Here is a possibility:
> test <- expand.grid(id = 1:2, sex = c('male', 'female'))
> sapply(test, class)
id sex
"integer" "factor"
> test <- transform(test, sex = as.character(sex))
> sapply(test, class)
id sex
"integer" "character"
But I am surprised at the reason you give for needing it as a character
vector, because factors often act as character vectors under matching
anyway.
> sexf <- factor(test[[2]])
> sexf
[1] male male female female
Levels: female male
> which(sexf %in% "male")
[1] 1 2
> which(sexf == "male")
[1] 1 2
Bill Venables
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Alexander Nervedi
Sent: Tuesday, 12 December 2006 10:09 AM
To: r-help@stat.math.ethz.ch
Subject: [R] strings as factors
Hi,
To be able to match cases with a benchmark I need to have a data.frame
with
a character id variable. however, I am surprised why this seems to be so
hard. In fact I was unable to succeed. Here is what I tried:
>test1 <-expand.grid(ID = 1:2, sex = c("male","female"))
>is(test1[,2])
[1] "factor" "oldClass"
>test2 <-expand.grid(ID = 1:2, sex = c('male','female'))
>is(test2[,2])
[1] "factor" "oldClass"
>test3 <-expand.grid(ID = 1:2, sex = I(c("male","female")))
>is(test3[,2])
[1] "factor" "oldClass"
>test4 <-expand.grid(ID = 1:2, sex = I(c('male','female')))
>is(test4[,2])
[1] "factor" "oldClass"
>options(stringsAsFactors = FALSE)
>options("stringsAsFactors")
$stringsAsFactors
[1] FALSE
>test5 <-expand.grid(ID = 1:2, sex = I(c('male','female')))
>is(test5[,2])
[1] "factor" "oldClass"
is there anyway I can get sex to be a character?
Arnab
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