optimizing two-step xquery for group and count

Question

In my database (Xquery 3.1, eXist-db 4.7) I have 12,000+ TEI XML documents (examples), each of which can have multiple references to a single stand-alone TEI document of keywords.

Each of these 12,000 example documents like the following, with variable number of keyword references:

<TEI type="example" group="X">
  <teiHeader>some content</teiHeader>
  <text>
    <front>
      <div type="keywords">
          <list type="keywords">
            <item type="keyword" corresp="KW0002"/>
            <item type="keyword" corresp="KW0034"/>
            <item type="keyword" corresp="KW0349"/>
            <item type="keyword" corresp="KW0670"/>
            <item type="keyword" corresp="KW1987"/>
          </list>
      </div>
    </front>
  </text>
</TEI>

The keyword document contains 2000+ xml:id references, each containing 5 language translations:

<category xml:id="KW0001">
  <desc xml:lang="de">geliebter</desc>
  <desc xml:lang="en">lover</desc>
  <desc xml:lang="es">amante</desc>
  <desc xml:lang="fr">amant</desc>
  <desc xml:lang="it">amante</desc>
</category>
<category xml:id="KW0002">
  <desc xml:lang="de">bischof</desc>
  <desc xml:lang="en">bishop</desc>
  <desc xml:lang="es">obispo</desc>
  <desc xml:lang="fr">évêque</desc>
  <desc xml:lang="it">vescovo</desc>
</category>

The objective of my query is to get all keywords in a selection (@group) of examples, then group them and count them for HTML.

My current solution takes a long time, despite having indexed all the elements and attributes. I suspect there is a more efficient way for putting this together, but I can't see it.

let $cols := collection($mydatabase)//TEI[@group="X"]
let $kwdoc := doc("keywords.xml")//category
let $kws := distinct-values($cols//item[@type="keyword"]/data(@corresp))
let $lis := for $kw in $kws
            let $count := count($cols//item[@type="keyword" and @corresp=$kw])
            order by $count descending
            return 
                <li>
                  <a href="{concat("www.example.com/keywords/",$kw)}">                                                                             
                    {for $x in $kwdoc[@xml:id=$kw]/tei:desc
                    return <span class="{@xml:lang}">{$x/text()}</span>} 
                    ({$count})
                 </a>
               </li>
 return <ul>{$lis}</ul>

This produces HTML items that look like this:

<ul>
  <li>
    <a href="www.example.com/keywords/KW0001">
     <span class="de">geliebter</span>
     <span class="en">lover</span>
     <span class="es">amante</span>
     <span class="fr">amant</span>
     <span class="it">amante</span>
    </a>
    (64)
  </li>
  <li>
    <a href="www.example.com/keywords/KW0002">
     <span class="de">bischof</span>
     <span class="en">bishop</span>
     <span class="es">obispo</span>
     <span class="fr">évêque</span>
     <span class="it">vescovo</span>
    </a>
    (64)
  </li>
</ul>

Many thanks in advance.

Answer 1

I think in XQuery 3 you should do that grouping with group by, hopefully that also performs better:

let $cols := collection($mydatabase)//TEI[@group="X"]
let $kwdoc := doc("keywords.xml")//category

let $lis := 
  for $group in $cols//item[@type = "keyword"]
  group by $keyword := $group/@corresp
  order by count($group) descending

            return 
                <li>
                  <a href="{concat("www.example.com/keywords/",$keyword )}">                                                                             
                    {for $desc in id($keyword, $kwdoc)/desc
                    return <span class="{$desc/@xml:lang}">{$desc/text()}</span>} 
                    ({count($group)})
                 </a>
               </li>

 return <ul>{$lis}</ul>

The only issue I haven't quite understood is whether the TEI documents in $cols can reference keywords that are not in the keyword document, with the code I have shown above that check is not made.