[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Using divide-and-conquer recursion to create a cu
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/
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