scala - Scala-如何从Tuple2s的集合构建一个不变的 map ？

Question

在Python中，可以从可重复的元组集合构造字典:

>>> listOfTuples = zip(range(10), [-x for x in range(10)])
>>> listOfTuples
[(0, 0), (1, -1), (2, -2), (3, -3), (4, -4), (5, -5), (6, -6), (7, -7), (8, -8), (9, -9)]
>>> theDict = dict(listOfTuples)
>>> theDict
{0: 0, 1: -1, 2: -2, 3: -3, 4: -4, 5: -5, 6: -6, 7: -7, 8: -8, 9: -9}
>>> 

是否有等效的Scala语法？我看到您可以使用varargs类型的Tuple2s来构建 map ，例如

scala> val theMap = Map((0,0),(1,-1))
theMap: scala.collection.immutable.Map[Int,Int] = Map((0,0), (1,-1))

scala> theMap(0)
res4: Int = 0

scala> theMap(1)
res5: Int = -1

scala> val tuplePairs = List((0,0),(1,-1))
tuplePairs: List[(Int, Int)] = List((0,0), (1,-1))

scala> val mapFromIterable = Map(tuplePairs)
<console>:6: error: type mismatch;
 found   : List[(Int, Int)]
 required: (?, ?)
       val mapFromIterable = Map(tuplePairs)
                                 ^

我可以遍历并手动分配每个值，但是似乎必须有一个更好的方法。

scala> var theMap:scala.collection.mutable.Map[Int,Int] = scala.collection.mutable.Map()   
theMap: scala.collection.mutable.Map[Int,Int] = Map()

scala> tuplePairs.foreach(x => theMap(x._1) = x._2)                                     

scala> theMap
res13: scala.collection.mutable.Map[Int,Int] = Map((1,-1), (0,0))

Answer 1

使用Scala 2.8.0 final，您可以这样做:

scala> val tuplePairs = List((0,0),(1,-1))
tuplePairs: List[(Int, Int)] = List((0,0), (1,-1))

scala> tuplePairs.toMap
res0: scala.collection.immutable.Map[Int,Int] = Map((0,0), (1,-1))

如果您使用的是Scala 2.7.7，则可以执行以下操作，以替代您使用的方法:

scala> val tuplePairs = List((0,0),(1,-1))
tuplePairs: List[(Int, Int)] = List((0,0), (1,-1))

scala> Map(tuplePairs: _*)
res2: scala.collection.immutable.Map[Int,Int] = Map(0 -> 0, 1 -> -1)

但是正如您所看到的，在2.8.0中，已有很多改进。