for $d in distinct-values($seq) return $d[count($seq[. eq $d]) ge $i]
equivalent? I think it is, and probably a lot more efficient,
although it is longer.
What I have prepared as an optimization is:
distinct-values($vSeq)[exists(index-of($vSeq,.)[2])]
it is still O(N^2) in the general case but is many times faster for
long sequences with only a small number of different items, such as a
long bit sequence.
--
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
-------------------------------------
You've achieved success in your field when you don't know whether what
you're doing is work or play
On Tue, Nov 18, 2008 at 6:55 AM, Andrew Welch
<andrew(_dot_)j(_dot_)welch(_at_)gmail(_dot_)com> wrote:
2008/11/18 Dimitre Novatchev <dnovatchev(_at_)gmail(_dot_)com>:
On Mon, Nov 17, 2008 at 12:53 PM, Michael Kay <mike(_at_)saxonica(_dot_)com>
wrote:
I quite like the fact that this generalizes.
$vSeq[index-of($vSeq,.)[$i]]
Yes, I intentionally omitted this in order not to overload the
readers, nevertheless anyone who tries the five problems will
naturally notice this generalization.
Is:
for $d in distinct-values($seq) return $d[count($seq[. eq $d]) ge $i]
equivalent? I think it is, and probably a lot more efficient,
although it is longer.
I can't see how to avoid the for expression...
--
Andrew Welch
http://andrewjwelch.com
Kernow: http://kernowforsaxon.sf.net/
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