try this:
> dftest<-data.frame(x=1:12, y=(1:12)%%4, z=(1:12)%%2)
> aggregate(cbind(x,y)~z, data=dftest, FUN=mean)
z x y
1 0 7 1
2 1 6 2
> x_test=c("x","y")
> a <- formula(paste0('cbind('
+ , x_test[1]
+ , ','
+ , x_test[2]
+ , ') ~ z'
+ ))
> a
cbind(x, y) ~ z
> aggregate(a, data = dftest, FUN = mean)
z x y
1 0 7 1
2 1 6 2
>
On Thu, May 23, 2013 at 3:05 AM, jpm miao <miao...@gmail.com> wrote:
> Hi,
> From time to time I need to do the aggregation. To illustrate, I present
> a toy example as below. In this example, the task is to aggregate x and y
> by z with the function mean.
> Could I call the aggregation function with x_test, where
> x_test=c("x","y")? Thanks
>
> Miao
>
>
> > dftest<-data.frame(x=1:12, y=(1:12)%%4, z=(1:12)%%2)
> > dftest
> x y z
> 1 1 1 1
> 2 2 2 0
> 3 3 3 1
> 4 4 0 0
> 5 5 1 1
> 6 6 2 0
> 7 7 3 1
> 8 8 0 0
> 9 9 1 1
> 10 10 2 0
> 11 11 3 1
> 12 12 0 0
> > aggregate(cbind(x,y)~z, data=dftest, FUN=mean)
> z x y
> 1 0 7 1
> 2 1 6 2
> > x_test=c("x","y")
> > aggregate(cbind(x_test)~z, data=dftest, FUN=mean)
> Error in model.frame.default(formula = cbind(x_test) ~ z, data = dftest) :
> variable lengths differ (found for 'z')
> a1aggregate(cbind(factor(x_test))~z, data=dftest, FUN=mean)
> Error in model.frame.default(formula = cbind(factor(x_test)) ~ z, data =
> dftest) :
> variable lengths differ (found for 'z')
> > aggregate(factor(x_test)~z, data=dftest, FUN=mean)
> Error in model.frame.default(formula = factor(x_test) ~ z, data = dftest) :
> variable lengths differ (found for 'z')
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
>
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
[[alternative HTML version deleted]]
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