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
>
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>   


-- 
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