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Re: XPath expression to check that there are no inter

Subject: Re: XPath expression to check that there are no intervening elements?
From: "Dimitre Novatchev dnovatchev@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 19 Jul 2016 17:45:24 -0000
Re:  XPath expression to check that there are no  inter
To summarize, we seemingly have this:

Shortest (XPath 2.0):

 B and not(*[. >> ../B[1] and . << ../B[last()]][not(self::B)])


Claimed as fastest (XPath 1.0):

B and not(B[1]/following-sibling::*[not(self::B)][1]/following-sibling::B)


Both of these have the same worst-case execution time -- when there is
no B, or where the block of Bs is the tail of the sequence of
siblings.

The latter expression will be significantly faster in cases of long
sequence of siblings and short block of Bs at the start of the
sequence.


On Tue, Jul 19, 2016 at 9:44 AM, G. Ken Holman g.ken.holman@xxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> Okay ... I wonder if this might be fastest of all:
>
>   B  and  not(
> B[1]/following-sibling::*[not(self::B)][1]/following-sibling::B )
>
> for "there is a B, and starting from it there is no following sibling non-B
> that has a following sibling B.  No counting and no looking backwards.
>
> . . . . . . Ken
>
> At 2016-07-19 16:36 +0000, Dimitre Novatchev dnovatchev@xxxxxxxxx wrote:
>
>> To avoid the false true when there are no Bs at all:
>>
>>
>> XPath 1.0:
>> B and not(B[1]/following-sibling::node()[not(position()  >
>> ../count(B))][not(self::B)])
>>
>>
>> XPath 2.0:
>> B and not(node()[. >> ../B[1] and . << ../B[last()]][not(self::B)])
>>
>> On Tue, Jul 19, 2016 at 9:29 AM, Dimitre Novatchev <dnovatchev@xxxxxxxxx>
>> wrote:
>> > XPath 1.0:
>> >
>> >  not(B[1]/following-sibling::node()[not(position() >
>> > ../count(B))][not(self::B)])
>> >
>> > XPath 2.0:
>> >
>> > not(node()[. >> ../B[1] and . << ../B[last()]][not(self::B)])
>> >
>> > Cheers,
>> > Dimitre
>> >
>> > On Tue, Jul 19, 2016 at 9:16 AM, G. Ken Holman g.ken.holman@xxxxxxxxx
>> > <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>> >> I like Mike's answer for XSLT 2.
>> >>
>> >> Your XSLT 1 solution can be improved by reducing the number of counting
>> >> actions.  Your solution has three uses of count() ... this has only
>> >> two:
>> >>
>> >>   count(B[last()]/preceding-sibling::*[not(self::B)]) =
>> >>   count(B[1]/preceding-sibling::*[not(self::B)])
>> >>
>> >> Avoiding the use of counting you could do the following:
>> >>
>> >>   string( generate-id( B[1]/preceding-sibling::*[not(self::B)][1] ) ) =
>> >>   string( generate-id( B[last()]/preceding-sibling::*[not(self::B)][1]
>> >> ) )
>> >>
>> >> The use of string() is necessary in case the sequence of children
>> >> starts
>> >> with B.
>> >>
>> >> Come to think of it, neither your answer, Mike's answer nor my answer
>> >> will
>> >> work if there are no B children at all.  You will get a false positive.
>> >> Does your test need to accommodate that situation?
>> >>
>> >> . . . . . . . . Ken
>> >>
>> >> At 2016-07-19 15:44 +0000, Costello, Roger L. costello@xxxxxxxxx wrote:
>> >>>
>> >>> Hi Folks,
>> >>>
>> >>> This XML has a solid block of <B> elements:
>> >>>
>> >>> <Document>
>> >>>     <A/>
>> >>>     <B/>
>> >>>     <B/>
>> >>> </Document>
>> >>>
>> >>> This XML has an intervening <C> element:
>> >>>
>> >>> <Document>
>> >>>     <A/>
>> >>>     <B/>
>> >>>     <C/>
>> >>>     <B/>
>> >>> </Document>
>> >>>
>> >>> I need an XPath expression to return a Boolean value:
>> >>>
>> >>>         Return true if there is one solid block of <B> elements
>> >>>                 (no intervening elements).
>> >>>         Otherwise, return false.
>> >>>
>> >>> I created a horrendous XPath expression to solve it:
>> >>>
>> >>> count(B) eq (B[last()]/count(preceding-sibling::*)+1 -
>> >>> B[1]/count(preceding-sibling::*))
>> >>>
>> >>> Can you provide a better (simpler, more efficient) XPath expression?
>> >>>
>> >>> /Roger
>> >>>
>
>
>
> --
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>
>
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-- 
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
---------------------------------------
To invent, you need a good imagination and a pile of junk
-------------------------------------
Never fight an inanimate object
-------------------------------------
To avoid situations in which you might make mistakes may be the
biggest mistake of all
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Quality means doing it right when no one is looking.
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You've achieved success in your field when you don't know whether what
you're doing is work or play
-------------------------------------
To achieve the impossible dream, try going to sleep.
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Facts do not cease to exist because they are ignored.
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Typing monkeys will write all Shakespeare's works in 200yrs.Will they
write all patents, too? :)
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Sanity is madness put to good use.
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I finally figured out the only reason to be alive is to enjoy it.

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