将函数映射到元组序列

Question

给定以下代码：

def map1[A,B,C](s: Seq[(A, B)])(f: A => C) : Seq[(C, B)] = 
  s.map { case (a, b) => (f(a), b) }

是否有更好的编码方式（也许 scalaz 中存在一些东西）？

你能帮我找个更好的名字吗？

是否有更通用的抽象要使用 (Iterable, TraversableOnce)？

Answer 1

您可以定义一个扩展方法：

import scala.collection.GenTraversable
import scala.collection.GenTraversableLike
import scala.collection.generic.CanBuildFrom

implicit class WithMapKeys[A, B, Repr](val self: GenTraversableLike[(A, B), Repr]) extends AnyVal {
  def mapKeys[C, That](f: A => C)(implicit bf: CanBuildFrom[Repr, (C, B), That]) = {
    self.map(x => f(x._1) -> x._2)
  }
}

然后：

Vector(1->"a", 2->"b").mapKeys(_+2)
// res0: scala.collection.immutable.Vector[(Int, String)] = Vector((3,a), (4,b))

Map(1->"a", 2->"b").mapKeys(_+2)
// res1: scala.collection.immutable.Map[Int,String] = Map(3 -> a, 4 -> b)

与迭代器类似：

implicit class WithMapKeysItr[A, B](val self: Iterator[(A, B)]) extends AnyVal {
  def mapKeys[C](f: A => C): Iterator[(C, B)] = {
    self.map(x => f(x._1) -> x._2)
  }
}

val i2 = Iterator(1->"a", 2->"b").mapKeys(_+2)
// i2: Iterator[(Int, String)] = non-empty iterator
i2.toVector
// res2: Vector[(Int, String)] = Vector((3,a), (4,b))

val to: TraversableOnce[(Int, String)] = Vector(1->"a", 2->"b")
val i3 = to.toIterator.mapKeys(_+2)
// i3: Iterator[(Int, String)] = non-empty iterator
i3.toMap
// res3: scala.collection.immutable.Map[Int,String] = Map(3 -> a, 4 -> b)

Answer 2

Shapeless 可以让它更漂亮一点（没有模式匹配）：

import shapeless.syntax.std.tuple._
def mapKeys[A,B,C](s: Seq[(A, B)])(f: A => C) : Seq[(C, B)] = 
   s.map(x => x.updatedAt(0, f(x.head)))

将函数映射到元组序列

map a function on a sequence of tuples

collections

scala