R help - Jan 2010 - 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(1,2,10,20,100,200))

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"
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

On 1/20/2010 5:37 PM, Dimitri Liakhovitski wrote:
> 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(1,2,10,20,100,200))
> 
> 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"
> 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?
> with(x, ave(x=values, factor, FUN=function(x){x/mean(x)}))
[1] 0.6666667 1.3333333 0.6666667 1.3333333 0.6666667 1.3333333

?ave
> Thanks a lot!
-- 
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894

Thanks a lot for your helpful suggestion!
Dimitri

On Wed, Jan 20, 2010 at 5:56 PM, Chuck Cleland <ccleland at optonline.net>
wrote:
> On 1/20/2010 5:37 PM, Dimitri Liakhovitski wrote:
>> 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(1,2,10,20,100,200))
>>
>> 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"
>> 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?
>
>> with(x, ave(x=values, factor, FUN=function(x){x/mean(x)}))
> [1] 0.6666667 1.3333333 0.6666667 1.3333333 0.6666667 1.3333333
>
> ?ave
>
>> Thanks a lot!
>
> --
> Chuck Cleland, Ph.D.
> NDRI, Inc. (www.ndri.org)
> 71 West 23rd Street, 8th floor
> New York, NY 10010
> tel: (212) 845-4495 (Tu, Th)
> tel: (732) 512-0171 (M, W, F)
> fax: (917) 438-0894
>


-- 
Dimitri Liakhovitski
Ninah.com
Dimitri.Liakhovitski at ninah.com

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 <wdunlap at tibco.com>
wrote:
>
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>> -----Original Message-----
>> From: r-help-bounces at r-project.org
>> [mailto:r-help-bounces at 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 at 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.Liakhovitski at ninah.com

Dimitri Liakhovitski <ld7631 <at> gmail.com> writes:
> 
> 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?
I think you are asking how to have the first observation in the ave solution be
calculated as 1 instead of NA.  

As you noted, the ave solution is currently using the default na.rm=F 
(see ?mean). Simply pass na.rm=T in to your custom function in the ave solution
for it to remove the NA and you will get 1 as the average using the ave
approach:

x$std.via.ave<-ave(x$values, x$factor, FUN=function(x)x/mean(x,na.rm=T))

--jason

On Wed, Jan 20, 2010 at 4:37 PM, Dimitri Liakhovitski <ld7631 at
gmail.com> wrote:
> 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(1,2,10,20,100,200))
>
> 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"
> 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?
Here's one way with the plyr package:

library(plyr)
ddply(x, "factor", transform, scaled = values / mean(values, na.rm =
T))

Hadley

-- 
http://had.co.nz/