XSLT – creating a network from all children elements - xslt

Question

With XSLT 2.0, I am trying to create a list of relations between all children of given elements, in a document such as:
<doc>
<part1>
<name>John</name>
<name>Paul</name>
<name>George</name>
<name>Ringo</name>
<place>Liverpool</place>
</part1>
<part2>
<name>Romeo</name>
<name>Romeo</name>
<name>Juliet</name>
<fam>Montague</fam>
<fam>Capulet</fam>
</part2>
</doc>
The result I would like to obtain, ideally by conflating and weighing the identical relations, would be (in whatever order) something like:
<doc>
<part1>
<rel><name>John</name><name>Paul</name></rel>
<rel><name>John</name><name>George</name></rel>
<rel><name>John</name><name>Ringo</name></rel>
<rel><name>Paul</name><name>George</name></rel>
<rel><name>Paul</name><name>Ringo</name></rel>
<rel><name>George</name><name>Ringo</name></rel>
<rel><name>John</name><place>Liverpool</place></rel>
<rel><name>Paul</name><place>Liverpool</place></rel>
<rel><name>George</name><place>Liverpool</place></rel>
<rel><name>Ringo</name><place>Liverpool</place></rel>
</part1>
<part2>
<rel weight="2"><name>Romeo</name><name>Juliet</name></rel>
<rel weight="2"><name>Romeo</name><fam>Montague</fam></rel>
<rel weight="2"><name>Romeo</name><fam>Capulet</fam></rel>
<rel><name>Juliet</name><fam>Montague</fam></rel>
<rel><name>Juliet</name><fam>Capulet</fam></rel>
<rel><fam>Montague</fam><fam>Capulet</fam></rel>
</part2>
</doc>
—but I'm not sure how to proceed. Many thanks in advance for your help.

Answer 1

You still haven't explained the logic that needs to be applied here, so this is based largely on a guess:
XSLT 2.0
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<!-- identity transform -->
<xsl:template match="/">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="doc/*">
<!-- first pass-->
<xsl:variable name="unique-items">
<xsl:for-each-group select="*" group-by="concat(name(), '|', .)">
<item name="{name()}" count="{count(current-group())}" value="{.}"/>
</xsl:for-each-group>
</xsl:variable>
<!-- output -->
<xsl:copy>
<xsl:for-each select="$unique-items/item">
<xsl:variable name="left" select="."/>
<xsl:for-each select="following-sibling::item">
<xsl:variable name="weight" select="$left/#count * #count" />
<rel>
<xsl:if test="$weight gt 1">
<xsl:attribute name="weight" select="$weight"/>
</xsl:if>
<xsl:apply-templates select="$left | ." />
</rel>
</xsl:for-each>
</xsl:for-each>
</xsl:copy>
</xsl:template>
<xsl:template match="item">
<xsl:element name="{#name}">
<xsl:value-of select="#value"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
The idea here is to remove duplicates in the first pass, then enumerate all combinations in the second (final) pass. The weight is computed by multiplying the number of occurrences of each member of a combination pair and shown only when it exceeds 1.

Answer 2

At least the combinatoric part of your problem could be solved with the following XSLT script. It does not solve the elimination of duplicates, but that could possibly be done in a second transformation.
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<!-- standard copy template -->
<xsl:template match="node()|#*">
<xsl:copy>
<xsl:apply-templates select="node()|#*" />
</xsl:copy>
</xsl:template>
<xsl:template match="doc/*">
<xsl:copy>
<xsl:variable name="l" select="./*"/>
<xsl:for-each select="$l">
<xsl:variable name="a" select="."/>
<xsl:variable name="posa" select="position()"/>
<xsl:variable name="namea" select="name()"/>
<xsl:for-each select="$l">
<xsl:if test="position() > $posa and (. != $a or name() != $namea)">
<rel>
<xsl:copy-of select="$a"/>
<xsl:copy-of select="."/>
</rel>
</xsl:if>
</xsl:for-each>
</xsl:for-each>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
When applied to the first part of your example, this produces:
<part1>
<rel><name>John</name><name>Paul</name></rel>
<rel><name>John</name><name>George</name></rel>
<rel><name>John</name><name>Ringo</name></rel>
<rel><name>John</name><place>Liverpool</place></rel>
<rel><name>Paul</name><name>George</name></rel>
<rel><name>Paul</name><name>Ringo</name></rel>
<rel><name>Paul</name><place>Liverpool</place></rel>
<rel><name>George</name><name>Ringo</name></rel>
<rel><name>George</name><place>Liverpool</place></rel>
<rel><name>Ringo</name><place>Liverpool</place></rel>
</part1>
Which seems about correct. If have no idea if the duplicate elimination (or weighting, as you call it) could be done in the same transformation.