模式匹配Scala Map类型

发布时间：2020-12-16 09:20:15 所属栏目：安全来源：网络整理

导读：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. 这样的东西应该是可能的 val record = Map("amenity" - "restaurant","cuisine" - "chinese","name" - "Golden Palace")record match { case Map("amenity" - "restaurant",

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.

这样的东西应该是可能的

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：

error：value Map不是一个case类构造函数,也没有一个unapply / unapplySeq方法

目前在地图中为键值组合模式匹配的最佳方式是什么？

解决方法

模式匹配不是你想要的.你想找到A是否完全包含B

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

编辑：添加匹配条件

这样,您可以轻松添加匹配条件

val record = Map("amenity" -> "restaurant","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

得到：

名单(中餐馆,che che)

（编辑：李大同）

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