Recursively handle nested lists in scala

Question

I'm teaching myself scala and trying to fatten my FP skills.

One of my references, Essentials of Programming Languages (available here), has a handy list of easy recursive functions. On page page 27/50, we are asked to implement swapper() function.

(swapper s1 s2 slist) returns a list the same as slist, but
with all occurrences of s1 replaced by s2 and all occurrences of s2 replaced by s1.


> (swapper ’a ’d ’(a b c d))
(d b c a)
> (swapper ’a ’d ’(a d () c d))
(d a () c a)
> (swapper ’x ’y ’((x) y (z (x))))
((y) x (z (y)))

In scala, this is:

swapper("a", "d", List("a","b","c","d"))
swapper("a", "d", List("a","d",List(),"c","d"))
swapper("x", "y", List( List("x"), "y", List("z", List("x"))))

My scala version handles all versions save the final x.

def swapper(a: Any, b: Any, lst: List[Any]): List[Any] ={
   def r(subList :List[Any], acc : List[Any]): List[Any] ={
     def swap (x :Any, xs: List[Any]) =
       if(x == a){
         r(xs, acc :+ b)
       } else if (x == b) {
         r(xs, acc :+ a)
       } else {
         r(xs, acc :+ x)
       }
     subList match {
     case Nil =>
       acc
     case List(x) :: xs =>
       r(xs, r(List(x), List()) +: acc)
     case x :: xs =>
       swap(x,xs)
     //case List(x) :: xs =>
   }
   }
  r(lst, List())
}

Instinctively, I think this is because I have no swap on the section "case List(x) :: xs" but I'm still struggling to fix it.

More difficult, still, this case breaks the tail-call optimization. How can I do this and where can I go to learn more about the general solution?

Nyavro · Accepted Answer · 2016-09-06 04:18:14Z

2

You can use this foldRight with pattern match approach:

def swapper(a:Any, b:Any, list:List[Any]):List[Any] = 
  list.foldRight(List.empty[Any]) {
    case (item, acc) if item==a => b::acc
    case (item, acc) if item==b => a::acc
    case (item:List[Any], acc) => swapper(a, b, item)::acc
    case (item, acc) => item::acc
  }

or even simplier (thanks to @marcospereira):

def swapper(a:Any, b:Any, list:List[Any]):List[Any] = 
  list.map {
    case item if item==a => b
    case item if item==b => a
    case item:List[Any] => swapper(a, b, item)
    case item => item
  }

edited Sep 6, 2016 at 4:18

answered Sep 6, 2016 at 1:30

Nyavro

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1 Comment

peoplemerge Over a year ago

OK! Thanks for your insight, this sure simplifies things. Though adding the annotation @tailrec does complain the recursive call is not a tail position.

marcospereira · Accepted Answer · 2016-09-06 03:56:11Z

1

A simpler way to solve this is just use map:

def swapper[T](a: T, b: T, list: List[T]): List[T] = list.map { item =>
  if (item == a) b
  else if (item == b) a
  else item
}

answered Sep 6, 2016 at 3:56

marcospereira

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1 Comment

jwvh Over a year ago

Doesn't appear to handle OP's 3rd test case.

jwvh · Accepted Answer · 2016-09-06 09:29:11Z

1

This seems to work.

def swapper[T](a: T, b: T, lst: List[_]): List[_] = {
  val m = Map[T, T](a -> b, b -> a).withDefault(identity)
  def swap(arg: List[_]): List[_] = arg.map{
    case l: List[_] => swap(l)
    case x: T => m(x)
  }
  swap(lst)
}

The List elements are inconsistent because it might be an element or it might be another List, so the type is List[Any], which is a sure sigh that someone needs to rethink this data representation.

edited Sep 6, 2016 at 9:29

answered Sep 6, 2016 at 4:14

jwvh

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