Question

当我尝试将一个案例类构造函数映射到元组列表时，这是有效的：

scala> case class MyClas(x:Int, y:String, z:String)
defined class MyClas

scala> Seq((1,"hey","you"), (2, "blue","shoe")) map Function.tupled(MyClas)
res1: Seq[MyClas] = List(MyClas(1,hey,you), MyClas(2,blue,shoe))

但是当类构造函数中存在默认参数时：

scala> case class MyClas(x:Int, y:String, z:String, zz:String="blue")
defined class MyClas

scala> Seq((1,"hey","you"), (2, "blue","shoe")) map Function.tupled(MyClas)
<console>:10: error: type mismatch;
 found   : MyClas.type
 required: (Int, String, String) => ?
              Seq((1,"hey","you"), (2, "blue","shoe")) map Function.tupled(MyClas)

以下作品，但感觉应该有一个更简单的方法：

scala> Seq((1,"hey","you"), (2, "blue","shoe")) map Function.tupled((x,y,z)=>MyClas(x,y,z))
res3: Seq[MyClas] = List(MyClas(1,hey,you,blue), MyClas(2,blue,shoe,blue))

编辑：哦，我忘了澄清这是Scala 2.10.4，但根据@ mohit的评论，这现在适用于Scala 2.11。有趣。

Answer 1

你可以添加一个伴随对象并为Tuple4和Tuple3定义一个tupled方法：

case class MyClas(x:Int, y:String, z:String, zz: String = "blue") {
}

object MyClas {
   def tupled(a: (Int, String, String)) = new MyClas(a._1, a._2, a._3)
   def tupled(a: (Int, String, String, String)) = new MyClas(a._1, a._2, a._3, a._4)
}

然后这样称呼：

Seq((1, "hey", "you"), (2, "blue", "shoe")).map(MyClas.tupled)
Seq((1, "hey", "you", "blah"), (2, "blue", "shoe", "blah")).map(MyClas.tupled)

使用默认参数映射Scala元组列表

1 个答案: