How to deep compare maps to filter distinct items in XQuery?

Question

I wish I could filter a list of maps to remove those that are identical.

The union operator of XQuery only works on nodes. The

How can I do a deep comparison of maps to output distinct maps ? I can suppose that the function deep-equal() could be useful there with higher order function but I couldn’t cope with it.

let $map1 := map{ 
  'a' : '1',
  'b' : '2'
}
let $map2 := map{ 
  'c' : '3',
  'd' : '4'
}
let $map3 := map{ 
  'a' : '1',
  'b' : '2'
}
return local:getDistinctMap( ($map1, $map2, $map3) )

That should return

map{ 
  'a' : '1',
  'b' : '2'
}
map{ 
  'c' : '3',
  'd' : '4'
}

Answer 1

You can use fold-left:

declare function local:getDistinctMaps($maps as map(*)*) as map(*)* {
  fold-left(
    $maps, 
    (),
    function($distinct-maps, $new-map) { 
      if (some $map in $distinct-maps satisfies deep-equal($map, $new-map))
      then $distinct-maps else ($distinct-maps, $new-map)
    }
  )
};

let $map1 := map{ 
  'a' : '1',
  'b' : '2'
}
let $map2 := map{ 
  'c' : '3',
  'd' : '4'
}
let $map3 := map{ 
  'a' : '1',
  'b' : '2'
}
return local:getDistinctMaps( ($map1, $map2, $map3) )