[xquery-talk] Count of Distinct elements performance problem
Kusunam, Srinivas
SKusunam at rlpt.com
Wed Aug 23 15:46:33 PDT 2006
Mike,
Thanks for the code for grouping. I have fixed this Xquery and now i am getting "java.lang.StackOverflowError" with this approach even on half file (with 222500 Model Years). Is there anything wrong with this XQuery or 222500 is too much for it?
Here is the XQuery:
declare function local:groups($seq as xs:string*, $s as xs:string, $count as xs:integer) {
if (empty($seq))
then <gp value="{$s}" count="{$count}"/>
else
if ($seq[1] eq $s)
then local:groups($seq[position() > 1], $s, $count+1)
else
(<gp value="{$s}" count="{$count}"/>, local:groups($seq[position() > 1], $seq[1], 1))
};
let $mdoc := doc('sampleXML.xml')/Body
let $sourModelYear := $mdoc/Title/Group/ModelYear
return
<Elements>
<Element name="ModelYear">
<frequencyDistribution>
{
let $sortedYears :=
for $dvalue in $sourModelYear
order by $dvalue
return string($dvalue)
return
local:groups($sortedYears[position()>1], $sortedYears[1], 1)
}
</frequencyDistribution>
</Element>
</Elements>
________________________________
From: Michael Kay [mailto:mhk at mhk.me.uk]
Sent: Tue 8/22/2006 4:14 PM
To: Kusunam, Srinivas; talk at xquery.com
Subject: RE: [xquery-talk] Count of Distinct elements performance problem
Something like this:
let $sortedYears :=
for $dvalue in $sourModelYear
order by $dvalue
return string($dvalue)
(: use string() rather than /text() - strings are simpler and likely to be
smaller, and /text() is fragile in the face of comments :)
return
f:groups($sortedYears[position()>1], $sortedYears[1], 1);
declare function f:groups($seq as xs:string()*, $s as xs:string, $count as
xs:integer)
as xs:integer {
if (empty($seq))
then ()
else
if ($seq[1] eq $s)
then f:count-group($seq[position() > 1], $s, $count+1)
else (<gp value="{$s}" count="{$count}"/>,
f:count-group($seq[position() > 1], $s, 1))
}
Not tested.
This is likely to perform well if there are a small number of large groups,
less well if there's a large number of small groups - but it depends on how
good the implementation is at recursion.
Michael Kay
http://www.saxonica.com/
> -----Original Message-----
> From: Kusunam, Srinivas [mailto:SKusunam at rlpt.com]
> Sent: 22 August 2006 20:44
> To: Michael Kay; talk at xquery.com
> Subject: RE: [xquery-talk] Count of Distinct elements
> performance problem
>
> Michael,
> Thanks a lot for your reply.
>
> I understand what you are suggesting and it is a
> good idea. But I am not clear about doing positional
> grouping. Here is how I am assuming my modified XQuery would like:
>
> let $mdoc := doc('input.xml')/Body
> let $sourModelYEAR := $mdoc/Title/ModelYear return <Elements>
> <Element name="ModelYear">
> <Distribution>
> {
> for $dvalue in $sourModelYEAR/text() ------ Extract
> order by $dvalue ------ Sort
> ............
> How do we do Grouping of $dvalue i.e.
> ModelYear's (1991, 1992, 1995, 1997 etc)????????
> .............
> return
> <distribution>
> <value>{ $dvalue }</value>
> <count>{ $eachcount }</count>
> </distribution>
> }
> </Distribution>
> </Element>
> </Elements>
>
> Thanks,
> Srinivas
>
> -----Original Message-----
> From: Michael Kay [mailto:mhk at mhk.me.uk]
> Sent: Tuesday, August 22, 2006 2:20 PM
> To: Kusunam, Srinivas; talk at xquery.com
> Subject: RE: [xquery-talk] Count of Distinct elements
> performance problem
>
> A performance question like this can only be answered with
> respect to a specific product.
>
> There aren't many XQuery engines that will handle an 8Gb
> file, so you're doing quite well.
>
> You might find that a multi-pass approach works faster:
>
> (a) extract the values
>
> (b) sort them
>
> (c) use a recursive scan to do positional grouping - depends
> on your product supporting tail call optimization
>
>
> That's likely to have O(n*log(n)) performance rather than O(n^2).
>
> Michael Kay
> http://www.saxonica.com/
>
>
>
> > -----Original Message-----
> > From: talk-bounces at xquery.com
> > [mailto:talk-bounces at xquery.com] On Behalf Of Kusunam, Srinivas
> > Sent: 22 August 2006 18:36
> > To: talk at xquery.com
> > Subject: [xquery-talk] Count of Distinct elements
> performance problem
> >
> > I am trying to find count of distinct elements (Model Year).
> > Here is my XQuery. It takes 4 hrs to get the count from 8GB file.
> > There are around 3000 distinct Model years in this file.
> >
> > let $mdoc := doc('input.xml')/Body
> > let $sourModelYEAR := $mdoc/Title/ModelYear return <Elements>
> > <Element name="ModelYear">
> > <Distribution>
> > {
> > for $dvalue in fn:distinct-values($sourModelYEAR)
> > let $eachcount := count($mdoc/Title[ModelYear=$dvalue])
> > return
> > <distribution>
> > <value>{ $dvalue }</value>
> > <count>{ $eachcount }</count>
> > </distribution>
> > }
> > </Distribution>
> > </Element>
> > </Elements>
> >
> > This Query seems to loop through the document for each value i.e.
> > overall 3000 times. I know this should be easily achievable
> if we have
> > Group-by in XQuery. Do any XQuery engine supports
> > (custom) Group-By now?
> > Or is there any other way to make this query efficient?
> >
> > Where as if I add one more element to find the pattern of
> the data it
> > finishes the job within 40 minutes? Why is this odd behavior?
> >
> > let $mdoc := doc('input.xml')/Body
> > let $sourModelYEAR := $mdoc/Title/ModelYear return <Elements>
> > <Element name="ModelYear">
> > <Distribution>
> > {
> > for $dvalue in fn:distinct-values($sourModelYEAR)
> > let $eachcount := count($mdoc/Title[ModelYear=$dvalue])
> > return
> > <distribution>
> > <value>{ $dvalue }</value>
> > <count>{ $eachcount }</count>
> > </distribution>
> > }
> > </Distribution>
> > <PatternDistribution>
> > {
> > for $phonenum in
> > distinct-values($sourModelYEAR/translate(.,
> > '0123456789','9999999999'))
> > return
> > <pattern>
> > <type>{ $phonenum }</type>
> > <count>{count($sourModelYEAR[translate(.,
> > '0123456789', '9999999999') eq $phonenum])}</count>
> > </pattern>
> > }
> > </PatternDistribution>
> > </Element>
> > </Elements>
> >
> >
> > Thanks,
> > Srini
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RLPTechnologies sends various types of email
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potential licensing of an RLPT product or service, and
you do not wish to receive further emails regarding Polk
products, forward this email to Do_Not_Send at rlpt.com
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The email and any files transmitted with it are confidential
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