XSLT remove duplicate and select higher value - xslt

Question

I am new in XSLT and looking for help to remove duplicates of <EMP> from an xml document on the basis of their children's combined value. From each group of elements with the same value for this, the one with highest value for AIB_Position/AIB must be output. Below is my sample xml document and the corresponding desired output.
<Row_entry>
<Employees>
<Emp>
<Emp_id>E1</Emp_id>
<Emp_Name>Name1</Emp_Name>
<Country>C1</Country>
<AIB_Position>
<AIB>1500</AIB>
</AIB_Position>
</Emp>
<Emp>
<Emp_id>E2</Emp_id>
<Emp_Name>Name2</Emp_Name>
<Country>C2</Country>
<AIB_Position>
<AIB>1700</AIB>
</AIB_Position>
</Emp>
<Emp>
<Emp_id>E2</Emp_id>
<Emp_Name>Name2</Emp_Name>
<Country>C2</Country>
<AIB_Position>
<AIB>1800</AIB>
</AIB_Position>
</Emp>
</Employees>
</Row_entry>
Desired output(Removed duplicate Emp elements based on the combined <Emp_id>, <Emp_Name>, <Country> value):
<Row_entry>
<Employees>
<Emp>
<Emp_id>E1</Emp_id>
<Emp_Name>Name1</Emp_Name>
<Country>C1</Country>
<AIB_Position>
<AIB>1500</AIB>
</AIB_Position>
</Emp>
<Emp>
<Emp_id>E2</Emp_id>
<Emp_Name>Name2</Emp_Name>
<Country>C2</Country>
<AIB_Position>
<AIB>1800</AIB>
</AIB_Position>
</Emp>
</Employees>
</Row_entry>

Answer 1

I think you want this (directly using the XPath 2.0 max() function):
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"
xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xsl:output indent="yes"/>
<xsl:template match="Employees">
<xsl:copy>
<xsl:for-each-group select="Emp" group-by="concat(Emp_id, '+', Emp_Name, '+', Country)">
<xsl:copy-of select="current-group()
[AIB_Position/AIB/number() = max(current-group()/AIB_Position/AIB/number())][1]"/>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
And if you suspect your XSLT processor of idiocy, such as calculating the max() more than once, use this more precisely directing transformation:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"
xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xsl:output indent="yes"/>
<xsl:template match="Employees">
<xsl:copy>
<xsl:for-each-group select="Emp"
group-by="concat(Emp_id, '+', Emp_Name, '+', Country)">
<xsl:copy-of select=
"for $max in max(current-group()/AIB_Position/AIB/number())
return
current-group()[AIB_Position/AIB/number() = $max][1]"/>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>

Answer 2

In XSLT 2 or later, use for-each-group, for instance in XSLT 3 with a composite grouping key, then sort each group and output the maximum value:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="#all"
version="3.0">
<xsl:output indent="yes"/>
<xsl:mode on-no-match="shallow-copy"/>
<xsl:template match="Employees">
<xsl:copy>
<xsl:for-each-group select="Emp" composite="yes" group-by="Emp_id, Emp_Name, Country">
<xsl:for-each select="current-group()">
<xsl:sort select="AIB_Position/AIB" order="descending"/>
<xsl:if test="position() = 1">
<xsl:copy-of select="."/>
</xsl:if>
</xsl:for-each>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
With an XSLT 3 processor supporting the higher order sort function you could shorten that code to use
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="#all"
version="3.0">
<xsl:output indent="yes"/>
<xsl:mode on-no-match="shallow-copy"/>
<xsl:template match="Employees">
<xsl:copy>
<xsl:for-each-group select="Emp" composite="yes" group-by="Emp_id, Emp_Name, Country">
<xsl:sequence select="sort(current-group(), (), function($emp) { xs:integer($emp/AIB_Position/AIB) })[last()]"/>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
https://stackoverflow.com/tags/xslt-grouping/info has some details on how to implement a composite grouping key of XSLT 3 in XSLT 2 by string-joining the components of the key, if you are limited to XSLT 2.