Re: [xsl] distinct-values() optimization, sorting by frequency

[David Carlisle <davidc@xxxxxxxxx>]

without knowing really what the input or output looks like its a bit
hard, but "want to sort by frequency of distinct-value " makes me think
you want to be using xsl:for-each-group rather than distinct-values.
for-each-group is much more powerful as it not only gives you the
sequence of distinct values but (in curent-group()) the sequence of
nodes that correspond to that value, so you can easily count them.

> I have to have it for the sort, and then I have to re-do it 
Youshouldn't need to sort twice, you can store the sorted list using
xsl:perform sort.

> Actually I don't think that <xsl:sort> is doing what I want at all! :-(
    <xsl:for-each select="$distinct-persNames">
  <xsl:sort select="count($persNames[normalize-space(lower-case(.))
        = .])"/>

te sort expression does not depend on the current node so it will be the
same foir each item, so no sorting will be done.
I suspect you intended current() rather than . but also there's no need
to work out  $persNames[normalize-space(lower-case(.)) for every item
as you've saved these normalised results earlier haven't you?
Unless saxon takes this expression out of the loop just calculating this
sort expression is going to be O(n^2) before you even start sorting (I
think)

David

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