Tough one. If you're not worried about O(n^2) performance, you can start by
building a data structure that captures the memberships of the groups by
doing
<xsl:for-each select="$in">
<xsl:variable name="x" select="."/>
<item id="generate-id()">
<xsl:for-each select="$in[deep-equal(., current())]">
<duplicate id="{generate-id()}"/>
</xsl:for-each>
</item>
</xsl:for-each>
You can then use conventional grouping to identify the distinct groups (not
trivial, but I assume you can solve that one), and use
<xsl:key name="gid" match="*" use="generate-id()"/>
to get back from the generated ids to the original nodes.
If you are worried about O(n^2) performance, but don't want to resort to
extensions, then you can try and define a hash function that will give the
same result for two nodes if they are deep-equal. You can then modify the
above to start by doing value-based grouping on the hash key, and then apply
the O(n^2) logic only within each of these groups. A simple but quite
effective hash key might be something like concat(count(.//*), string(.)).
(Of course this still has O(n^2) performance in the worst case where all the
input nodes are deep-equal to each other, but one assumes that case is
unlikely).
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Vladimir Nesterovsky [mailto:vladimir(_at_)nesterovsky-bros(_dot_)com]
Sent: 22 October 2008 07:50
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] Group on deep equal criterion
Hello!
What is the best way to group elements by deep-equal() criterion?
Thanks.
--
Vladimir Nesterovsky
http://www.nesterovsky-bros.com
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