Thank you!

Am 26.09.2012 13:31, schrieb Jim Lemon [via R]:
> On 09/26/2012 08:53 PM, Tagmarie wrote:
>
> > Hello,
> > I tried for about three hours now to solve this problem but I can't 
> figure
> > it out. I am sure someone knows how do it. At least I hope so.
> >
> > I have a data frame somewhat like this:
> >
> > myframe<- data.frame (ID=c("Ernie", "Ernie", "Bert", "Bert"),
> > Timestamp=c("24.09.2012 09:00", "24.09.2012 09:00", "24.09.2012 10:00",
> > "25.09.2012 10:00"), Hunger=c("1","5","2","2"), 
> Temp=c("25","30","27","28")
> > )
> > myframe
> >
> > As you can see for Ernie I do have different data for 24.09.2012 
> 9:00. Now I
> > would like to average the Hunger and Temp value for this timestamp 
> to get a
> > data frame without duplicated Times and the respective average Temp and
> > Hunger.
> >
> > I tried something like
> > Meanframe<-  by(myframe[, 3:4], duplicated(myframe$ID,
> > Zusatzdaten3$Timestamp) == TRUE, mean)
> > but it doesn't work and I guess that it is also totally crap ;-)
> >
> Hi Tagmarie,
> Your problem is that both Hunger and Temp are read in as factors. If you
> try it like this:
>
> by(as.numeric(as.character(myframe[,3])),myframe[,"ID"],mean)
> by(as.numeric(as.character(myframe[,4])),myframe[,"ID"],mean)
>
> You might get what you want. The "as.character" call is necessary,
> otherwise you will get the wrong mean values.
>
> Jim
>
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