Question

I’ve got a sequence of values. They can all be equal… or not. So with XQuery I want to get the most frequent item in the sequence.

let $counter := 0, $index1 := 0 
for $value in $sequence 
if (count(index-of($value, $sequence))) 
then 
{ 
$counter := count(index-of($value, $sequence)) $index1 := index-of($value) 
} else {} 

I can’t make this work, so I suppose I’m doing something wrong.

Thanks in advance for any help you could give me.

Answer 1

Use:

  for $maxFreq in 
           max(for $val in distinct-values($sequence)
                     return count(index-of($sequence, $val))
               )
   return
      distinct-values($sequence)[count(index-of($sequence, .)) eq $maxFreq]

---

Update, Dec. 2015:

This is notably shorter, though may not be too-efficient:

$pSeq[index-of($pSeq,.)[max(for $item in $pSeq return count(index-of($pSeq,$item)))]]

The shortest expression can be constructed for XPath 3.1:

And even shorter and copyable — using a one-character name:

$s[index-of($s,.)[max($s ! count(index-of($s, .)))]]