Question

想象一下，我在Scala中有一个Map[String, String]。

我想匹配地图中的全套键值配对。

这样的事情应该是可能的

val record = Map("amenity" -> "restaurant", "cuisine" -> "chinese", "name" -> "Golden Palace")
record match {
    case Map("amenity" -> "restaurant", "cuisine" -> "chinese") => "a Chinese restaurant"
    case Map("amenity" -> "restaurant", "cuisine" -> "italian") => "an Italian restaurant"
    case Map("amenity" -> "restaurant") => "some other restaurant"
    case _ => "something else entirely"
}

编译器抱怨说：

error: value Map is not a case class constructor, nor does it have an unapply/unapplySeq method

目前哪种模式匹配Map

中的键值组合的最佳方式

Answer 1

您可以使用flatMap提取您感兴趣的值，然后与之匹配：

List("amenity","cuisine") flatMap ( record get _ ) match {
  case "restaurant"::"chinese"::_ => "a Chinese restaurant"
  case "restaurant"::"italian"::_ => "an Italian restaurant"
  case "restaurant"::_            => "some other restaurant"
  case _                          => "something else entirely"
}

请参阅this snippets page上的＃1。

您可以检查键的任意列表是否具有特定的值，如下所示：

if ( ( keys flatMap ( record get _ ) ) == values ) ...

请注意，即使地图上没有关键字，上述方法仍然有效，但如果关键字共享某些值，您可能希望使用map代替flatMap，并明确Some您的值列表中的/ None。例如。在这种情况下，如果“美食”可能不存在且“美食”的价值可能是“餐馆”（这个例子很愚蠢，但也许不在另一个上下文中），那么case "restaurant"::_将是模棱两可的。

另外，值得注意的是case "restaurant"::"chinese"::_比case List("restaurant","chinese")略高一些，因为后者不必要地检查在这两者之后没有更多的元素。

Answer 2

你可以查看有问题的值，将它们粘贴在一个元组中，并在其上进行模式匹配：

val record = Map("amenity" -> "restaurant", "cuisine" -> "chinese", "name" -> "Golden Palace")
(record.get("amenity"), record.get("cuisine")) match {
    case (Some("restaurant"), Some("chinese")) => "a Chinese restaurant"
    case (Some("restaurant"), Some("italian")) => "an Italian restaurant"
    case (Some("restaurant"), _) => "some other restaurant"
    case _ => "something else entirely"
}

或者，你可以做一些嵌套的匹配，这可能会更清洁：

val record = Map("amenity" -> "restaurant", "cuisine" -> "chinese", "name" -> "Golden Palace")
record.get("amenity") match {
  case Some("restaurant") => record.get("cuisine") match {
    case Some("chinese") => "a Chinese restaurant"
    case Some("italian") => "an Italian restaurant"
    case _ => "some other restaurant"
  }
  case _ => "something else entirely"
}

请注意map.get(key)返回Option[ValueType]（在这种情况下，ValueType将是String），因此它将返回None而不是如果该键中不存在该键则抛出异常地图。

Answer 3

模式匹配不是你想要的。您想要查找A是否完全包含B

val record = Map("amenity" -> "restaurant", "cuisine" -> "chinese", "name" -> "Golden Palace")
val expect = Map("amenity" -> "restaurant", "cuisine" -> "chinese")
expect.keys.forall( key => expect( key ) == record( key ) )

修改：添加匹配条件

这样您就可以轻松添加匹配条件

val record = Map("amenity" -> "restaurant", "cuisine" -> "chinese", "name" -> "Golden Palace")

case class FoodMatcher( kv: Map[String,String], output: String )

val matchers = List( 
    FoodMatcher(  Map("amenity" -> "restaurant", "cuisine" -> "chinese"), "chinese restaurant, che che" ),
    FoodMatcher(  Map("amenity" -> "restaurant", "cuisine" -> "italian"), "italian restaurant, mama mia" )
)

for {
    matcher <- matchers if matcher.kv.keys.forall( key => matcher.kv( key ) == record( key ) )
} yield matcher.output

给出：

List(chinese restaurant, che che)

Answer 4

我发现以下解决方案使用与case类最相似的提取器。它虽然主要是语法上的肉汁。

object Ex {
   def unapply(m: Map[String, Int]) : Option[(Int,Int) = for {
       a <- m.get("A")
       b <- m.get("B")
   } yield (a, b)
}

val ms = List(Map("A" -> 1, "B" -> 2),
    Map("C" -> 1),
    Map("C" -> 1, "A" -> 2, "B" -> 3),
    Map("C" -> 1, "A" -> 1, "B" -> 2)
    )  

ms.map {
    case Ex(1, 2) => println("match")
    case _        => println("nomatch")
}

Answer 5

另一个需要您指定要提取的键并允许您匹配值的版本如下：

SIZE

Answer 6

因为尽管同意所有其他答案都非常明智，但我有兴趣看看实际上是否有使用地图进行模式匹配的方法，我将以下内容放在一起。它使用与最佳答案相同的逻辑来确定匹配。

class MapSubsetMatcher[Key, Value](matcher: Map[Key, Value]) {
  def unapply(arg: Map[Key, Value]): Option[Map[Key, Value]] = {
    if (matcher.keys.forall(
      key => arg.contains(key) && matcher(key) == arg(key)
    ))
      Some(arg)
    else
      None
  }
}

val chineseRestaurant = new MapSubsetMatcher(Map("amenity" -> "restaurant", "cuisine" -> "chinese"))
val italianRestaurant = new MapSubsetMatcher(Map("amenity" -> "restaurant", "cuisine" -> "italian"))
val greatPizza = new MapSubsetMatcher(Map("pizza_rating" -> "excellent"))

val record = Map("amenity" -> "restaurant", "cuisine" -> "chinese", "name" -> "Golden Palace")
val frankies = Map("amenity" -> "restaurant", "cuisine" -> "italian", "name" -> "Frankie's", "pizza_rating" -> "excellent")


def matcher(x: Any): String = x match {
  case greatPizza(_) => "It's really good, you should go there."
  case chineseRestaurant(matchedMap) => "a Chinese restaurant called " +
    matchedMap.getOrElse("name", "INSERT NAME HERE")
  case italianRestaurant(_) => "an Italian restaurant"
  case _ => "something else entirely"
}

matcher(record)
// a Chinese restaurant called Golden Palace
matcher(frankies)
// It's really good, you should go there.

与Scala Map类型匹配的模式

6 个答案: