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Re: [xsl] [Summary] Three ways to express in XPath that there are no duplicates in a list of items


Subject: Re: [xsl] [Summary] Three ways to express in XPath that there are no duplicates in a list of items
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxx>
Date: Fri, 2 Nov 2012 10:00:02 -0400

Hi,

for $n in (count(distinct-values(Websites/*)) + 1) return
empty(Websites/*[$n])

(Or in 3.0 using a let.)

I know, it's bad. (But maybe in a good way?)

Cheers,
Wendell

On Fri, Nov 2, 2012 at 9:30 AM, Michael Kay <mike@xxxxxxxxxxxx> wrote:
>
> On 02/11/2012 13:01, Dimitre Novatchev wrote:
>>>
>>> count(Websites/*) = count(distinct-values(Websites/*))
>>
>> A more efficient version (for an XPath processor with weak optimizer)
>> of this is:
>>
>> not(Websites/*[count(distinct-values(Websites/*))+1])
>>
>> So, we only count up to:
>>       count(distinct-values(Websites/*))+1
>>
>> and don't need to count all children of Websites and then compare them
>> to the count of distinct values.
>
> That solution is wrong because the context for child::Websites is different
> inside the predicate.
>
> In some circumstances it would work in the form
>
> not(//Websites/*[count(distinct-values(//Websites/*))+1])
>
>
> But that solution is more dependent on a smart optimizer than the original,
> because a naive engine will reevaluate the predicate once for every node
> selected; and the performance of an unoptimized implementation of the
> original is still (almost) linear, whereas an unoptimized implementation of
> the latter is quadratic.
>
> Michael Kay
> Saxonica
>



-- 
Wendell Piez | http://www.wendellpiez.com
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