[R] making zoo objects with zoo without format argument?
Gabor Grothendieck
ggrothendieck at gmail.com
Tue Jul 8 20:43:59 CEST 2008
There is no data in your data frame, just index info, so I assume you
want a zero width time series:
zoo(, as.yearmon(x$Yearmonth, "%Y-%m"))
This also works but then you are left with a character date which you
may not want:
zoo(, x$Yearmonth)
On Tue, Jul 8, 2008 at 1:43 PM, stephen sefick <ssefick at gmail.com> wrote:
> #this is a subset of a larger data frame and I am okay with subsetting it as
> there are redundant time stamps, but I would like to create a zoo object out
> of this and I am having a hard #time figuring out how to do this the date
> structure is year and then month
>
> x <- structure(list(Yearmonth = structure(c(12L, 24L, 1L, 13L, 14L,
> 3L, 15L, 4L, 16L, 5L, 17L, 6L, 18L, 7L, 19L, 8L, 20L, 9L, 21L,
> 10L, 22L, 11L, 23L), .Label = c("2006-02", "2006-03", "2006-04",
> "2006-05", "2006-06", "2006-07", "2006-08", "2006-09", "2006-10",
> "2006-11", "2006-12", "2007-01", "2007-02", "2007-03", "2007-04",
> "2007-05", "2007-06", "2007-07", "2007-08", "2007-09", "2007-10",
> "2007-11", "2007-12", "2008-01"), class = "factor"), Month = c(1L,
> 1L, 2L, 2L, 3L, 4L, 4L, 5L, 5L, 6L, 6L, 7L, 7L, 8L, 8L, 9L, 9L,
> 10L, 10L, 11L, 11L, 12L, 12L)), .Names = c("Yearmonth", "Month"
> ), class = "data.frame", row.names = c(NA, 23L))
>
> #thanks Stephen
>
> --
> Let's not spend our time and resources thinking about things that are so
> little or so large that all they really do for us is puff us up and make us
> feel like gods. We are mammals, and have not exhausted the annoying little
> problems of being mammals.
>
> -K. Mullis
>
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>
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