R help - Nov 2006 - Why the factor levels returned by cut() are not ordered?

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
> help(factor)
    ...
    If 'ordered' is 'TRUE', the factor levels are assumed to be
ordered.
    ...

Wolfram

Hi

it is not stated that the cut shall return ordered factor. If you 
want you can use

ordered(cut( breaks=3, sample(10 ) ))
or modify code for cut.default to accept ordered switch.

HTH
Petr




On 29 Nov 2006 at 9:59, Wolfram Fischer wrote:

Date sent:      	Wed, 29 Nov 2006 09:59:36 +0100
From:           	Wolfram Fischer <wolfram at fischer-zim.ch>
To:             	r-help at stat.math.ethz.ch
Subject:        	[R] Why the factor levels returned by cut() are not ordered?
> 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
> 
> > help(factor)
>     ...
>     If 'ordered' is 'TRUE', the factor levels are assumed
to be
>     ordered. ...
> 
> Wolfram
> 
> ______________________________________________
> 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.
Petr Pikal
petr.pikal at precheza.cz

Wolfram Fischer wrote:
> What is the reason, that the levels of the factor
> returned by cut() are not marked as ordered levels?
I don't know, but you can always make it ordered with

ordered(cut(breaks = 3, sample(10)))
>> help(factor)
>     ...
>     If 'ordered' is 'TRUE', the factor levels are assumed
to be ordered.
>     ...
The help file for factor() probably doesn't tell you much about how
cut() works. :-)

-- 
Bj?rn-Helge Mevik

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.)
>> help(factor)
>>     
>     ...
>     If 'ordered' is 'TRUE', the factor levels are assumed
to be ordered.
>     ...
>
> Wolfram
>
> ______________________________________________
> 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.
>   

-- 
   O__  ---- Peter Dalgaard             ?ster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark          Ph:  (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45) 35327907

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 > >
> ______________________________________________ >
> 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. >   
> 
> 
> -- 
>    O__  ---- Peter Dalgaard             ?ster Farimagsgade 5, Entr.B
>   c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
>  (*) \(*) -- University of Copenhagen   Denmark          Ph:  (+45)
>  35327918
> ~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45)
> 35327907
> 
> ______________________________________________
> 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.
Petr Pikal
petr.pikal at precheza.cz