[R] aggregation question

[Liaw, Andy]

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
> > 
> 
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