Count consecutive repetitive element in a sequence using XQuery

Question

If the Sequence = [a a b c c c a d d e e e f g h h]

then the Output = [1 2 1 1 2 3 1 1 2 1 2 3 1 1 1 2]

Have tried to use recursion but no luck...Please Help..Thanks in Anticipation

Note: Using XQuery implementation 1.0

One of my failed implementation looks like:

declare function local:test($sequence,$count){

for $counter in (1 to count($sequence))

let $maxIndex := count($sequence)

return

if (matches(subsequence($sequence,1,$maxIndex)[$counter],subsequence($sequence,1,$maxIndex)[$counter + +1])) then let $count := $count + 1 return $count[last()]

else let $count := 1 return $count[last()]


};

Answer 1

I got a working solution for my Question. Credits: odie_63 @ http://odieweblog.wordpress.com/

declare namespace xf = "http://tempuri.org/OSBTestProject/Resources/XQuery/test/";

declare function local:sequence-group($seq as item()*) as item()*
{
   let $start-of-group :=
   fn:index-of(
     for $i in 1 to count($seq)
     let $prev := $seq[$i - 1]
     return if ($prev != $seq[$i] or not($prev)) then 1 else 0
   , 1
   )
   return
     for $i in 1 to count($seq)
     return $i - $start-of-group[. le $i][last()] + 1
};

declare function xf:test($test as xs:string) as xs:integer*
 {
let $test1 := tokenize($test, ',')
 return  local:sequence-group($test1)
};

declare variable $test as xs:string external;

xf:test($test)

Input: a,a,b,c,c,c,a,d,d,e,e,e,f,g,h,h

Output: 1 2 1 1 2 3 1 1 2 1 2 3 1 1 1 2