對(duì)于Java，如果要獲取一個(gè)map的所有key可以通過keySet()，獲取所有的value的值調(diào)用values()

• 那么對(duì)于scala，有更加統(tǒng)一的方式

scala> val scores = Map("Alice" -> 10, "Bob" -> 3, "Cindy" -> 8)
scores: scala.collection.immutable.Map[String,Int] = Map(Alice -> 10, Bob -> 3, Cindy -> 8)

//獲取所有的key
scala> val nameList = scores.map(_._1)
nameList: scala.collection.immutable.Iterable[String] = List(Alice, Bob, Cindy)

//獲取所有的value
scala> val scoreList = scores.map(_._2)
scoreList: scala.collection.immutable.Iterable[Int] = List(10, 3, 8)

• 同樣對(duì)于Tuples可以使用一樣的方法

scala> val scores = List((1, "Alice", 10), (2, "Bob", 30), (3, "Cindy", 50))
scores: List[(Int, String, Int)] = List((1,Alice,10), (2,Bob,30), (3,Cindy,50))

//獲取所有Tuples的第3個(gè)元素
scala> scores.map(_._3)
res1: List[Int] = List(10, 30, 50)

• 反向操作可以使用zip，將兩個(gè)list轉(zhuǎn)換為一個(gè)map，其中一個(gè)list作為key，另一個(gè)作為value

scala> val keyList = List("Alice", "Bob", "Cindy")
keyList: List[String] = List(Alice, Bob, Cindy)

scala> val valueList = List(10, 3, 8)
valueList: List[Int] = List(10, 3, 8)

scala> val scores = keyList.zip(valueList).toMap
scores: scala.collection.immutable.Map[String,Int] = Map(Alice -> 10, Bob -> 3, Cindy -> 8)