RE: Re: Grouping problem?

Hi Lars. I think you're right about the correct definition of the problem.
So I still didn't have my conditions right.

Here's my 3rd version of the recursive solution, which now produces the same
output as Benjamin's.

I would guess that this recursive solution would be quicker (for documents
larger than a certain size), but on the other hand perhaps it could produce
stack-overflow problems on really large documents? As I understand it, using
tail-recursion is supposed to allow the XSLT interpreter to optimise the
recursion by converting to it a loop, without pushing stack frames ... but I
don't know how many processors actually implement this? Does anyone know? Is
my code below appropriate for this kind of optimisation?

http://www-106.ibm.com/developerworks/xml/library/x-xslrecur/?dwzone=xml#opt
4
http://info.astrian.net/jargon/terms/t/tail_recursion.html :-)

<xsl:template match="root">
	<xsl:copy>
		<xsl:call-template name="group-ele">
			<xsl:with-param name="ele-list" select="ele"/>
		</xsl:call-template>
	</xsl:copy>
</xsl:template>

<xsl:template name="group-ele">
	<xsl:param name="ele-list" select="/.."/>
	<xsl:param name="count" select="0"/>
	<xsl:if test="$ele-list">
		<xsl:if test="$count = 10">
			<br/>
		</xsl:if>
		<xsl:variable name="first-ele" select="$ele-list[1]"/>
		<xsl:variable name="new-count" select="$count mod 10 + $first-ele/@sum"/>
		<xsl:if test="$new-count &gt; 10">
			<br/>
		</xsl:if>
		<xsl:copy-of select="$first-ele"/>
		<xsl:call-template name="group-ele">
			<xsl:with-param name="ele-list" select="$ele-list[position()&gt;1]"/>
			<xsl:with-param name="count" select="$new-count"/>
		</xsl:call-template>
	</xsl:if>
</xsl:template>

> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Lars Huttar
> Sent: Wednesday, 23 April 2003 10:10
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: RE:  Re: Grouping problem?
>
>
> Dimitre Novatchev wrote:
> > Unfortunately, this does not produce the right result in the
> > general case.
>
> I believe it does in the example data you gave, at least
> as I understand Benjamin's definition of the problem.
>
> > The result produced by your transformation is:
> >
> > <root>
> >   <ele sum="3"></ele>
> >   <ele sum="4"></ele>
> >   <ele sum="2"></ele>
> >   <br />
> >   <ele sum="10"></ele>
> >   <ele sum="1"></ele>
> >   <br />
> >   <ele sum="5"></ele>
> >   <ele sum="1"></ele>
> >   <ele sum="2"></ele>
> > </root>
> >
> > The second group above has a sum of 11.
>
> Benjamin's requirement was:
>
> >I'm trying to break apart the ele element when the sum total
> of preceding
> >siblings and self is greater than the increment of 10 by
> putting an element
> >to denote the break.
>
> E.g. in the above data, a <br/> appears just before the cumulative sum
> would exceed 10, and again just before the cumulative sum would exceed
> 20.
> Benjamin was not saying that each group should total 10 or less.
>
> Not to discount the rest of your message...
>
> Lars
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>


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