[R] subsetting, aggregating and zoo

Gabor Grothendieck ggrothendieck at gmail.com
Sun Oct 29 19:23:40 CET 2006


Sorry, the line starting idx <- should have time(z) in place of z.
That is,

year <- as.Date(c(
"1988-01-13", "1988-01-14", "1988-01-16", "1988-01-20", "1988-01-21",
"1988-01-22", "1988-01-25", "1988-01-26", "1988-01-27", "1988-01-28"))

x <- c(
 7.973946,  9.933518,  7.978227,  7.512960,  6.641862,  5.667780,  5.721358,
 6.863729,   9.600000,   9.049846)

z <- zoo(x, year)

idx <- cumsum(c(1, diff(time(z)) != 1))

starts <- time(z)[match(idx, idx)]
ends <- time(z)[cumsum(table(idx))[idx]]

aggregate(z, starts, mean)


By the way, dput(v, control = "all") will output variable v
in a form easily pastable by someone else into their session.

On 10/29/06, antonio rodriguez <antonio.raju at gmail.com> wrote:
> Gabor Grothendieck escribió:
> > Try this:
> >
> > # test data
> > x <- c(1:4, 6:8, 10:14)
> > z <- zoo(x, as.Date(x))
> >
> > # idx is 1 for first run, 2 for second run, etc.
> > idx <- cumsum(c(1, diff(z) != 1))
> >
> > # starts replaces each time with the start time of that run
> > # ends is similar but for ends
> > starts <- time(z)[match(idx, idx)]
> > ends <- time(z)[cumsum(table(idx))[idx]]
> >
> > # average over each run using the time of the end of run for the result
> > # replace ends with starts if that is preferred
> > aggregate(z, ends, mean)
> >
> Yes it's OK in your example, but when I try to do it with my data I
> don't get the same figure.
>
> is.zoo(z)
> [1]TRUE
>
> atributes(z)
> $index
>   [1] "1988-01-13" "1988-01-14" "1988-01-16" "1988-01-20" "1988-01-21"
> ..................................................................................................
> [3861] "2005-12-20" "2005-12-23" "2005-12-24" "2005-12-25" "2005-12-26"
> [3866] "2005-12-27" "2005-12-30"
>
> $class
> [1] "zoo"
>
> z[1:10]
>
> 1988-01-13 1988-01-14 1988-01-16 1988-01-20 1988-01-21 1988-01-22 1988-01-25
>  7.973946   9.933518   7.978227   7.512960   6.641862   5.667780   5.721358
> 1988-01-26 1988-01-27 1988-01-28
>  6.863729   9.600000   9.049846
>
> If I follow your instructions,
>
> idx <- cumsum(c(1, diff(z) != 1))
> starts <- time(z)[match(idx, idx)]
> ends <- time(z)[cumsum(table(idx))[idx]]
>
> s1 <- aggregate(z, starts, mean)
> s1[1:10]
>
> 1988-01-13 1988-01-14 1988-01-16 1988-01-20 1988-01-21 1988-01-22 1988-01-25
>  7.973946   9.933518   7.978227   7.512960   6.641862   5.667780   5.721358
> 1988-01-26 1988-01-27 1988-01-28
>  6.863729   9.600000   9.049846
>
> s2 <- aggregate(z, starts, mean)
> s2[1:10]
>
> 1988-01-13 1988-01-14 1988-01-16 1988-01-20 1988-01-21 1988-01-22 1988-01-25
>  7.973946   9.933518   7.978227   7.512960   6.641862   5.667780   5.721358
> 1988-01-26 1988-01-27 1988-01-28
>  6.863729   9.600000   9.049846
>
>
> Always the same. Don't know why (there are not NA's in the series)
>
> Antonio
>
>
>
>
>
>
>



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