[no subject]

<lq>
Fold-left

Processes the supplied sequence from left to right, applying the supplied
function repeatedly to each item in turn, together with an accumulated result
value.
</lq>

So given this expression:

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

I understand this to be iterating over $nodes from left to right, applying the
function function($a, $n) to each node, where $a is the next node and $n is
the accumulated value (being the result returned by the function on each
invocation).

The $a, $n except $a in the function body constructs a new sequence of ($a,
$n), excluding $a if it is already in $n. This sequence is then passed as the
second parameter of the next invocation of the function. This has the effect
of preserving the order of input node list.

If I have this right then its essentially equivalent to my original recursive
solution but much, much simpler and of course usable directly as an XPath
expression.

Cheers,

E.


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From: Michael Kay mike@xxxxxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Wednesday, December 29, 2021 at 4:37 AM
To: xsl-list <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Subject: Re:  XQuery/XPath 3.1: Node List to Node Set ("distinct nodes")
[External Email]

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<mailto:mike@xxxxxxxxxxxx>
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx<mailto:xsl-list-service@xxxxxxxxxxxx
rytech.com>> 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