Scala: Remove duplicates in list of objects

I've got a list of objects List[Object] which are all instantiated from the same class. This class has a field which must be unique Object.property. What is the cleanest way to iterate the list of objects and remove all objects(but the first) with the same property?

list.groupBy(_.property).map(_._2.head)

Explanation: The groupBy method accepts a function that converts an element to a key for grouping. _.property is just shorthand for elem: Object => elem.property (the compiler generates a unique name, something like x$1). So now we have a map Map[Property, List[Object]]. A Map[K,V] extends Traversable[(K,V)]. So it can be traversed like a list, but elements are a tuple. This is similar to Java's Map#entrySet(). The map method creates a new collection by iterating each element and applying a function to it. In this case the function is _._2.head which is shorthand for elem: (Property, List[Object]) => elem._2.head. _2 is just a method of Tuple that returns the second element. The second element is List[Object] and head returns the first element

To get the result to be a type you want:

import collection.breakOut
val l2: List[Object] = list.groupBy(_.property).map(_._2.head)(breakOut)

To explain briefly, map actually expects two arguments, a function and an object that is used to construct the result. In the first code snippet you don't see the second value because it is marked as implicit and so provided by the compiler from a list of predefined values in scope. The result is usually obtained from the mapped container. This is usually a good thing. map on List will return List, map on Array will return Array etc. In this case however, we want to express the container we want as result. This is where the breakOut method is used. It constructs a builder (the thing that builds results) by only looking at the desired result type. It is a generic method and the compiler infers its generic types because we explicitly typed l2 to be List[Object] or, to preserve order (assuming Object#property is of type Property):

list.foldRight((List[Object](), Set[Property]())) {
  case (o, cum@(objects, props)) => 
    if (props(o.property)) cum else (o :: objects, props + o.property))
}._1

foldRight is a method that accepts an initial result and a function that accepts an element and returns an updated result. The method iterates each element, updating the result according to applying the function to each element and returning the final result. We go from right to left (rather than left to right with foldLeft) because we are prepending to objects - this is O(1), but appending is O(N). Also observe the good styling here, we are using a pattern match to extract the elements.

In this case, the initial result is a pair (tuple) of an empty list and a set. The list is the result we're interested in and the set is used to keep track of what properties we already encountered. In each iteration we check if the set props already contains the property (in Scala, obj(x) is translated to obj.apply(x). In Set, the method apply is def apply(a: A): Boolean. That is, accepts an element and returns true / false if it exists or not). If the property exists (already encountered), the result is returned as-is. Otherwise the result is updated to contain the object (o :: objects) and the property is recorded (props + o.property)

Update: @andreypopp wanted a generic method:

import scala.collection.IterableLike
import scala.collection.generic.CanBuildFrom

class RichCollection[A, Repr](xs: IterableLike[A, Repr]){
  def distinctBy[B, That](f: A => B)(implicit cbf: CanBuildFrom[Repr, A, That]) = {
    val builder = cbf(xs.repr)
    val i = xs.iterator
    var set = Set[B]()
    while (i.hasNext) {
      val o = i.next
      val b = f(o)
      if (!set(b)) {
        set += b
        builder += o
      }
    }
    builder.result
  }
}

implicit def toRich[A, Repr](xs: IterableLike[A, Repr]) = new RichCollection(xs)

to use:

scala> list.distinctBy(_.property)
res7: List[Obj] = List(Obj(1), Obj(2), Obj(3))

Also note that this is pretty efficient as we are using a builder. If you have really large lists, you may want to use a mutable HashSet instead of a regular set and benchmark the performance.

Starting Scala 2.13, most collections are now provided with a distinctBy method which returns all elements of the sequence ignoring the duplicates after applying a given transforming function:

list.distinctBy(_.property)

For instance:

List(("a", 2), ("b", 2), ("a", 5)).distinctBy(_._1) // List((a,2), (b,2))
List(("a", 2.7), ("b", 2.1), ("a", 5.4)).distinctBy(_._2.floor) // List((a,2.7), (a,5.4))

Here is a little bit sneaky but fast solution that preserves order:

list.filterNot{ var set = Set[Property]()
    obj => val b = set(obj.property); set += obj.property; b}

Although it uses internally a var, I think it is easier to understand and to read than the foldLeft-solution.

A lot of good answers above. However, distinctBy is already in Scala, but in a not-so-obvious place. Perhaps you can use it like

def distinctBy[A, B](xs: List[A])(f: A => B): List[A] =
  scala.reflect.internal.util.Collections.distinctBy(xs)(f)

With preserve order:

def distinctBy[L, E](list: List[L])(f: L => E): List[L] =
  list.foldLeft((Vector.empty[L], Set.empty[E])) {
    case ((acc, set), item) =>
      val key = f(item)
      if (set.contains(key)) (acc, set)
      else (acc :+ item, set + key)
  }._1.toList

distinctBy(list)(_.property)

One more solution

@tailrec
def collectUnique(l: List[Object], s: Set[Property], u: List[Object]): List[Object] = l match {
  case Nil => u.reverse
  case (h :: t) => 
    if (s(h.property)) collectUnique(t, s, u) else collectUnique(t, s + h.prop, h :: u)
}

I found a way to make it work with groupBy, with one intermediary step:

def distinctBy[T, P, From[X] <: TraversableLike[X, From[X]]](collection: From[T])(property: T => P): From[T] = {
  val uniqueValues: Set[T] = collection.groupBy(property).map(_._2.head)(breakOut)
  collection.filter(uniqueValues)
}

Use it like this:

scala> distinctBy(List(redVolvo, bluePrius, redLeon))(_.color)
res0: List[Car] = List(redVolvo, bluePrius)

Similar to IttayD's first solution, but it filters the original collection based on the set of unique values. If my expectations are correct, this does three traversals: one for groupBy, one for map and one for filter. It maintains the ordering of the original collection, but does not necessarily take the first value for each property. For example, it could have returned List(bluePrius, redLeon) instead.

Of course, IttayD's solution is still faster since it does only one traversal.

My solution also has the disadvantage that, if the collection has Cars that are actually the same, both will be in the output list. This could be fixed by removing filter and returning uniqueValues directly, with type From[T]. However, it seems like CanBuildFrom[Map[P, From[T]], T, From[T]] does not exist... suggestions are welcome!

With a collection and a function from a record to a key this yields a list of records distinct by key. It's not clear whether groupBy will preserve the order in the original collection. It may even depend on the type of collection. I'm guessing either head or last will consistently yield the earliest element.

collection.groupBy(keyFunction).values.map(_.head)

When will Scala get a nubBy? It's been in Haskell for decades.

If you want to remove duplicates and preserve the order of the list you can try this two liner:

val tmpUniqueList = scala.collection.mutable.Set[String]()
val myUniqueObjects = for(o <- myObjects if tmpUniqueList.add(o.property)) yield o