CODEWITHSUNDEEP
scalagenericsshapeless
pichsenmeister
pichsenmeister

Reputation: 2132

Different types in Map Scala

I need a Map where I put different types of values (Double, String, Int,...) in it, key can be String.

Is there a way to do this, so that I get the correct type with map.apply(k) like

val map: Map[String, SomeType] = Map()
val d: Double = map.apply("double")
val str: String = map.apply("string")

I already tried it with a generic type

class Container[T](element: T) {
    def get: T = element
}

val d: Container[Double] = new Container(4.0)
val str: Container[String] = new Container("string")
val m: Map[String, Container] = Map("double" -> d, "string" -> str)

but it's not possible since Container takes an parameter. Is there any solution to this?

Upvotes: 18

Views: 15373

Answers (7)

pichsenmeister
pichsenmeister

Reputation: 2132

I finally found my own solution, which worked best in my case:

case class Container[+T](element: T) {
    def get[T]: T = {
        element.asInstanceOf[T]
    }
}

val map: Map[String, Container[Any]] = Map("a" -> Container[Double](4.0), "b" -> Container[String]("test"))
val double: Double = map.apply("a").get[Double]
val string: String = map.apply("b").get[String]

Upvotes: 4

Miles Sabin
Miles Sabin

Reputation: 23056

This is now very straightforward in shapeless,

scala> import shapeless._ ; import syntax.singleton._ ; import record._
import shapeless._
import syntax.singleton._
import record._

scala> val map = ("double" ->> 4.0) :: ("string" ->> "foo") :: HNil
map: ... <complex type elided> ... = 4.0 :: foo :: HNil

scala> map("double")
res0: Double with shapeless.record.KeyTag[String("double")] = 4.0

scala> map("string")
res1: String with shapeless.record.KeyTag[String("string")] = foo

scala> map("double")+1.0
res2: Double = 5.0

scala> val map2 = map.updateWith("double")(_+1.0)
map2: ... <complex type elided> ... = 5.0 :: foo :: HNil

scala> map2("double")
res3: Double = 5.0

This is with shapeless 2.0.0-SNAPSHOT as of the date of this answer.

Upvotes: 18

R&#252;diger Klaehn
R&#252;diger Klaehn

Reputation: 12565

This is not straightforward.

The type of the value depends on the key. So the key has to carry the information about what type its value is. This is a common pattern. It is used for example in SBT (see for example SettingsKey[T]) and Shapeless Records (Example). However, in SBT the keys are a huge, complex class hierarchy of its own, and the HList in shapeless is pretty complex and also does more than you want.

So here is a small example of how you could implement this. The key knows the type, and the only way to create a Record or to get a value out of a Record is the key. We use a Map[Key, Any] internally as storage, but the casts are hidden and guaranteed to succeed. There is an operator to create records from keys, and an operator to merge records. I chose the operators so you can concatenate Records without having to use brackets.

sealed trait Record {

  def apply[T](key:Key[T]) : T

  def get[T](key:Key[T]) : Option[T]

  def ++ (that:Record) : Record
}

private class RecordImpl(private val inner:Map[Key[_], Any]) extends Record {

  def apply[T](key:Key[T]) : T = inner.apply(key).asInstanceOf[T]

  def get[T](key:Key[T]) : Option[T] = inner.get(key).asInstanceOf[Option[T]]

  def ++ (that:Record) = that match {
    case that:RecordImpl => new RecordImpl(this.inner ++ that.inner)
  }
}

final class Key[T] {
  def ~>(value:T) : Record = new RecordImpl(Map(this -> value))
}

object Key {

  def apply[T] = new Key[T]
}

Here is how you would use this. First define some keys:

val a = Key[Int]
val b = Key[String]
val c = Key[Float]

Then use them to create a record

val record = a ~> 1 ++ b ~> "abc" ++ c ~> 1.0f

When accessing the record using the keys, you will get a value of the right type back

scala> record(a)
res0: Int = 1

scala> record(b)
res1: String = abc

scala> record(c)
res2: Float = 1.0

I find this sort of data structure very useful. Sometimes you need more flexibility than a case class provides, but you don't want to resort to something completely type-unsafe like a Map[String,Any]. This is a good middle ground.

Edit: another option would be to have a map that uses a (name, type) pair as the real key internally. You have to provide both the name and the type when getting a value. If you choose the wrong type there is no entry. However this has a big potential for errors, like when you put in a byte and try to get out an int. So I think this is not a good idea.

import reflect.runtime.universe.TypeTag

class TypedMap[K](val inner:Map[(K, TypeTag[_]), Any]) extends AnyVal {
  def updated[V](key:K, value:V)(implicit tag:TypeTag[V]) = new TypedMap[K](inner + ((key, tag) -> value))

  def apply[V](key:K)(implicit tag:TypeTag[V]) = inner.apply((key, tag)).asInstanceOf[V]

  def get[V](key:K)(implicit tag:TypeTag[V]) = inner.get((key, tag)).asInstanceOf[Option[V]]
}

object TypedMap {
  def empty[K] = new TypedMap[K](Map.empty)
}

Usage:

scala> val x = TypedMap.empty[String].updated("a", 1).updated("b", "a string")
x: TypedMap[String] = TypedMap@30e1a76d

scala> x.apply[Int]("a")
res0: Int = 1

scala> x.apply[String]("b")
res1: String = a string

// this is what happens when you try to get something out with the wrong type.
scala> x.apply[Int]("b")
java.util.NoSuchElementException: key not found: (b,Int)

Upvotes: 23

Richard Sitze
Richard Sitze

Reputation: 8463

(a) Scala containers don't track type information for what's placed inside them, and

(b) the return "type" for an apply/get method with a simple String parameter/key is going to be static for a given instance of the object the method is to be applied to.

This feels very much like a design decision that needs to be rethought.

Upvotes: 3

overthink
overthink

Reputation: 24473

I don't think there's a way to get bare map.apply() to do what you'd want. As the other answers suggest, some sort of container class will be necessary. Here's an example that restricts the values to be only certain types (String, Double, Int, in this case):

sealed trait MapVal
case class StringMapVal(value: String) extends MapVal
case class DoubleMapVal(value: Double) extends MapVal
case class IntMapVal(value: Int) extends MapVal

val myMap: Map[String, MapVal] =                                                               
  Map("key1" -> StringMapVal("value1"),
      "key2" -> DoubleMapVal(3.14),
      "key3" -> IntMapVal(42))

myMap.keys.foreach { k =>
  val message =
    myMap(k) match { // map.apply() in your example code
      case StringMapVal(x) => "string: %s".format(x)
      case DoubleMapVal(x) => "double: %.2f".format(x)
      case IntMapVal(x) => "int: %d".format(x)
    }
  println(message)
}

The main benefit of the sealted trait is compile-time checking for non-exhaustive matches in pattern matching.

I also like this approach because it's relatively simple by Scala standards. You can go off into the weeds for something more robust, but in my opinion you're into diminishing returns pretty quickly.

Upvotes: 2

wheaties
wheaties

Reputation: 35990

There is a way but it's complicated. See Unboxed union types in Scala. Essentially you'll have to type the Map to some type Int |v| Double to be able to hold both Int and Double. You'll also pay a high price in compile times.

Upvotes: 1

dhg
dhg

Reputation: 52701

If you want to do this you'd have to specify the type of Container to be Any, because Any is a supertype of both Double and String. 

val d: Container[Any] = new Container(4.0)
val str: Container[Any] = new Container("string")
val m: Map[String, Container[Any]] = Map("double" -> d, "string" -> str)

Or to make things easier, you can change the definition of Container so that it's no longer type invariant:

class Container[+T](element: T) {
  def get: T = element
  override def toString = s"Container($element)"
}

val d: Container[Double] = new Container(4.0)
val str: Container[String] = new Container("string")
val m: Map[String, Container[Any]] = Map("double" -> d, "string" -> str)

Upvotes: 1

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