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

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

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
<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/>
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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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