[R] A More efficient method?
S Ellison
S.Ellison at lgc.co.uk
Wed Jul 4 17:57:35 CEST 2007
#Given
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
#and defining
coding<-array(c(-1,1), dimnames=list(unique(Cat) ))
#(ie an array of values corresponding to your character array levels, and with names set to those levels)
coding[Cat]
#does what you want.
>>> Keith Alan Chamberlain <Keith.Chamberlain at Colorado.EDU> 04/07/2007 14:44:44 >>>
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this, but one
which can also handle an arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] "a" "a" "a" "b" "b" "b" "a" "a" "b"
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
*******************************************************************
This email and any attachments are confidential. Any use, co...{{dropped}}
More information about the R-help
mailing list