Re: Efficient XPath 2.0 expression to return each <row

Efficiency depends on the processor, and you're going to need a processor like
Saxon-EE that does join optimization.

My best attempt in XPath 2.0 is

//row[some $q in //row satisfies (./navaid eq $q/navaid and not(. is $q))]

but I haven't checked whether it gets suitably optimized.

Move to more powerful technology. In XQuery 3.1 you can do

for $r in //rows
group by $id := navaid
where count($r) gt 1
return $r

which is almost certain to be implemented efficiently by any self-respecting
processor


Michael Kay
Saxonica


> On 13 Dec 2018, at 18:07, Costello, Roger L. costello@xxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> Hi Folks,
>
> I have a large XML document containing data about airports around the
world:
>
> <airports>
>    <row>
>        <navaid>A</navaid>
>    </row>
>    <row>
>        <navaid>B</navaid>
>    </row>
>    <row>
>        <navaid>A</navaid>
>    </row>
> </airports>
>
> Notice that there is only one <row> element having the B navaid, but two
<row> elements having the A navaid.
>
> I want an XPath 2.0 expression to return each <row> element for which there
are other <row> elements having the same navaid. For the above example, I want
the XPath expression to return the first and third <row> elements.
>
> Here is one way to do it:
>
> //row[navaid = (preceding-sibling::row/navaid,
following-sibling::row/navaid)]
>
> Eek! That is horribly inefficient. I ran that XPath expression on my XML
document and it took a long time to finish.
>
> Is there an efficient XPath 2.0 expression to solve this problem?
>
> Note: I am running the XPath expression from Oxygen's XPath evaluator, not
from an XSLT program.
>
> /Roger