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At 2008-12-10 18:42 -0600, Quinn Dombrowski wrote:
Sorry, that would be beyond my volunteer allocation for this morning ... too much going on right now.
But for you and the archive, be aware that your concept of intersect above wouldn't help in the case of XSLT 2.0:
$a intersect $b
For this reason an "Aa" node in place 1 would not be recognized as being the same as the "Aa" node in place 2 and would not be returned in an intersection.
I hope this steers you away from an unsuccessful approach.
. . . . . . . . . Ken
Re: [xsl] Calculating groups of repeating elements
Subject: Re: [xsl] Calculating groups of repeating elements From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Thu, 11 Dec 2008 08:41:25 -0500 |
At 2008-12-10 18:42 -0600, Quinn Dombrowski wrote:
Thanks for looking at this! Sorry for not being as clear as I could've been with what I'm looking for. For the example data set, I'm trying to automatically generate an output something like this:
Aa + C + Qqq: 2 places (1, 3) Aa + C: 3 places (1, 2, 3) Aa + Zz: 2 places (2, 3) C + Qqq: 2 places (1, 3)
So it lists all the groups of 2+ words that appear together in 2+ places. This list is sorted by length of the group (3 words is the maximum number of words that occurs in 2+ places in the sample data), but it'd be nice to also be able to sort by number of places:
Aa + C: 3 places (1, 2, 3) Aa + Zz: 2 places (2, 3) etc.
Sorry, that would be beyond my volunteer allocation for this morning ... too much going on right now.
I was using intersects to get places with Aa AND C AND Qqq (<xsl:value-of select="count(atlas/place/place_number[../words/word='Aa'] intersect atlas/place/place_number[../words/word='C'])"/>), but got overwhelmed by the number of ways I'd have to plug in all the different words to go through the data exhaustively-- the real data has 250+ places and 75+ words.
But for you and the archive, be aware that your concept of intersect above wouldn't help in the case of XSLT 2.0:
$a intersect $b
returns those tree nodes in node set $a that are the same tree nodes found in node set $b ... not by value, but the nodes themselves.
For this reason an "Aa" node in place 1 would not be recognized as being the same as the "Aa" node in place 2 and would not be returned in an intersection.
I hope this steers you away from an unsuccessful approach.
. . . . . . . . . Ken
-- Upcoming XSLT/XSL-FO, UBL and code list hands-on training classes: : Sydney, AU 2009-01/02; Brussels, BE 2009-03; Prague, CZ 2009-03 Training tools: Comprehensive interactive XSLT/XPath 1.0/2.0 video Video sample lesson: http://www.youtube.com/watch?v=PrNjJCh7Ppg Video course overview: http://www.youtube.com/watch?v=VTiodiij6gE G. Ken Holman mailto:gkholman@xxxxxxxxxxxxxxxxxxxx Crane Softwrights Ltd. http://www.CraneSoftwrights.com/s/ Male Cancer Awareness Nov'07 http://www.CraneSoftwrights.com/s/bc Legal business disclaimers: http://www.CraneSoftwrights.com/legal
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