Scala:接受List[People]并将其转换为Map[String，List[People]]

Question

我对Scala非常陌生，我正在努力研究如何使用集合。如果我有一个ListPeople (所有人都有名字)，那么是否有办法将其转换为Map[Person.getName，ListPeople]。

我试过了

list map (t => t.getName() -> t) toMap, 

但这只会返回一个MapName，伙计们。我该怎么做？

提前谢谢。

Answer 1

groupBy将获取一个List[People]，并根据您提供的“判别器函数”将其转换为Map[String, List[People]]，在本例中，该函数将从Person中选择名称。

scala> case class Person(name: String, age:Int)
defined class Person

scala> val people = List(Person("Alice", 42), Person("Bob", 42))
people: List[Person] = List(Person(Alice,42), Person(Bob,42))

scala> people.groupBy(p => p.name)
res0: scala.collection.immutable.Map[String,List[Person]] = Map(Bob -> List(Person(Bob,42)), Alice -> List(Person(Alice,42)))

Answer 2

如果您想自己动手并学习它的工作原理，您可以尝试如下：

scala> case class Person(name: String, surname: String, age:Int)
defined class Person

scala> val people = List(Person("Alice", "Wonderland", 42), Person("Bob", "Father", 42), Person("Bob", "Son", 20))
people: List[Person] = List(Person(Alice,Wonderland,42), Person(Bob,Father,42), Person(Bob,Son,20))

scala> val names = people.map(_.name).distinct
names: List[String] = List(Alice, Bob)

scala> names.map(n => n -> people.filter(_.name == n)).toMap
res0: scala.collection.immutable.Map[String,List[Person]] = Map(Alice -> List(Person(Alice,Wonderland,42)), Bob -> List(Person(Bob,Father,42), Person(Bob,Son,20)))

干杯