Scala: extracting a repeated value from a list

Question

I have often the need to check if many values are equal and in case extract the common value. That is, I need a function that will work like follows:

extract(List()) // None
extract(List(1,2,3)) // None
extract(List(2,2,2)) // Some(2)

Assuming one has a pimp that will add tailOption to seqs (it is trivial to write one or there is one in scalaz), one implementation looks like

def extract[A](l: Seq[A]): Option[A] = {

  def combine(s: A)(r: Seq[A]): Option[A] =
    r.foldLeft(Some(s): Option[A]) { (acc, n) => acc flatMap { v =>
      if (v == n) Some(v) else None
    } }

  for {
    h <- l.headOption
    t <- l.tailOption
    res <- combine(h)(t)
  } yield res
}

Is there something like that - possibly more general - already in Scalaz, or some simpler way to write it?

So you basically want to know, if all the values in the list are equal? — drexin, Sep 26 '12 at 13:40

score 3 · Accepted Answer · answered Sep 26 '12 at 13:26

3

This seems like a really complicated way to write

def extract[A](l:Seq[A]):Option[A] = l.headOption.flatMap(h =>
  if (l.tail.forall(h==)) Some(h) else None)

You don't need tailOption, since the anonymous function that gets passed as an argument to flatMap is only executed if l is not empty.

answered Sep 26 '12 at 13:26

Kim Stebel

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2

With `headOption.filter` it can be a little shorter. – incrop Sep 26 '12 at 13:29
Yes, but that's not how filter is usually used, so I think it would make it less readable. – Kim Stebel Sep 26 '12 at 13:30
Yes, it was quite a complicated way to write that :-) – Andrea Sep 26 '12 at 13:43

score 2 · Answer 2 · answered Sep 26 '12 at 13:45

2

If you only want to delete duplicates toSet is enough:

def equalValue[A](xs: Seq[A]): Option[A] = {
  val set = xs.toSet
  if (set.size == 1) Some(set.head) else None
}

scala> equalValue(List())
res8: Option[Nothing] = None

scala> equalValue(List(1,2,3))
res9: Option[Int] = None

scala> equalValue(List(2,2,2))
res10: Option[Int] = Some(2)

answered Sep 26 '12 at 13:45

kiritsuku

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or xs.distinct instead of xs.toSet. – The Archetypal Paul Sep 26 '12 at 17:10

score 2 · Answer 3 · answered Sep 26 '12 at 14:41

2

This is a fluent solution

yourSeq.groupBy(x => x) match {case m if m.size==1 => m.head._1; case _ => None}

answered Sep 26 '12 at 14:41

Federico Dal Maso

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3
6

score -1 · Answer 4 · answered Sep 26 '12 at 13:28

-1

You could use a map to count the number of occurrences of each element in the list and then return only those that occur more than once:

def extract[T](ts: Iterable[T]): Iterable[T] = {
  var counter: Map[T, Int] = Map()

  ts.foreach{t =>
    val cnt = counter.get(t).getOrElse(0) + 1
    counter = counter.updated(t, cnt)
  }

  counter.filter(_._2 > 1).map(_._1)
}

println(extract(List())) // List()
println(extract(List(1,2,3))) // List()
println(extract(List(2,2,2))) // List(2)
println(extract(List(2,3,2,0,2,3))) // List(2,3)

You can also use a foldLeft instead of foreach and use the empty map as the initial accumulator of foldLeft.

answered Sep 26 '12 at 13:28

Malte Schwerhoff

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The downvote is not mine, but I can only guess that it is there because this does not answer my question - which is to return the common value if all values are equal and None otherwise – Andrea Sep 26 '12 at 13:40
@Andrea Ok, then I misinterpreted your question (and answered it before the clarifying comments where made). Sorry. – Malte Schwerhoff Sep 26 '12 at 13:47

Scala: extracting a repeated value from a list

4 Answers4