Question

我对Scala很新，并且正在努力弄清楚如何使用Collections。如果我有一个List [People]，所有人都有名字，有没有办法把它变成Map [Person.getName，List [People]]。

我试过

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

但只返回Map [Name，People]。我该怎么做？

提前致谢。

Answer 1

groupBy将根据＆＃34;鉴别器功能＆＃34;将List[People]转换为Map[String, List[People]]。在这种情况下，您提供的是从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

如果你想自己动手并了解它是如何运作的，你可以试试这个：

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

Scala：列出[人物]并将其转换为地图[String，List [People]]

2 个答案: