Re: [xsl] XQuery/XPath 3.1: Node List to Node Set ("distinct nodes")

["Michael Kay mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Here is an improved (and shorter) version of the fold-left() solution:

fold-left($nodes, (), function($a, $n) { $a, $n except $a })

Michael Kay
Saxonica

> On 28 Dec 2021, at 23:47, Michael Kay mike@xxxxxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> For a solution that delivers distinct nodes in order of first appearance, my
preference would be
>
> $nodes => fold-left((), function($all, $this) {if ($all intersect $this)
then $all else ($all, $this)})
>
> It's likely to be O(n^2) in most implementations, whereas Martin Honnen's
solution is probably O(n log n) -- but this one is XPath rather than XQuery,
and feels more elegant.
>
> Michael Kay
> Saxonica
>
>> On 28 Dec 2021, at 21:56, Michael Kay mike@xxxxxxxxxxxx
<mailto:mike@xxxxxxxxxxxx> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx
<mailto:xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>> wrote:
>>
>> You might consider
>>
>> $nodes | ()
>>
>> a bit more intuitive.
>>
>> Michael Kay
>> Saxonica
>>
>>> On 28 Dec 2021, at 19:23, Eliot Kimber eliot.kimber@xxxxxxxxxxxxxx
<mailto:eliot.kimber@xxxxxxxxxxxxxx> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx
<mailto:xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>> wrote:
>>>
>>> Hmph.
>>>
>>> That is certainly much more efficient p
 but is not necessarily obvious
(at least not to me).
>>>
>>> Thanks!
>>>
>>> E.
>>> _____________________________________________
>>>
>>> Eliot Kimber
>>> Sr Staff Content Engineer
>>> O: 512 554 9368
>>> M: 512 554 9368
>>> servicenow.com <https://www.servicenow.com/>
>>> LinkedIn <https://www.linkedin.com/company/servicenow> | Twitter
<https://twitter.com/servicenow> | YouTube
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>>>
>>> From: Martin Honnen martin.honnen@xxxxxx <mailto:martin.honnen@xxxxxx>
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx
<mailto:xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>>
>>> Date: Tuesday, December 28, 2021 at 1:15 PM
>>> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
<mailto:xsl-list@xxxxxxxxxxxxxxxxxxxxxx> <xsl-list@xxxxxxxxxxxxxxxxxxxxxx
<mailto:xsl-list@xxxxxxxxxxxxxxxxxxxxxx>>
>>> Subject: Re: [xsl] XQuery/XPath 3.1: Node List to Node Set ("distinct
nodes")
>>>
>>> [External Email]
>>>
>>>
>>> On 28.12.2021 20:10, Eliot Kimber eliot.kimber@xxxxxxxxxxxxxx
<mailto:eliot.kimber@xxxxxxxxxxxxxx> wrote:
>>> > I couldnbt find an answer in my google and markmail searching so I
>>> > thought Ibd ask here:
>>> >
>>> > Given an arbitrary list of nodes that may contain duplicates, what is
>>> > the most efficient way to reduce the node list to a set?
>>> >
>>> > The solution I came up with is a recursive function:
>>> >
>>> > (:
>>> >
>>> > Get the unique nodes from the supplied sequence
>>> >
>>> > @param nodes The sequence of nodes to evaluate
>>> >
>>> > @return A sequence of nodes such that each node in $nodes exists
exactly
>>> > once.
>>> >
>>> > :)
>>> >
>>> > declare function dutils:distinctNodes($nodes as node()*) as node()* {
>>> >
>>> >    dutils:_getDistinctNodes($nodes, ())
>>> >
>>> > };
>>> >
>>> > declare function dutils:_getDistinctNodes($nodes as node()*,
$resultList
>>> > as node()*) as node()* {
>>> >
>>> >    if (exists($nodes))
>>> >
>>> >    then
>>> >
>>> >    let $node := head($nodes)
>>> >
>>> >    return dutils:_getDistinctNodes(tail($nodes), ($resultList | $node))
>>> >
>>> >    else $resultList
>>> >
>>> > };
>>> >
>>> > Which works but I feel like Ibm missing some obvious way to do this
more
>>> > directly, but Ibm not seeing it.
>>> >
>>> > Am I missing a better solution?
>>>
>>> $nodes/.
>>>
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