On Wed, Dec 29, 2021 at 2:37 AM Michael Kay mike(_at_)saxonica(_dot_)com <
xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Here is an improved (and shorter) version of the fold-left() solution:
fold-left($nodes, (), function($a, $n) { $a, $n except $a })
Let's compare the two:
$nodes => fold-left((), function($all, $this) {if ($all intersect $this)
then $all else ($all, $this)})
$nodes => fold-left((), function($all, $this) {$all, $this except $all})
Shorter, yes.
As for performance, doing the same measurement as earlier gives us 0.46ms
-- that is almost exactly the same as before (0.44ms) with a very small
additional time taken (0.02ms) which can safely be ignored.
Also, at least to me, subtracting a set from a singleton set seems
counter-intuitive, but I understand why this is done.
Cheers,
Dimitre
Michael Kay
Saxonica
On 28 Dec 2021, at 23:47, Michael Kay mike(_at_)saxonica(_dot_)com <
xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)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
On 28 Dec 2021, at 21:56, Michael Kay mike(_at_)saxonica(_dot_)com <
xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
You might consider
$nodes | ()
a bit more intuitive.
Michael Kay
Saxonica
On 28 Dec 2021, at 19:23, Eliot Kimber
eliot(_dot_)kimber(_at_)servicenow(_dot_)com <
xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Hmph.
That is certainly much more efficient 😊 but is not necessarily obvious
(at least not to me).
Thanks!
E.
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*From: *Martin Honnen martin(_dot_)honnen(_at_)gmx(_dot_)de <
xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com>
*Date: *Tuesday, December 28, 2021 at 1:15 PM
*To: *xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
<xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
*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(_dot_)kimber(_at_)servicenow(_dot_)com wrote:
I couldn’t find an answer in my google and markmail searching so I
thought I’d 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 I’m missing some obvious way to do this more
directly, but I’m not seeing it.
Am I missing a better solution?
$nodes/.
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