Remove duplicates in List specifying equality function

Using the Foo and customEquals from misingFaktor's answer:

  case class Foo(a: Int, b: Int)
  val (a, b, c, d) = (Foo(3, 4), Foo(3, 1), Foo(2, 5), Foo(2, 5))
  def customEquals(x: Foo, y: Foo) = x.a == y.a

  (Seq(a, b, c, d).foldLeft(Seq[Foo]()) {
    (unique, curr) => {
      if (!unique.exists(customEquals(curr, _)))
        curr +: unique
      else
        unique
    }
  }).reverse

If result ordering is important but the duplicate to be removed is not, then foldRight is preferable

  Seq(a, b, c, d).foldRight(Seq[Foo]()) {
    (curr, unique) => {
      if (!unique.exists(customEquals(curr, _)))
        curr +: unique
      else
        unique
    }
  }

There is a simple (simpler) way to do this:

list.groupBy(_.key).mapValues(_.head)

If you want you can use the resulting map instantly by replacing _.head by a function block like:

sameElements => { val observedItem = sameElements.head
                  new A (var1 = observedItem.firstAttr,
                         var2 = "SomethingElse") }

to return a new A for each distinct element.

There is only one minor problem. The above code (list.groupBy(_.key).mapValues(_.head)) didnt explains very well the intention to remove duplicates. For that reason it would be great to have a function like distinctIn[A](attr: A => B) or distinctBy[A](eq: (A, A) -> Boolean).