Omit rep. You just want a <- zoo(1:9, ...). To get the last day of
the month you don`t need b since as.Date.yearmon will give it with the
argument frac = 1:
> aggregate(a, as.Date(as.yearmon(time(a)), frac = 1), tail, 1)
2009-03-31 2009-04-30 2009-05-31 2009-06-30
2 5 8 9
On Thu, Apr 8, 2010 at 11:18 AM, Sergey Goriatchev <serg...@gmail.com> wrote:
> Hello, everyone
>
> I have the following problem:
> Say I have an irregular zoo timeseries like this:
>
> a <- zoo(rep(1:9), as.Date(c("2009-03-20", "2009-03-27", "2009-04-24",
> "2009-04-25", "2009-04-30", "2009-05-15", "2009-05-22", "2009-05-29",
> "2009-06-26")))
>
> and I have regular zoo timeseries like this:
>
> b <- zoo(rep(1:4), as.Date(c("2009-03-31", "2009-04-30", "2009-05-31",
> "2009-06-30")))
>
> >From "a" I need to extract those rows that hold values for the last
> day of each month (creating series "c"). Then I have to merge these
> values with "b", such that the result has the index of "c".
>
> How could I do this most efficiently?
>
> Thank you in advance!
>
> Best,
> Sergey
>
> --
> Simplicity is the last step of art./Bruce Lee
>
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> and provide commented, minimal, self-contained, reproducible code.
>
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