Re: [xsl] Calculating cumulative values

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

> If I needed to sort the data before I determined the cumulative numbers, how
> do I ensure this happens before aggregating the data?

This is a well-known technique, which requires the use of (in your
case msxsl:node-set() ) the extension function exsl:node-set() or the
XSLT-processor-implemented xxx:node-set() extension function.

A smal blueprint is below:

 <xsl:variable name="vrtfOtputTree">
     <!-- Whatever processing required to produce a new tree -->
 </xsl:variable>

 <xsl:variable name="vOutputTree" select="xxx:node-set($vrtfOtputTree)"/>

<!-- Then continue with using the temporary tree that was obtained
from the RTF -->

  <!-- For example: -->

  <xsl:value-of select="$vOutputTree/*/*[1]"/>

Of course, one can use FXSL for XSLT 1.0 and use the same approach,
shown in the FXSL 2.0 solution.


--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
---------------------------------------
To invent, you need a good imagination and a pile of junk
-------------------------------------
You've achieved success in your field when you don't know whether what
you're doing is work or play



On 2/3/07, Simon Shutter <simon@xxxxxxxxxxx> wrote:
> I think I may go with 'conventional' solution provided by Andrew, because it
> is natively supported by .Net 2 and my data sets are relatively small and
> tranformed pretty quickly.  I will look more into FXSL for the longer term.
>
> If I needed to sort the data before I determined the cumulative numbers, how
> do I ensure this happens before aggregating the data?
>
> Thanks,
>
> Simon
>
>
> -----Original Message-----
> From: Dimitre Novatchev [mailto:dnovatchev@xxxxxxxxx]
> Sent: February 3, 2007 4:48 PM
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] Calculating cumulative values
>
> On 2/3/07, Simon Shutter <simon@xxxxxxxxxxx> wrote:
> > Thanks, Dimitre.
> >
> > Two questions -
> >
> > a) Will this work with .Net 2.0?
>
> No, the two XSLT processors that come as part of  .NET 2.0 (the classes
> XslTransform and XslCompiledTransform) only support XSLT 1.0.
>
> However, in a .NET environment one can use Saxon.NET, which is a port of
> Saxon 8.x and implements XSLT 2.0.
>
>
> >
> > b) What are the external files : func-scanlDVC.xsl and func-Operators.xsl?
>
> They are part of the FXSL library.
>
> The latest version of FXSL can be checked out from the CVS of the project
> at:
>
>   http://sourceforge.net/cvs/?group_id=53841
>
>
> One can learn more about FXSL from the presentations made at the Extreme
> Markup Languages Conference in Montreal 2006 (FXSL 2 -- for XSLT 2.0, and
> FXSL 1 -- for XSLT 1.0) here:
>
>
> http://www.idealliance.org/papers/extreme/proceedings/xslfo-pdf/2006/Novatch
> ev01/EML2006Novatchev01.pdf
>
> and here:
>
>
> http://www.idealliance.org/papers/extreme/proceedings/xslfo-pdf/2003/Novatch
> ev01/EML2003Novatchev01.pdf
>
>
> Other papers about different important features of FXSL can be found at the
> home page of FXSL:
>
>   http://fxsl.sf.net
>
>
>
> --
> Cheers,
> Dimitre Novatchev
> ---------------------------------------
> Truly great madness cannot be achieved without significant intelligence.
> ---------------------------------------
> To invent, you need a good imagination and a pile of junk
> -------------------------------------
> You've achieved success in your field when you don't know whether what
> you're doing is work or play
>
>
>
>
>
> >
> > Simon
> >
> > -----Original Message-----
> > From: Dimitre Novatchev [mailto:dnovatchev@xxxxxxxxx]
> > Sent: February 3, 2007 2:50 PM
> > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > Subject: Re: [xsl] Calculating cumulative values
> >
> > On 2/3/07, Andrew Welch <andrew.j.welch@xxxxxxxxx> wrote:
> >
> > >
> > > You can just use XPath here, no need for recursion:
> > >
> > > <xsl:template match="point">
> > >  <xsl:copy>
> > >   <xsl:copy-of select="@*"/>
> > >   <xsl:attribute name="y2">
> > >     <xsl:value-of select="sum(./@y1|preceding-sibling::point[@x =
> > > current()/@x]/@y1)"/>
> > >   </xsl:attribute>
> > >  </xsl:copy>
> > > </xsl:template>
> >
> >
> > However evaluating this XPath expression repeatedly is O(N^2) and will
> > probably be slow for long lists.
> >
> > Here is a solution, using the FXSL function
> >
> >      f:scanl1()
> >
> > This transformation:
> >
> > <xsl:stylesheet version="2.0"
> > xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
> >  xmlns:f="http://fxsl.sf.net/";
> >  exclude-result-prefixes="f"
> > >
> >  <xsl:import href="../f/func-scanlDVC.xsl"/>
> >  <xsl:import href="../f/func-Operators.xsl"/>
> >
> >  <!-- To be applied on testFunc-scanlDVC3.xml -->
> >  <xsl:output omit-xml-declaration="yes" indent="yes"/>
> >
> >
> >  <xsl:template match="node()|@*">
> >    <xsl:copy>
> >      <xsl:apply-templates select="node()|@*"/>
> >    </xsl:copy>
> >  </xsl:template>
> >
> >  <xsl:template match="set">
> >    <xsl:copy>
> >      <xsl:copy-of select="@*"/>
> >      <xsl:for-each-group select="point" group-by="@x">
> >         <xsl:for-each select="f:scanl1(f:add(), current-group()/@y1)">
> >            <point x="{current-group()[1]/@x}" y="{.}"/>
> >         </xsl:for-each>
> >     </xsl:for-each-group>
> >    </xsl:copy>
> >  </xsl:template>
> > </xsl:stylesheet>
> >
> > When applied against the originally provided xml document:
> >
> > <root id="theroot">
> >        <set id="1">
> >                <point x="1" y1="2" />
> >                <point x="1" y1="3" />
> >                <point x="1" y1="0" />
> >                <point x="1" y1="2" />
> >                <point x="1" y1="2" />
> >                <point x="2" y1="3" />
> >                <point x="2" y1="0" />
> >                <point x="2" y1="2" />
> >                <point x="3" y1="2" />
> >                <point x="3" y1="3" />
> >                <point x="3" y1="1" />
> >                <point x="3" y1="2" />
> >                <point x="3" y1="2" />
> >        </set>
> >        <set id="2">
> >                <point x="1" y1="2" />
> >                <point x="1" y1="3" />
> >                <point x="1" y1="0" />
> >                <point x="1" y1="2" />
> >                <point x="2" y1="2" />
> >                <point x="3" y1="2" />
> >                <point x="3" y1="2" />
> >                <point x="3" y1="2" />
> >        </set>
> >        <set id="n">
> >                <point x="1" y1="2" />
> >                <point x="1" y1="3" />
> >                <point x="1" y1="2" />
> >                <point x="2" y1="3" />
> >                <point x="2" y1="0" />
> >                <point x="2" y1="2" />
> >                <point x="3" y1="3" />
> >        </set>
> > </root>
> >
> > produces the wanted result:
> >
> > <root id="theroot">
> >   <set id="1">
> >      <point x="1" y="2"/>
> >      <point x="1" y="5"/>
> >      <point x="1" y="5"/>
> >      <point x="1" y="7"/>
> >      <point x="1" y="9"/>
> >      <point x="2" y="3"/>
> >      <point x="2" y="3"/>
> >      <point x="2" y="5"/>
> >      <point x="3" y="2"/>
> >      <point x="3" y="5"/>
> >      <point x="3" y="6"/>
> >      <point x="3" y="8"/>
> >      <point x="3" y="10"/>
> >   </set>
> >   <set id="2">
> >      <point x="1" y="2"/>
> >      <point x="1" y="5"/>
> >      <point x="1" y="5"/>
> >      <point x="1" y="7"/>
> >      <point x="2" y="2"/>
> >      <point x="3" y="2"/>
> >      <point x="3" y="4"/>
> >      <point x="3" y="6"/>
> >   </set>
> >   <set id="n">
> >      <point x="1" y="2"/>
> >      <point x="1" y="5"/>
> >      <point x="1" y="7"/>
> >      <point x="2" y="3"/>
> >      <point x="2" y="3"/>
> >      <point x="2" y="5"/>
> >      <point x="3" y="3"/>
> >   </set>
> > </root>
> >
> >
> > --
> > Cheers,
> > Dimitre Novatchev
> > ---------------------------------------
> > Truly great madness cannot be achieved without significant intelligence.
> > ---------------------------------------
> > To invent, you need a good imagination and a pile of junk
> > -------------------------------------
> > You've achieved success in your field when you don't know whether what
> > you're doing is work or play