Re: [xsl] How to Do Random "Shuffle"?

["David Rudel fwqhgads@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

When I have to do this (essentially create a permutation of the numbers
from 1 to N), I combine random:random-sequence with saxon:sort

I'm away right now so I'm not working on a machine with XSLT, so the
following syntax may be off, but I use:

<xsl:variable name="rand" select="random:random-sequence(N)"/>

<xsl:variable name="sorted.rand" select="saxon:sort($rand)"/>

<xsl:variable name="permutation" select="$rand!index-of($sorted.rand,.)"/>

The select attribute of the last can also be written as "for $i in $rand
return index-of($sorted.rand,$i)"  .


On Saturday, September 13, 2014, Eliot Kimber ekimber@xxxxxxxxxxxx <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> Using XSLT 2 I need to implement rendering of "match table" questions
> where you have two sets of items, the match item and the thing it matches
> to. I want to present this as a literal table, where the first column is
> the match-from items in source order and the second column is the match-to
> items, in random order.
>
> I think this is best characterized as a "shuffle" problem, where you want
> to reorder a list randomly but all items in the list must be accounted
> for.
>
> I can think of a recursive algorithm: given a list, generate a random
> integer between 1 and the list length, select that item and add it to the
> result list, then call this function on the original list minus the node
> you just selected.
>
> Is there an easier or more efficient way to do it?
>
> Thanks,
>
> Eliot
> bbbbb
> Eliot Kimber, Owner
> Contrext, LLC
> http://contrext.com
>
>
>


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