As always the question seems silly after you've learned the answer!
Thanks a lot,
baptiste
On 3 February 2010 15:06, Petr PIKAL wrote:
> # order levels
> f.t<-factor(f, levels=disorder)
> # change levels
> levels(f.t) <- new.lev
> all.equal(f.t,f2)
> [1] TRUE
>
>
> Regards
> Petr
>
>
>>
>> Bes
Dear list,
I cannot find an elegant solution to this problem. I have a factor f
containing several levels (5) and I wish to create a new factor of the
same length with fewer levels (2). This new factor should therefore
group together some levels of the original data. Ideally this grouping
would be
2 matches
Mail list logo