Subject: RE: [xsl] How the other half live From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Tue, 18 Nov 2008 15:21:43 -0000 |
> for $d in distinct-values($seq) return $d[count($seq[. eq $d]) ge $i] > > equivalent? I think it is, and probably a lot more > efficient, although it is longer. > They are both O(n^2). What about: <xsl:for-each-group select="$seq" group-by="."> <xsl:sequence select="current-group()[$i]"/. </xsl:for-each-group> which also scores quite well on brevity, I think - in syntax tree form, it has 6 nodes which is the same as Dimitre's expression; and I think it rates higher on both efficiency and clarity. Michael Kay http://www.saxonica.com/
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