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(_at_)mitre(_dot_)org
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> 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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