Regression stars

I thought that the default was the way it was for performance reasons. For large data.frames or repeated applications, using factors should be faster for non-trivial strings.

fs <- c('apple','peach','watermelon','spinach','persimmon','potato','kale')
n <- 1000000

a1 <- data.frame(f=sample(fs,n,replace=TRUE), x1=rnorm(n), x2=rnorm(n), stringsAsFactors=TRUE)
a2 <- data.frame(f=sample(fs,n,replace=TRUE), x1=rnorm(n), x2=rnorm(n), stringsAsFactors=FALSE)

fn <- function(i,x) x[x$f %in% c('kale','spinach'),]
system.time(z <- sapply(1:100, fn, a1))

  user  system elapsed 
19.614   4.037  24.649 

system.time(z <- sapply(1:100, fn, a2))

  user  system elapsed 
19.726   7.715  36.761 

system.time(z <- sapply(1:100, fn, a1))

rm(z)
gc()

system.time(z <- sapply(1:100, fn, a2))

On Feb 12, 2013, at 10:40 AM, Ben Bolker <bbolker at gmail.com> wrote:

Thanks, Uwe.
Now let me go one step farther.

Can you (or anyone) give a good argument **other than backward
compatibility** for keeping the stringAsFactors=TRUE argument on
data.frame()?

I appreciate your distinction between data.frame() and read.table()'s
use of stringAsFactors, and I can see that there is some point for
quick-and-dirty interactive use in setting all non-numeric variables to
factors (arguing that wanting non-numerics as factors is somewhat more
common than wanting them as strings).

It might be nice to add an optional stringsAsFactors (and check.names)
argument to transform(): I've had to write my own Transform() function
to allow the defaults to be overridden, since transform() calls
data.frame() with the defaults.  (Setting the stringsAsFactors option
globally would work, although not for check.names.)

Ben BOlker

What I will likely do is
make a few changes so that character vectors are automatically changed
to factors in modelling functions, so that operating with
stringsAsFactors=FALSE doesn't trigger silly warnings.

Duncan Murdoch

[apologies for snipping context: "gmane made me do it"]