scala - 按键集过滤 map

Question

是否有过滤 Map 的快捷方式，仅保留给定 Set 中包含键的条目？

这是一些示例代码

scala> val map = Map("1"->1, "2"->2, "3"->3)
map: scala.collection.immutable.Map[java.lang.String,Int] = Map(1 -> 1, 2 -> 2, 3 -> 3)

scala> map.filterKeys(Set("1","2").contains)
res0: scala.collection.immutable.Map[java.lang.String,Int] = Map(1 -> 1, 2 -> 2)

我正在寻找比这更短的东西。

Answer 1

回答问题

您可以利用以下事实:Set[A]是谓词；即 A => Boolean

map filterKeys set

这是在工作:

scala> val map = Map("1" -> 1, "2" -> 2, "3" -> 3)
map: scala.collection.immutable.Map[java.lang.String,Int] = Map(1 -> 1, 2 -> 2, 3 -> 3)

scala> val set = Set("1", "2")
set: scala.collection.immutable.Set[java.lang.String] = Set(1, 2)

scala> map filterKeys set
res0: scala.collection.immutable.Map[java.lang.String,Int] = Map(1 -> 1, 2 -> 2)

或者，如果您更喜欢:

scala> map filterKeys Set("1", "2")
res1: scala.collection.immutable.Map[java.lang.String,Int] = Map(1 -> 1, 2 -> 2)

谓词

在谓词周围有一些包装器实际上非常有用。像这样:

scala> class PredicateW[A](self: A => Boolean) {
   | def and(other: A => Boolean): A => Boolean = a => self(a) && other(a)
   | def or(other: A => Boolean): A => Boolean = a => self(a) || other(a)
   | def unary_! : A => Boolean = a => !self(a)
   | }
defined class PredicateW

还有一个隐式转换:

scala> implicit def Predicate_Is_PredicateW[A](p: A => Boolean) = new PredicateW(p)
Predicate_Is_PredicateW: [A](p: A => Boolean)PredicateW[A]

然后你可以使用它:

scala> map filterKeys (Set("1", "2") and Set("2", "3"))
res2: scala.collection.immutable.Map[java.lang.String,Int] = Map(2 -> 2)

scala> map filterKeys (Set("1", "2") or Set("2", "3"))
res3: scala.collection.immutable.Map[java.lang.String,Int] = Map(1 -> 1, 2 -> 2, 3 -> 3)

scala> map filterKeys !Set("2", "3")
res4: scala.collection.immutable.Map[java.lang.String,Int] = Map(1 -> 1)

这可以扩展到 xor , nand等等等等，如果你包含符号 unicode 可以使代码变得非常可读:

val mustReport = trades filter (uncoveredShort ∨ exceedsDollarMax)

val european = { 
  val Europe = (_ : Market).exchange.country.region == Region.EU
  trades filter (_.market ∈: Europe)
}