Question

如果符合特定条件，我想复制列表并在复制时修改项目。是否有更好/更快/更惯用的方法来实现这一目标？

class Animal(val t: String, val n : String) {}

val farm =  List[Animal](
  new Animal("pig","squeeky"),
  new Animal("chicken","clucker"),
  new Animal("fox","grr"),
  new Animal("chicken","clucky"),
  new Animal("chicken","cluckster"))

// i want to replace 'fox' with 'deadfox'
val safefarm = farm.map(n => {
  if (n.t == "fox")
    new Animal("deadfox","")
  else n
})

farm.map(_.t)
// res4: List[String] = List(pig, chicken, fox, chicken, chicken)

safefarm.map(_.t)
// res5: List[String] = List(pig, chicken, deadfox, chicken, chicken)

Answer 1

通过Animal个案例类并在map中使用模式匹配，您可以更加优雅。

case class Animal(t: String, n : String)

val farm =  List[Animal](
  Animal("pig","squeeky"),
  Animal("chicken","clucker"),
  Animal("fox","grr"),
  Animal("chicken","clucky"),
  Animal("chicken","cluckster")
)

val safefarm = farm map {
  case Animal("fox", _) => Animal("deadfox", "")
  case a => a
}

map只能通过List，因此速度最快。但我认为模式匹配使其更具可读性。

Answer 2

对于给定case class的更多旁注，copy方法会创建一个新的case class实例，其中所选字段已更改;考虑这个例子：

scala> case class Animal(t: String, n : String)
defined class Animal

scala> val a = Animal("fox","grr")
a: Animal = Animal(fox,grr)

scala> val b = a.copy(t="deadfox")
b: Animal = Animal(deadfox,grr)

此处b: Animal是a: Animal的副本，其中字段t: String已更改。

如何在复制时修改集合中与谓词匹配的元素？

2 个答案: