Hi Juli,
What you can do is to make your outlier remover into a function like this:
remove_outlier_by_sd<-function(x,nsd=3) {
meanx<-mean(x,na.rm=TRUE)
sdx<-sd(x,na.rm=TRUE)
return(x[abs(x-xmean) < nsd*sdx])
}
Then apply the function to your data frame ("table")
newDA<-sapply(DA,remove_outlier_by_sd)
newDA will be a list, as it is likely that its elements will be of
different lengths. You may be told that you really shouldn't remove
outliers and learn to love them, but I will leave that to others.
Jim
On Sat, Sep 12, 2015 at 12:15 AM, Juli <julianeleusch...@web.de> wrote:
> Hey,
>
> i want to remove outliers so I tried do do this:
>
> # 1 define mean and sd
> sd.AT_ZU_SPAET <- sd(AT_ZU_SPAET)
> mitt.AT_ZU_SPAET <- mean(AT_ZU_SPAET)
> #
> sd.Anzahl_BAF <- sd(Anzahl_BAF)
> mitt.Anzahl_BAF <- mean(Anzahl_BAF)
> #
> sd.Änderungsintervall <- sd(Änderungsintervall)
> mitt.Änderungsintervall <- mean(Änderungsintervall)
> #
> # 2 identify outliers
> DA[ abs(AT_ZU_SPAET - mitt.AT_ZU_SPAET) > ( 3 * sd.AT_ZU_SPAET) , ]
> DA[ abs(Anzahl_BAF - mitt.Anzahl_BAF) > ( 3 * sd.Anzahl_BAF) , ]
> DA[ abs(Änderungsintervall - mitt.Änderungsintervall) > ( 3 *
> sd.Änderungsintervall) , ]
> #
> # 3 remove outliers
> AT_ZU_SPAET.clean <- DA[ (abs(AT_ZU_SPAET - mitt.AT_ZU_SPAET) <
> (3*sd.AT_ZU_SPAET)), ]
> Anzahl_BAF.clean <- DA[ (abs(Anzahl_BAF - mitt.Anzahl_BAF) <
> (3*sd.Anzahl_BAF)), ]
> Änderungsintervall.clean <- DA[ (abs(Änderungsintervall -
> mitt.Änderungsintervall) <
> (3*sd.Änderungsintervall)), ]
>
> My problem ist, that I am only able to remove the outliers of one column of
> my table, but I want to remove the outliers of every column of the table.
>
> Could anybody help me?
>
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/removing-outlier-tp4712137.html
> Sent from the R help mailing list archive at Nabble.com.
>
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