From: Christoph Lehmann <christoph.lehmann_at_gmx.ch>

Date: Sat 16 Apr 2005 - 05:23:54 EST

Date: Sat 16 Apr 2005 - 05:23:54 EST

thanks a lot for your kind help.

aggregate(data$meas, list(id = data$id), sum)

*>
**>
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> 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
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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
**>
*

-- ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.htmlReceived on Sat Apr 16 05:28:23 2005

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