[S] Summary: Changing default ordering of factors

["Gérald Jean" <Gerald.Jean@spgdag.ca>]

Hello Susers,

yesterday I posted the following message regarding the default ordering of
factors in S+.

<Hello everyone,

< I am a statistician working  in an actuarian group, I mostly develop
GLM's
<for them and they like to pick their own reference level in each factor
<rather than having Splus pick it for them.  By default Splus will pick the
<first level in the alphabetically sorted list of levels.  My question is
<how do we change this default behaviour or if this is not possible how to
<change the default ordering of factor levels.  I know how to do this
<manually, but since I have a whole bunch to do with a lot of levels and
<most likely I'll have to do it often I would like to have a function do it
<for me.  Here is a little example of what I'd like to automate.
<
<> ttt.age
<[1] 19 20 30 30 19 20 20 19 19
<> levels(ttt.age)
<[1] "19" "30" "20"
<> table(ttt.age)
< 19 30 20
<  4  2  3
<> levels(ttt.age)_list('30'='30','20'='20','19'='19')
<> levels(ttt.age)
<[1] "30" "20" "19"
<> table(ttt.age)
< 30 20 19
<  2  3  4
<> ttt.age
<[1] 19 20 30 30 19 20 20 19 19
<
<ttt.age have not been changed, but now if it was an explanatory variable
in
< a data.frame Splus would use level "30" as the reference
<rather than the default level  "19".

Thanks to all who responded.  Bert Gunter and Nick Ellis suggested using
the S+ "ordered" function to order the levels of the factors the way I
wanted them.  Victor Moreno and Brad Biggerstaff suggested using Brian
Ripley's function "relevel" which I ended up doing.  It worked fine.  Here
it is.

x   : is the factor for which we want to change the reference level
ref : is either the character name of the new reference level or its index
in "levels(x)"

relevel <- function(x, ref, ...) UseMethod("relevel")
relevel.default <- function(x, ref, ...) stop("relevel only for factors")
relevel.ordered <- function(x, ref, ...) stop("relevel only for factors")
relevel.factor <- function(x, ref, ...)
{
  lev <- levels(x); nlev <- length(lev)
  if(is.character(ref)) ref <- match(ref, lev)
  if(is.na(ref)) stop("ref must be an existing level")
  if(ref < 1 || ref > nlev) stop(paste("ref =", ref, "must be in 1 :",
nlev))
  factor(x, levels = lev[c(ref, seq(along=lev)[-ref])] )
}

Thanks again,

Gérald Jean
Analyste-conseil (statistiques), Actuariat
télephone            : (418) 835-8839
télecopieur          : (418) 835-5865
courrier électronique: gerald.jean@spgdag.ca

"In God we trust all others must bring data"


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