Re: [xsl] Efficient XPath 2.0 expression to return each <row> element for which there are other <row> elements having the same navaid?

Subject: Re: [xsl] Efficient XPath 2.0 expression to return each <row> element for which there are other <row> elements having the same navaid?
From: "Liam R. E. Quin liam@xxxxxxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Thu, 13 Dec 2018 19:54:24 -0000
On Thu, 2018-12-13 at 18:51 +0000, Martin Honnen martin.honnen@xxxxxx
wrote:
> On 13.12.2018 19:40, Eliot Kimber ekimber@xxxxxxxxxxxx wrote:
> > If this is a real task support effort and not just a learning
> > exercise and you're using Oxygen then I would follow Graydon's lead
> > and use XQuery 3 to build a map that can then give you a quick
> > answer.
> 
> If you move from XPath 2 to XQuery 3 you can as well solve it with
> grouping

Yes, and that's likely to use the element value index if there is one.

Note also that Roger's original solution,
//row[navaid = (preceding-sibling::row/navaid, following-sibling::row/navaid)]
might not be as slow is it appears in an XQuery system with an element
value index. Depending on the database,
    //row[ count(//row/navaid = navaid) gt 1) ]
might be even more likely to trigger an efficient use of the index (i
didn't check, though), and is closer to the way the original problem
was stated.

Performance in an interpreted language is very often a case of doing
measurements and considering the trade-off between speed and
maintainability. However, it's easy to write O(NB2) expressions (or
worse) that run fine for a small test suite and bog down unacceptably
in production.

Liam


-- 
Liam Quin, https://www.holoweb.net/liam/
Available for XML/Document/Information Architecture/
XSL/XQuery/Web/Text Processing/A11Y consulting & training.
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