your
> series but if every day is included then even simpler solutions
> are possible.
>
> On 8/29/07, Shubha Vishwanath Karanth <shubhak at ambaresearch.com> wrote:
> > Hi R users,
> >
> >
> >
> > Is there a function in R, which does some calculation only for the month
> > end in a daily data?... In other words, is there a command in R,
> > equivalent to "last." function in SAS?
> >
> >
> >
> > BR, Shubha
> >
> >
> >        [[alternative HTML version deleted]]
> >
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