Is there a way to get a Scala HashMap to automatically initialize values?' - scala

Question

I thought it could be done as follows
val hash = new HashMap[String, ListBuffer[Int]].withDefaultValue(ListBuffer())
hash("A").append(1)
hash("B").append(2)
println(hash("B").head)
However the above prints the unintuitive value of 1. I would like
hash("B").append(2)
To do something like the following behind the scenes
if (!hash.contains("B")) hash.put("B", ListBuffer())

Answer 1

Use getOrElseUpdate to provide the default value at the point of access:
scala> import collection.mutable._
import collection.mutable._
scala> def defaultValue = ListBuffer[Int]()
defaultValue: scala.collection.mutable.ListBuffer[Int]
scala> val hash = new HashMap[String, ListBuffer[Int]]
hash: scala.collection.mutable.HashMap[String,scala.collection.mutable.ListBuffer[Int]] = Map()
scala> hash.getOrElseUpdate("A", defaultValue).append(1)
scala> hash.getOrElseUpdate("B", defaultValue).append(2)
scala> println(hash("B").head)
2

Answer 2

withDefaultValue uses exactly the same value each time. In your case, it's the same empty ListBuffer that gets shared by everyone.
If you use withDefault instead, you could generate a new ListBuffer every time, but it wouldn't get stored.
So what you'd really like is a method that would know to add the default value. You can create such a method inside a wrapper class and then write an implicit conversion:
class InstantiateDefaults[A,B](h: collection.mutable.Map[A,B]) {
def retrieve(a: A) = h.getOrElseUpdate(a, h(a))
}
implicit def hash_can_instantiate[A,B](h: collection.mutable.Map[A,B]) = {
new InstantiateDefaults(h)
}
Now your code works as desired (except for the extra method name, which you could pick to be shorter if you wanted):
val hash = new collection.mutable.HashMap[
String, collection.mutable.ListBuffer[Int]
].withDefault(_ => collection.mutable.ListBuffer())
scala> hash.retrieve("A").append(1)
scala> hash.retrieve("B").append(2)
scala> hash("B").head
res28: Int = 2
Note that the solution (with the implicit) doesn't need to know the default value itself at all, so you can do this once and then default-with-addition to your heart's content.