Re: [xsl] Finding the position of node in foreign node list

[Dimitre Novatchev <dnovatchev@xxxxxxxxx>]

David C. and Joris already showed possible solutions. 

The time complexity of these ssolutions would most probably be close to:

   O(count($A) * count($B))

If $A and $B are sufficiently large, more efficient algorithms exist,
e.g. with time complexity close to:

   O(log($A) * count($B) )


See for example the implementation of set:leading() (in EXSLT for MSXML4) at:

     http://www.xml.com/pub/a/2003/08/06/exslt.html?page=2


Cheers,

Dimitre Novatchev.


On Thu, 3 Mar 2005 22:18:31 +0200, Jan Rademan <jrademan@xxxxxxxxxxx> wrote:
> Hi
> 
> Here is a problem I'm playing around with:
> 
> I have two variables containing two sets of nodes.
> 
> $A contains 5 elements A,B,C,D,E of element type Service
> 
> $B contains a subset of A$ e.g 2 elements B and D
> 
> I am using them to create a table which looks like this
> 
>    A    B    C    D    E  (Table header, which comes from $A)
>          B          D        (First table row, which comes from $B)
> 
> $B is reset though a for-each loop, which then loops to fill the table:
> 
> A    B    C    D    E
>      B          D
> A          C    D
>                        E
> and so on.
> 
> The for each which runs through the table, builds the table by running
> though each element in $A and checking to see if it exists in $B. If so, the
> value written.
> 
> I need to determine the positional value of the element being written, as it
> occurs in the in the variable $B
> 
> If I just use the postion function, I get the position value from the
> superset variable $A
> 
> so I get output like this:
> 
> A    B    C    D    E
>       2          4
> 1           3    4
>                         5
> 
> The ouput I am looking for should rather list the position as it occurs in
> the subset variable $B:
> 
> A    B    C    D    E
>       1          2
> 1           2    3
>                         1
> 
> Any ideas how I can find out what the position of the current node is in a
> foreign node set? (As opposed to the position in the current node set)
> 
> Thanks.
> 
> Jan Rademan