[R] aggregation question

[Liaw, Andy]

> From: Christoph Lehmann
> 
> 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?

split() only split the data frame into a list of data frames, according to
the variable supplied as the second argument.  You can then use
sapply()/lapply() to apply the same operation on each piece, where each
piece contains all the variables.

Andy

 
> 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
> > > > 
> > > 
> > > -- 
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> > 
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