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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