[R] Why the factor levels returned by cut() are not ordered?

[Petr Pikal]

On 29 Nov 2006 at 12:08, Peter Dalgaard wrote:

Date sent:      	Wed, 29 Nov 2006 12:08:21 +0100
From:           	Peter Dalgaard <P.Dalgaard at biostat.ku.dk>
To:             	Wolfram Fischer <wolfram at fischer-zim.ch>
Copies to:      	r-help at stat.math.ethz.ch
Subject:        	Re: [R] Why the factor levels returned by cut() are not ordered?

> Wolfram Fischer wrote:
> > What is the reason, that the levels of the factor
> > returned by cut() are not marked as ordered levels?
> >
> >   
> >> is.ordered( cut( breaks=3, sample(10 ) ) )
> >>     
> > FALSE
> >
> >   
> It would arguably be the Right Thing, but there would be complications
> in modeling, where ordered factors result in polynomial contrast
> coding. (This, in my opinion, is a design mistake inherited from S,
> but it's not easy to change at this stage.)

Well

what about to change cut.default

 if (codes.only)
         code
     else factor(code, seq(labels), labels)

to

 if (codes.only)
         code
     else factor(code, seq(labels), labels, ...)

which enables to use ordered switch

> is.ordered( cut( breaks=3, sample(10 ), ordered=T ) )
[1] TRUE

Without this the result is same as before (I believe :-)

Petr

 >> help(factor) >>     >  
>   ... >     If 'ordered' is 'TRUE', the factor levels are assumed to
> be ordered. >     ... > > Wolfram > >
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> 
> -- 
>    O__  ---- Peter Dalgaard             Řster Farimagsgade 5, Entr.B
>   c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
>  (*) \(*) -- University of Copenhagen   Denmark          Ph:  (+45)
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> ~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45)
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Petr Pikal
petr.pikal at precheza.cz