Easiest way to decide if List contains duplicates?

Question

One way is this

list.distinct.size != list.size

Is there any better way? It would have been nice to have a containsDuplicates method

Answer 1

You can also write:

list.toSet.size != list.size

But the result will be the same because distinct is already implemented with a Set. In both case the time complexity should be O(n): you must traverse the list and Set insertion is O(1).

Answer 2

Assuming "better" means "faster", see the alternative approaches benchmarked in this question, which seems to show some quicker methods (although note that distinct uses a HashSet and is already O(n)). YMMV of course, depending on specific test case, scala version etc. Probably any significant improvement over the "distinct.size" approach would come from an early-out as soon as a duplicate is found, but how much of a speed-up is actually obtained would depend strongly on how common duplicates actually are in your use-case.

If you mean "better" in that you want to write list.containsDuplicates instead of containsDuplicates(list), use an implicit:

implicit def enhanceWithContainsDuplicates[T](s:List[T]) = new {
  def containsDuplicates = (s.distinct.size != s.size)
}

assert(List(1,2,2,3).containsDuplicates)
assert(!List("a","b","c").containsDuplicates)

Answer 3

I think this would stop as soon as a duplicate was found and is probably more efficient than doing distinct.size - since I assume distinct keeps a set as well:

@annotation.tailrec
def containsDups[A](list: List[A], seen: Set[A] = Set[A]()): Boolean = 
  list match {
    case x :: xs => if (seen.contains(x)) true else containsDups(xs, seen + x)
    case _ => false
}

containsDups(List(1,1,2,3))
// Boolean = true

containsDups(List(1,2,3))
// Boolean = false

I realize you asked for easy and I don't now that this version is, but finding a duplicate is also finding if there is an element that has been seen before:

def containsDups[A](list: List[A]): Boolean =  {
  list.iterator.scanLeft(Set[A]())((set, a) => set + a) // incremental sets
    .zip(list.iterator)
    .exists{ case (set, a) => set contains a }
}

Answer 4

@annotation.tailrec 
def containsDuplicates [T] (s: Seq[T]) : Boolean = 
  if (s.size < 2) false else 
    s.tail.contains (s.head) || containsDuplicates (s.tail)

I didn't measure this, and think it is similar to huynhjl's solution, but a bit more simple to understand.

It returns early, if a duplicate is found, so I looked into the source of Seq.contains, whether this returns early - it does.

In SeqLike, 'contains (e)' is defined as 'exists (_ == e)', and exists is defined in TraversableLike:

def exists (p: A => Boolean): Boolean = {
  var result = false
  breakable {
    for (x <- this)
      if (p (x)) { result = true; break }
  }
  result
}

I'm curious how to speed things up with parallel collections on multi cores, but I guess it is a general problem with early-returning, while another thread will keep running, because it doesn't know, that the solution is already found.

Answer 5

Summary: I've written a very efficient function which returns both List.distinct and a List consisting of each element which appeared more than once and the index at which the element duplicate appeared.

Note: This answer is a straight copy of the answer on a related question.

Details: If you need a bit more information about the duplicates themselves, like I did, I have written a more general function which iterates across a List (as ordering was significant) exactly once and returns a Tuple2 consisting of the original List deduped (all duplicates after the first are removed; i.e. the same as invoking distinct) and a second List showing each duplicate and an Int index at which it occurred within the original List.

Here's the function:

def filterDupes[A](items: List[A]): (List[A], List[(A, Int)]) = {
  def recursive(remaining: List[A], index: Int, accumulator: (List[A], List[(A, Int)])): (List[A], List[(A, Int)]) =
    if (remaining.isEmpty)
      accumulator
    else
      recursive(
          remaining.tail
        , index + 1
        , if (accumulator._1.contains(remaining.head))
            (accumulator._1, (remaining.head, index) :: accumulator._2)
          else
            (remaining.head :: accumulator._1, accumulator._2)
      )
  val (distinct, dupes) = recursive(items, 0, (Nil, Nil))
  (distinct.reverse, dupes.reverse)
}

An below is an example which might make it a bit more intuitive. Given this List of String values:

val withDupes =
  List("a.b", "a.c", "b.a", "b.b", "a.c", "c.a", "a.c", "d.b", "a.b")

...and then performing the following:

val (deduped, dupeAndIndexs) =
  filterDupes(withDupes)

...the results are:

deduped: List[String] = List(a.b, a.c, b.a, b.b, c.a, d.b)
dupeAndIndexs: List[(String, Int)] = List((a.c,4), (a.c,6), (a.b,8))

And if you just want the duplicates, you simply map across dupeAndIndexes and invoke distinct:

val dupesOnly =
  dupeAndIndexs.map(_._1).distinct

...or all in a single call:

val dupesOnly =
  filterDupes(withDupes)._2.map(_._1).distinct

...or if a Set is preferred, skip distinct and invoke toSet...

val dupesOnly2 =
  dupeAndIndexs.map(_._1).toSet

...or all in a single call:

val dupesOnly2 =
  filterDupes(withDupes)._2.map(_._1).toSet

This is a straight copy of the filterDupes function out of my open source Scala library, ScalaOlio. It's located at org.scalaolio.collection.immutable.List_._.

Answer 6

If you're trying to check for duplicates in a test then ScalaTest can be helpful.

import org.scalatest.Inspectors._
import org.scalatest.Matchers._
forEvery(list.distinct) { item =>
  withClue(s"value $item, the number of occurences") {
    list.count(_ == item) shouldBe 1
  }
}

// example:
scala> val list = List(1,2,3,4,3,2)
list: List[Int] = List(1, 2, 3, 4, 3, 2)

scala> forEvery(list) { item => withClue(s"value $item, the number of occurences") { list.count(_ == item) shouldBe 1 } }
org.scalatest.exceptions.TestFailedException: forEvery failed, because:
  at index 1, value 2, the number of occurences 2 was not equal to 1 (<console>:19),
  at index 2, value 3, the number of occurences 2 was not equal to 1 (<console>:19)
in List(1, 2, 3, 4)