On Nov 10, 2009, at 10:49 AM, baptiste auguie wrote:

Dear list,

subset has a 'drop' argument that I had often mistaken for the one in
[.factor which removes unused levels.
Clearly it doesn't work that way, as shown below,

d <- data.frame(x = factor(letters[1:15]), y = factor(LETTERS[1:3]))
s <- subset(d, y=="A", drop=TRUE)
str(s)
'data.frame':   5 obs. of  2 variables:
$ x: Factor w/ 15 levels "a","b","c","d",..: 1 4 7 10 13
$ y: Factor w/ 3 levels "A","B","C": 1 1 1 1 1

The subset still retains all the unused factor levels. I wonder how
people usually get rid of all unused levels in a data.frame after
subsetting? I came up with this but I may have missed a better
built-in solution,

dropit <- function (d, columns = names(d), ...)
{
   d[columns] = lapply(d[columns], "[", drop=TRUE, ...)
   d
}


If you are looking for a one-liner, then consider:

data.frame(lapply(s, function(x) if (is.factor(x)){ factor(x)} else {x}))

I added a numeric column to make sure I had not clobbered a non-factor variable.

> d <- data.frame(x = factor(letters[1:15]), y = factor(LETTERS[1:3]), N=1:15)
> s <- subset(d, y=="A", drop=TRUE)
> str( data.frame(lapply(s, function(x) if (is.factor(x)){ factor(x)} else {x})) )
'data.frame':   5 obs. of  3 variables:
 $ x: Factor w/ 5 levels "a","d","g","j",..: 1 2 3 4 5
 $ y: Factor w/ 1 level "A": 1 1 1 1 1
 $ N: int  1 4 7 10 13


str(dropit(s))
'data.frame':   5 obs. of  2 variables:
$ x: Factor w/ 5 levels "a","d","g","j",..: 1 2 3 4 5
$ y: Factor w/ 1 level "A": 1 1 1 1 1


Best regards,

baptiste

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