On Wed, 18 Jun 2008, Andrew Yee wrote:
Thanks for everyone's suggestions.
I think factor(foo, levels = unique(foo)) works best for my needs.
(By the way, I'm still trying to figure out how to use the ordered
option in factor().)
Some functions behave differently if the factor levels are in a meaningful
order, and to let them know that use ordered = TRUE
Observe the difference:
y <- rnorm(5)
lm( y ~ factor( letters[1:5] , ordered=FALSE ))
lm( y ~ factor( letters[1:5] , ordered=TRUE ))
HTH,
Chuck
Andrew
On Wed, Jun 18, 2008 at 12:32 AM, Peter Alspach
<[EMAIL PROTECTED]> wrote:
Andrew
levels(factor(foo, levels=c('b','a')))
should work. You can make foo an ordered factor too, but that is not
necessary.
HTH .......
Peter Alspach
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Andrew Yee
Sent: Wednesday, 18 June 2008 4:22 p.m.
To: r-help@r-project.org
Subject: [R] keeping original order in factor()
Apologies for the naieveness of this question, but I'm having
trouble figuring out to have factor() maintain original ordering.
For example,
foo <- c("b","b","a","a")
levels(factor(foo, ordered=T)) #I'd like this to return as "b" "a"
#not "a" "b"
I thought having ordered=T would do the trick.
Thanks,
Andrew
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