Re: [xsl] Using divide-and-conquer recursion to create a cumulative sequence

[Andrew Welch <andrew.j.welch@xxxxxxxxx>]

2009/12/11 Costello, Roger L. <costello@xxxxxxxxx>:
>
> Hi Folks,
>
> I wish to convert a sequence of N numbers:
>
>   (23, 41, 70, 103, 99, 6)
>
> Into a cumulative sequence, in which each number is the sum of the previous
numbers:
>
>   (23, 64, 134, 237, 336, 342)


Hi Roger,

Here's a function that recursively processes sequence, with linear
performance:


<xsl:stylesheet version="2.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
    xmlns:xs="http://www.w3.org/2001/XMLSchema";
    xmlns:f="f"
    exclude-result-prefixes="xs">

<xsl:variable name="seq" select="(23, 41, 70, 103, 99, 6)" as="xs:integer+"/>

<xsl:function name="f:sumSeq" as="xs:integer+">
	<xsl:param name="seq" as="xs:integer+"/>
	<xsl:param name="pos" as="xs:integer"/>

	<xsl:sequence select="if ($pos gt count($seq)) then $seq else
		f:sumSeq((subsequence($seq, 1, $pos), $seq[$pos + 1] + $seq[$pos],
subsequence($seq, $pos + 2)), $pos + 1)"/>
</xsl:function>

<xsl:template match="/">
	<xsl:value-of select="f:sumSeq($seq, 1)"/>
</xsl:template>

</xsl:stylesheet>


cheers
--
Andrew Welch
http://andrewjwelch.com
Kernow: http://kernowforsaxon.sf.net/