Arithmetic operations on R factors

Question

I have an R dataframe and I'm trying to subtract one column from another. I extract the columns using the $ operator but the class of the columns is 'factor' and R won't perform arithmetic operations on factors. Are there special functions to do this?

Answer 1

If you really want the levels of the factor to be used, you're either doing something very wrong or too clever for its own good.

If what you have is a factor containing numbers stored in the levels of the factor, then you want to coerce it to numeric first using as.numeric(as.character(...)):

dat <- data.frame(f=as.character(runif(10)))

You can see the difference between accessing the factor indices and assigning the factor contents here:

> as.numeric(dat$f)
 [1]  9  7  2  1  4  6  5  3 10  8
> as.numeric(as.character(dat$f))
 [1] 0.6369432 0.4455214 0.1204000 0.0336245 0.2731787 0.4219241 0.2910194
 [8] 0.1868443 0.9443593 0.5784658

Timings vs. an alternative approach which only does the conversion on the levels shows it's faster if levels are not unique to each element:

dat <- data.frame( f = sample(as.character(runif(10)),10^4,replace=TRUE) )
library(microbenchmark)
microbenchmark(
  as.numeric(as.character(dat$f)),
  as.numeric( levels(dat$f) )[dat$f] ,
  as.numeric( levels(dat$f)[dat$f] ),
  times=50
  )

                              expr     min      lq  median      uq     max
1  as.numeric(as.character(dat$f)) 7835865 7869228 7919699 7998399 9576694
2 as.numeric(levels(dat$f))[dat$f]  237814  242947  255778  270321  371263
3 as.numeric(levels(dat$f)[dat$f]) 7817045 7905156 7964610 8121583 9297819

Therefore, if length(levels(dat$f)) < length(dat$f), use as.numeric(levels(dat$f))[dat$f] for a substantial speed gain.

If length(levels(dat$f)) is approximately equal to length(dat$f), there is no speed gain:

dat <- data.frame( f = as.character(runif(10^4) ) )
library(microbenchmark)
microbenchmark(
  as.numeric(as.character(dat$f)),
  as.numeric( levels(dat$f) )[dat$f] ,
  as.numeric( levels(dat$f)[dat$f] ),
  times=50
  )

                              expr     min      lq  median      uq      max
1  as.numeric(as.character(dat$f)) 7986423 8036895 8101480 8202850 12522842
2 as.numeric(levels(dat$f))[dat$f] 7815335 7866661 7949640 8102764 15809456
3 as.numeric(levels(dat$f)[dat$f]) 7989845 8040316 8122012 8330312 10420161

Answer 2

You can define your own operators to do that, see ? Arith. Without group generics, you can define your own binary operators %operator%:

%-% <- function (factor1, factor2){
  # put in the code here to calculate difference 
  # of two factors (e.g. facor1 level cat - factor2 level mouse = ?)
}