不用MultiMap也可以创建:
import scala.collection.mutable._, JavaConverters._, java.util.TreeMap
val myMap = new mutable.HashMap[(Int, Int), mutable.Map[Date, Set[CashFlow]]]{
override def apply(k: (Int, Int)) =
getOrElseUpdate(k, new TreeMap[Date, mutable.Set[CashFlow]].asScala)
}
myMap: scala.collection.mutable.Map[(Int, Int),scala.collection.mutable.Map[java.util.Date,scala.collection.mutable.Set[CashFlow]]] = Map()
需要override def apply,因为withDefaultValue 将始终返回相同的值。
也可以使用标准包装器从 java 的 TreeMap 创建 MultiMap:
import scala.collection.convert.Wrappers._
//This wrapper will hold your Java's TreeMap inside, and delegate all operations to it
scala> def newTreeMultiMap[K, V]: MultiMap[K, V] = new JMapWrapper(new TreeMap[K, Set[V]]) with MultiMap[K, V]
newTreeMultiMap: [K, V]=> scala.collection.mutable.MultiMap[K,V]
scala> val myMap = new HashMap[(Int, Int), MultiMap[Int, String]]{ override def apply(k: (Int, Int)) = getOrElseUpdate(k, newTreeMultiMap[Int, String]) }
myMap: scala.collection.mutable.Map[(Int, Int),scala.collection.mutable.MultiMap[java.util.Date,CashFlow]] = Map()
我认为它应该在标准库中,但没有找到。
示例(我使用Int 而不是Date 来表明排序有效):
scala> val myMap = new HashMap[(Int, Int), MultiMap[Int, String]]{ override def apply(k: (Int, Int)) = getOrElseUpdate(k, newTreeMultiMap[Int, String]) }
myMap: scala.collection.mutable.Map[(Int, Int),scala.collection.mutable.MultiMap[Int,String]] = Map()
scala> myMap(0 -> 0).addBinding(4, "aaa") //no exceptions on myMap(0 -> 0) with empty Map
res27: scala.collection.mutable.MultiMap[Int,String] = Map(4 -> Set(aaa))
scala> myMap(0 -> 0).addBinding(2, "aaa") //should be before 4
res28: scala.collection.mutable.MultiMap[Int,String] = Map(2 -> Set(aaa), 4 -> Set(aaa))
scala> myMap(0 -> 0).addBinding(5, "aaa") //should be after 4
res29: scala.collection.mutable.MultiMap[Int,String] = Map(2 -> Set(aaa), 4 -> Set(aaa), 5 -> Set(aaa))
scala> myMap(0 -> 0).addBinding(4, "bbb")
res30: scala.collection.mutable.MultiMap[Int,String] = Map(2 -> Set(aaa), 4 -> Set(aaa, bbb), 5 -> Set(aaa))
scala> myMap(0 -> 0).toList //finally use the order
res31: List[(Int, scala.collection.mutable.Set[String])] = List((2,Set(aaa)), (4,Set(aaa, bbb)), (5,Set(aaa)))
scala> myMap(0 -> 1).addBinding(2, "aaa")
res18: scala.collection.mutable.MultiMap[Int,String] = Map(2 -> Set(aaa))