刪除不包含集合中任何值的密鑰

Question

val s = Set("blue", "orange") 
val m = Map("product_orange_123" -> 1, "prodoct_blue_123" -> 2, "product_green_123" -> 5, "product_blue_887" -> 7) 

我想刪除映射中不包含集合中任何值的任何密鑰。

預期輸出：

("product_orange_123" -> 1, "prodoct_blue_123" -> 2, "product_blue_887" -> 7) 

Answer 1

嗯，我想你的意思地圖

篩選鍵不包含集合中的任何值

m.filterKeys(key => s.exists(key.contains(_))) 

這會做

Answer 2

設置擴展功能1

收集在動作在動作

m.collect { case (k, v) if s(k.split("_")(1)) => k -> v } 

篩選

m.filterKeys(key => s(key.split("_")(1))) 

濾波器在動作

m.filter { case (k, _) => s(k.split("_")(1)) } 

說明

Set延伸Function1，並且set實例可以直接應用於一個密鑰以檢查它是否存在於該集合中。

scala> val s = Set("blue", "orange") 
s: scala.collection.immutable.Set[String] = Set(blue, orange) 

scala> s("blue") 
res0: Boolean = true 

scala> s("apple") 
res1: Boolean = false 

scala> val s = Set("blue", "orange") 
s: scala.collection.immutable.Set[String] = Set(blue, orange) 

scala> val m = Map("product_orange_123" -> 1, "prodoct_blue_123" -> 2, "product_green_123" -> 5, "product_blue_887" -> 7) 
m: scala.collection.immutable.Map[String,Int] = Map(product_orange_123 -> 1, prodoct_blue_123 -> 2, product_green_123 -> 5, product_blue_887 -> 7) 

scala> m.collect { case (k, v) if s(k.split("_")(1)) => k -> v } 
res2: scala.collection.immutable.Map[String,Int] = Map(product_orange_123 -> 1, prodoct_blue_123 -> 2, product_blue_887 -> 7) 

scala> m.filterKeys(key => s(key.split("_")(1))) 
res3: scala.collection.immutable.Map[String,Int] = Map(product_orange_123 -> 1, prodoct_blue_123 -> 2, product_blue_887 -> 7) 

scala> m.filter { case (k, _) => s(k.split("_")(1)) } 
res4: scala.collection.immutable.Map[String,Int] = Map(product_orange_123 -> 1, prodoct_blue_123 -> 2, product_blue_887 -> 7)