## Re: [xsl] Find the node with maximum elements

 Subject: Re: [xsl] Find the node with maximum elements From: "Mukul Gandhi" Date: Sun, 4 Nov 2007 18:15:45 +0530
```Hi Mike,

The following solution is likely more efficient than Ken's (because,
it calculates 'max' only once):

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; version="2.0">

<xsl:output method="text" />

<xsl:template match="/Sample">
<xsl:variable name="max-count" select="max(*/count(Car))" />
<xsl:for-each select="*">
<xsl:if test="count(Car) = \$max-count">
<xsl:value-of select="local-name()" /><xsl:text> </xsl:text>
</xsl:if>
</xsl:for-each>
</xsl:template>

</xsl:stylesheet>

On 11/4/07, Michael Kay <mike@xxxxxxxxxxxx> wrote:
> Yes, I read the requirement "return the node", not the example "return Honda
> and Mitsubishi".
>
> The solution *[count(*) = max(current()/*/count(*))] is easy to write, but
> it's very dependent on optimization. Saxon will move the condition
> max(current()/*/count(*)) out of the loop if it's written this way, but not
> if it's written *[count(*) = max(../*/count(*))]. Even if the max() is
> calculated outside the loop, you're visiting each node twice and calculating
> the "key" (count(*) twice for each node. Hence the slight preference for the
> sorting approach.
>
> The most efficient solution is probably a recursive function, but that's not
> the easiest to write. It really calls out for a higher-order function along
> the lines of saxon:highest().
>
> Michael Kay
> http://www.saxonica.com/

--
Regards,
Mukul Gandhi

```