Dear Sundar, dear Andy
manyt thanks for the length(unique(x)) hint. It solves of course my
problem in a very elegant way. Just of curiosity (or for potential future
problems): how could I solve it in a way, conceptually different, namely,
that the computation on 'meas' being dependent on the variable 'date'?,
means the computation on a variable x in the function passed to aggregate
is conditional on the value of another variable y? I hope you understand
what I mean, let's think of an example:
E.g for the example data.frame below, the sum shall be taken over the
variable meas only for all entries with a corresponding 'data' != 2
for this do I have to nest two aggregate statements, or is there a way
using sapply or similar apply-based commands?
thanks a lot for your kind help.
Cheers!
Christoph
aggregate(data$meas, list(id = data$id), sum)
>
>
> Christoph Lehmann wrote on 4/15/2005 9:51 AM:
> > Hi I have a question concerning aggregation
> >
> > (simple demo code S. below)
> >
> > I have the data.frame
> >
> > id meas date
> > 1 a 0.637513747 1
> > 2 a 0.187710063 2
> > 3 a 0.247098459 2
> > 4 a 0.306447690 3
> > 5 b 0.407573577 2
> > 6 b 0.783255085 2
> > 7 b 0.344265082 3
> > 8 b 0.103893068 3
> > 9 c 0.738649586 1
> > 10 c 0.614154037 2
> > 11 c 0.949924371 3
> > 12 c 0.008187858 4
> >
> > When I want for each id the sum of its meas I do:
> >
> > aggregate(data$meas, list(id = data$id), sum)
> >
> > If I want to know the number of meas(ures) for each id I do, eg
> >
> > aggregate(data$meas, list(id = data$id), length)
> >
> > NOW: Is there a way to compute the number of meas(ures) for each id
with
> > not identical date (e.g using diff()?
> > so that I get eg:
> >
> > id x
> > 1 a 3
> > 2 b 2
> > 3 c 4
> >
> >
> > I am sure it must be possible
> >
> > thanks for any (even short) hint
> >
> > cheers
> > Christoph
> >
> >
> >
> > --------------
> > data <- data.frame(c(rep("a", 4), rep("b", 4), rep("c", 4)),
> > runif(12), c(1, 2, 2, 3, 2, 2, 3, 3, 1, 2, 3, 4))
> > names(data) <- c("id", "meas", "date")
> >
> > m <- aggregate(data$meas, list(id = data$id), sum)
> > names(m) <- c("id", "cum.meas")
> >
>
>
> How about:
>
> m <- aggregate(data["date"], data["id"],
> function(x) length(unique(x)))
>
> --sundar
>
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