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regrouping factor levels

2 messages · ravi, PIKAL Petr

Original
ravi
# 
Hi all,
I had some trouble in?regrouping factor levels for a variable. After some experiments, I have figured out how I can recode to modify the factor levels. I would now like some help to understand why some methods work and others don't.

Here's my code :
rm(list=ls())
###some trials in recoding factor levels
char<-letters[1:10]
fac<-factor(char)
levels(fac)
print(fac)
##first method of recoding factors
fac1<-fac
levels(fac1)[c("a","b","c")]<-"A"
levels(fac1)[c("d","e","f")]<-"B"
levels(fac1)[c("g","h","i","j")]<-"C"
levels(fac1)
print(fac1)
##second method
fac2<-fac? 
levels(fac2)[c(1,2,3)]<-"A"
levels(fac2)[c(2,3,4)]<-"B" # not c(4,5,6)
levels(fac2)[c(3,4,5,6)]<-"C" # not c(7,8,9,10)
levels(fac2)
print(fac2)
#third method
fac3<-fac
levels(fac3)<-list("A"=c("a","b","c"),"B"=c("d","e","f"),"C"=c("g","h","i","j"))
levels(fac3)
print(fac3)

I first tried method 1 and had no luck with it at all. The levels A, B, and C just got added to the existing levels without affecting the fac variable.
After some time, I was able to figure out how I should use method 2.
After reading the help documentation, I?arrived at method 3.

I would appreciate help in understanding why?the first method does not work. In my application, I had long factor names and Tinn-R just would not accept?statements running to several lines. Partial substitution?was desirable then.?Having spent a considerable?amount of time on this, I would like to understand the underlying problem with method 1 as it is. The deeper understanding could be useful for me later. 
Thanking You,
Ravi?
2 days later
PIKAL Petr
# 
Hi

r-help-bounces at r-project.org napsal dne 22.05.2009 18:53:37:
some
levels.
others don't.
levels(fac3)<-list("A"=c("a","b","c"),"B"=c("d","e","f"),"C"=c("g","h","i","j"))
and C
variable.
work. 

See the difference in those 2 selection methods

levels(fac1)[c("a","b","c")]

and

levels(fac2)[c(1,2,3)]

a, b, c are vector items not names so you need to check their presence by 
%in% operator

modified method1
which.levels <- levels(fac1) %in% c("a","b","c")
levels(fac1)[which.levels] <- "A"

Method4
fac4<-fac
levels(fac4)<-c(rep("A",3), rep("B", 3), rep("C",4))

You can either to replace all levels at once or to select levels and 
replace them with correct number of items.

Regards
Petr
desirable
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