On Wed, 29 Nov 2006, Petr Pikal wrote:
>
>
> On 29 Nov 2006 at 12:08, Peter Dalgaard wrote:
>
> Date sent: Wed, 29 Nov 2006 12:08:21 +0100
> From: Peter Dalgaard <[EMAIL PROTECTED]>
> To: Wolfram Fischer <[EMAIL PROTECTED]>
> Copies to: [email protected]
> 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, ...)
Hmm, did you read the help for cut?
...: further arguments passed to or from other methods.
and so very likely factor() would barf at some of these. I'll add an
'ordered' argument to cut.default for R-devel.
>
> 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 :-)
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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