RE: [R] aggregation question

From: Christoph Lehmann <christoph.lehmann_at_gmx.ch>
Date: Sat 16 Apr 2005 - 08:31:11 EST

great, Andy! Thanks a lot- I didn't know split. So 'split' can be used as alternative for 'aggregate', with the advantage that in the passed self-defined function one can consider more than one variable of the to-be-aggregated data.frame?

Christoph
> If I understood you correctly, here's one way:
>

> > sumWO2 <- sapply(split(dat, dat$id), function(d) sum(d$meas[d$date !=
> 2]))

> > sumWO2
> a b c
> 0.9439614 0.4481582 1.6967618
>
> Andy
>
>
> > From: Christoph Lehmann
> >
> > 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
> > >
> >
> > --
> > +++ GMX - Die erste Adresse für Mail, Message, More +++
> >

> >
> >
> >
>
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Received on Sat Apr 16 08:38:43 2005

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