如果过滤器功能导致列表为空，则返回未过滤的列表

Question

如果调用原始未过滤列表，如果filter返回空列表，否则返回过滤列表，我该如何返回呢？

scala> val l = List(1,2,3)
scala> l.filter(_ == 4)
res1: List[Int] = List() // would like this to be List(1,2,3)
scala> l.filter(_ == 3)
res: List[Int] = List(3) // want to maintain this behavior

Answer 1

评论中已经提到了适当的答案。不过，如果您不想为子类烦恼，您可以为此乐趣编写隐式：

scala> val l = List(1,2,3)
l: List[Int] = List(1, 2, 3)

scala> case class ListFilter[T](list: List[T]) {
   def filterOrSelf(f: T => Boolean) = list.filter(f) match {
      case Nil => list 
      case l => l
   }
}
defined class ListFilter

scala> implicit def toListFilter[T](list: List[T]) = ListFilter(list)
toListFilter: [T](list: List[T])ListFilter[T]

scala> l.filterOrSelf(_ == 4)
res0: List[Int] = List(1, 2, 3)

scala> l.filterOrSelf(_ == 3)
res1: List[Int] = List(3)