Re: XPath to find duplicate elements

Leo,

> Dimitre, I am not sure whether I understand your two centsb&
> For each node you read, you have to compare all following nodes. How would
> you like to do that in the initial reading?

The initial loading will not compare any nodes -- it will establish
the value that is the same if two elements are deep-equal.

This value will be part of the data maintained for every node.

Then, whenever deep-equal comparison is needed, only the values stored
in the two nodes' data will be compared.

Is it clear now?


Cheers,
Dimitre



On Thu, Apr 5, 2018 at 1:13 PM, Leo Studer leo.studer@xxxxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> Dimitre, I am not sure whether I understand your two centsb&
> For each node you read, you have to compare all following nodes. How would
> you like to do that in the initial reading?
>
> Yours
> Leo
>
>
> On 5 Apr 2018, at 20:15, Dimitre Novatchev dnovatchev@xxxxxxxxx
> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> Just a thought on this.
>
> If we can implement deep-equal($n1, $n2)  as calculating some
> ValueIdentity (VI) for each of the nodes and returning
>
>     VI1 eq VI2 ...
>
> Then the VIs for each node can be calculated on the initial reading of
> the XML document (even in streaming mode) with almost no additional
> effort. If these VIs are made properties of the nodes, then we could
> just compare the VIs of the nodes and avoid any repeated zip-like
> comparison of the subtrees that are topped by these nodes. And this
> also seems more declarative/functional.
>
> Just my 2c.
>
> Cheers,
> Dimitre
>
>
> let $xml := <a>
>                 <b>
>                     <c>1</c>
>                     <d>1</d>
>                     <d>2</d>
>                 </b>
>                 <b>
>                     <c>2</c>
>                     <d>1</d>
>                     <d>3</d>
>                 </b>
>                 <b>
>                     <c>3</c>
>                     <d>2</d>
>                     <d>3</d>
>                 </b>
>                 <b>
>                     <c>3</c>
>                     <d>2</d>
>                     <d>3</d>
>                 </b>
>             </a>
>             return
>
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--
Cheers,
Dimitre Novatchev
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