Count consecutive repetitive element in a sequence using XQuery

KaushikBose · Feb 24, 2015 3:53AM

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()] };

It seems that as $counter is a sequence hence it returns a sequence in matches condition and hence this implementation fail

odie_63 · Feb 24, 2015 8:26AM

Interesting exercise, easily solved with SQL analytic functions but a little bit harder using XQuery.

Here's one way :

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
 };
 let $seq := ("a", "a", "b", "c", "c", "c", "a", "d", "d", "e", "e", "e", "f", "g", "h", "h")
 return local:sequence-group($seq)

The idea is to first find the starting index of each group, in this case :

$start-of-group := (1, 3, 4, 7, 8, 10, 13, 14, 15)

Then for each item in the sequence, we just have to retrieve the associated starting index and substract it from the current position.

odie_63 · Feb 24, 2015 8:26AM

Interesting exercise, easily solved with SQL analytic functions but a little bit harder using XQuery.

Here's one way :

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
 };
 let $seq := ("a", "a", "b", "c", "c", "c", "a", "d", "d", "e", "e", "e", "f", "g", "h", "h")
 return local:sequence-group($seq)

The idea is to first find the starting index of each group, in this case :

$start-of-group := (1, 3, 4, 7, 8, 10, 13, 14, 15)

Then for each item in the sequence, we just have to retrieve the associated starting index and substract it from the current position.

Marco Gralike · Feb 24, 2015 8:51AM

Thanks @odie_63. Nice example of function declaration as well.

> SELECT XMLQuery(
   '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
    };
    let $seq := ("a", "a", "b", "c", "c", "c", "a", "d", "d", "e", "e", "e", "f", "g", "h", "h")
    return local:sequence-group($seq)' 
    RETURNING CONTENT) as content
FROM dual

CONTENT                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       
------------------------------------
1 2 1 1 2 3 1 1 2 1 2 3 1 1 1 2

KaushikBose · Feb 25, 2015 2:42AM

Thanks that worked perfect

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