One follow up question - the proposed solution was (notice - this time
I am introducing one NA in data frame "x")
x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,NA,10,20,100,200))
x$std.via.ave<-ave(x$values, x$factor, FUN=function(x)x/mean(x))
I compared the result to my own clumsy solution:
factor.level.means<-as.data.frame(tapply(x$values,x$factor,mean, na.rm=T))
factor.level.means$factor<-row.names(factor.level.means)
names(factor.level.means)[1]<-"means"
factor.level.means
x$std<-NA
for(i in 1:nrow(x)){ #i<-1
x[i,"std"]<-factor.level.means[factor.level.means$factor==x[i,"factor"],"means"]
}
x$std<-x$values/x$std
If one compares x$std to x$std.via.ave - one notices that ave results
in an NA for the very first observation - because it seems to be using
na.rm=F when it calculates the means.
Is there a way to fix that in the ave solution?
Thank you!
Dimitri
On Wed, Jan 20, 2010 at 5:55 PM, William Dunlap <wdun...@tibco.com> wrote:
>
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>> -----Original Message-----
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri
>> Liakhovitski
>> Sent: Wednesday, January 20, 2010 2:38 PM
>> To: r-help
>> Subject: [R] standardizing one variable by dividing each
>> value by the mean -but within levels of a factor
>>
>> Hello!
>>
>> I have a data frame with a factor and a numeric variable:
>>
>> x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,1
>> 0,20,100,200))
>>
>> For each level of "factor" - I would like to divide each value of
>> "values" by the mean of "values" that corresponds to the level of
>> "factor"
>
> ave() can do it:
> > ave(x$values, x$factor, FUN=function(x)x/mean(x))
> [1] 0.6666667 1.3333333 0.6666667 1.3333333 0.6666667 1.3333333
> The plyr package has functions which extend this
> sort of thing.
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>
>> In other words, I would like to get a new variable that is equal to:
>> 1/1.5
>> 2/1.5
>> 10/15
>> 20/15
>> 100/150
>> 200/150
>>
>> I realize I could do it through tapply starting with:
>> factor.level.means<-tapply(x$values,x$factor,mean) ... etc.
>>
>>
>> But it seems clunky to me.
>> Is there a more elegant way of doing it?
>>
>> Thanks a lot!
>>
>>
>> --
>> Dimitri Liakhovitski
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
--
Dimitri Liakhovitski
Ninah.com
dimitri.liakhovit...@ninah.com
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