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Re: Re: how calculate sum(x*y)
- From: Dimitre Novatchev <dnovatchev at yahoo dot com>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Mon, 7 Oct 2002 07:33:34 -0700 (PDT)
- Subject: [xsl] Re: Re: how calculate sum(x*y)
- Reply-to: xsl-list at lists dot mulberrytech dot com
--- Andrey Solonchuk <solo at ibis dot odessa dot ua> wrote:
> Hello Dimitre,
>
> Monday, October 7, 2002, 12:54:36 PM, you wrote:
>
> Thanks, for advise? but I try both variant and
> explored that
>
> <xsl:variable name="x">
> <xsl:for-each select="//data">
> <xsl:element name="ddd">
> <xsl:value-of select="x * y"/>
> </xsl:element>
> </xsl:for-each>
> </xsl:variable>
> <xsl:value-of select="sum(xalan:nodeset($x)/ddd)"/>*
>
> this variant vork mauch more faster than using FXSLT
> may be I do something wrong, then correct me please
Hi Andrey,
No you're not doing anything wrong and your observations are correct.
What I meant in my reply was that the "foldl" solution would be faster
than the "zipWith" solution -- and this is true.
I have run a series of experiments with xml files having different
number of "data" elements. "foldl" consistently outperformed "zipWith",
even though in the "zipWith" solution I used the XPath sum() function
on the node-set containing the products.
For example, for a source xml file with 10000 "data" elements, with
MSXML4 on a 800MHz Pentium 3 the results are:
"foldl": 13.019sec.
"zipWith": 17.706sec.
The version using xsl:for-each is faster than the "foldl" solution,
because it uses explicit iteration and not recursion.
For cases where explicit iteration can be used, and having everything
approximately equal, the explicit iteration solution might be faster.
This advantage may not exist when compared to a recursive solution
using tail recursion (and run on an XSLT processor that recognises tail
recursion and implements it as iteration (e.g. Saxon)).
=====
Cheers,
Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL
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