Re: Efficient XPath 2.0 expression to return each <row

Hi Roger,
If using XSLT, use <xsl:key> and the key() function.

If not, the provided solution has time complexity O(N^2)

If XPath 3 is available one can use maps for probably the fatest
solution -- I believe it will be like O(N) -- depending on the values
distribution.

Without maps, one can write their own sorting function (sort() is a
standard function only in XPath 3.1) and then use a binary search-like
technique. This will be O(N*log(N))

Cheers,
Dimitre
On Thu, Dec 13, 2018 at 10:07 AM 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
> 



-- 
Cheers,
Dimitre Novatchev
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