压缩集合上的Scala函数文字和占位符语法

Question

我目前正在学习Scala,并且一直在努力在zipped集合上使用占位符语法.例如,我想从l2 [i]> = l1 [i]的项目中过滤压缩数组.如何使用显式函数文字或占位符语法执行此操作？我试过了:

scala> val l = List(3, 0, 5) zip List(1, 2, 3)
l: List[(Int, Int)] = List((3,1), (4,2), (5,3))

scala> l.filter((x, y) => x > y)
<console>:9: error: missing parameter type
Note: The expected type requires a one-argument function accepting a 2-Tuple.
      Consider a pattern matching anonymous function, `{ case (x, y) =>  ... }`
              l.filter((x, y) => x > y)
                        ^
<console>:9: error: missing parameter type
              l.filter((x, y) => x > y)

scala> l.filter((x:Int, y:Int) => x > y)
<console>:9: error: type mismatch;
     found   : (Int, Int) => Boolean
     required: ((Int, Int)) => Boolean
                  l.filter((x:Int, y:Int) => x > y)

尝试占位符语法:

scala> l.filter(_ > _)
      <console>:9: error: missing parameter type for expanded function ((x$1, x$2) => x$1.$greater(x$2))
  Note: The expected type requires a one-argument function accepting a 2-Tuple.
  Consider a pattern matching anonymous function, `{ case (x$1, x$2) =>  ... }`
        l.filter(_ > _)
            ^
<console>:9: error: missing parameter type for expanded function ((x$1: <error>, x$2) => x$1.$greater(x$2))
        l.filter(_ > _)

所以它似乎需要一个函数Pair:

scala> l.filter(_._1 > _._2)
<console>:9: error: missing parameter type for expanded function ((x$1, x$2) => x$1._1.$greater(x$2._2))
Note: The expected type requires a one-argument function accepting a 2-Tuple.
      Consider a pattern matching anonymous function, `{ case (x$1, x$2) =>  ... }`
              l.filter(_._1 > _._2)
                       ^
<console>:9: error: missing parameter type for expanded function ((x$1: <error>, x$2) => x$1._1.$greater(x$2._2))
              l.filter(_._1 > _._2)

那么我做错了什么？是match唯一一个吗？谢谢您的帮助.

Answer 1

用这个:

l.filter { case (x, y) => x > y }

要么

l.filter(x => x._1 > x._2)

同样在Scala中,类型信息不会从函数体传递到其参数.