Function.tupled和占位符语法

Question

我在另一个答案中看到了Function.tupled示例的这种用法:Map(1 -> "one", 2 -> "two") map Function.tupled(_ -> _.length).

有用:

scala> Map(1 -> "one", 2 -> "two") map Function.tupled(_ -> _.length)
<console>:5: warning: method tupled in object Function is deprecated: 
Use `f.tuple` instead
       Map(1 -> "one", 2 -> "two") map Function.tupled(_ -> _.length)
                                                ^
res0: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3, 2 -> 3)

如果我不想使用占位符语法,我似乎无法做到.

scala> Map(1 -> "one", 2 -> "two") map (x => x._1 -> x._2.length)
res1: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3, 2 -> 3)

直接使用占位符语法不起作用:

scala> Map(1 -> "one", 2 -> "two") map (_._1 -> _._2.length)
<console>:5: error: wrong number of parameters; expected = 1
       Map(1 -> "one", 2 -> "two") map (_._1 -> _._2.length)

Function.tupled如何工作？似乎发生了很多事情Function.tupled(_ -> _.length).另外我如何使用它来获取弃用警告？

Answer 1

更新今天,为了解决此问题, 已删除了弃用.

一个函数的元组只是适应FunctionN[A1, A2, ..., AN, R]一个Function1[(A1, A2, ..., AN), R]

Function.tuple不赞成使用FunctionN#tupled.(可能是非预期的)结果是类型推断器无法推断出以下参数类型:

scala> Map(1 -> "one", 2 -> "two") map (_ -> _.length).tupled                 
<console>:5: error: missing parameter type for expanded function ((x$1, x$2) => x$1.$minus$greater(x$2.length))
       Map(1 -> "one", 2 -> "two") map (_ -> _.length).tupled
                                        ^
<console>:5: error: missing parameter type for expanded function ((x$1: <error>, x$2) => x$1.$minus$greater(x$2.length))
       Map(1 -> "one", 2 -> "two") map (_ -> _.length).tupled

任何这些都可行:

scala> Map(1 -> "one", 2 -> "two") map { case (a, b)  => a -> b.length }
res8: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3, 2 -> 3)

scala> Map(1 -> "one", 2 -> "two") map ((_: Int) -> (_: String).length).tupled           
res9: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3, 2 -> 3)

scala> Map(1 -> "one", 2 -> "two") map ((p: (Int, String))  => p._1 -> p._2.length)
res12: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3, 2 -> 3)

我建议你阅读这个最近问题的答案,以便更深入地理解函数文字中的'_',以及类型推理如何工作:

在Scala中,使用`_`和使用命名标识符有什么区别？

UPDATE

在回答评论时,确实如此.

scala> val f = (x:Int, y:String) => x + ": " + y
f: (Int, String) => java.lang.String = <function2>

scala> f.tupled
res0: ((Int, String)) => java.lang.String = <function1>

scala> Map(1 -> "1") map f.tupled
res1: scala.collection.immutable.Iterable[java.lang.String] = List(1: 1)

这需要Scala 2.8.请注意,如果函数的返回类型是a Tuple2,则Map #map可以生成另一个映射,否则为List,如上所述.