在Scala中反转Map(String – > List)
发布时间:2020-12-16 18:19:20 所属栏目:安全 来源:网络整理
导读:我有一个Map [String,List [String]],我想反转它.例如,如果我有类似的东西 "1" - List("a","b","c") "2" - List("a","j","k") "3" - List("a","c") 结果应该是 "a" - List("1","2","3") "b" - List("1") "c" - List("1","3") "j" - List("2") "k" - List("2"
我有一个Map [String,List [String]],我想反转它.例如,如果我有类似的东西
"1" -> List("a","b","c") "2" -> List("a","j","k") "3" -> List("a","c") 结果应该是 "a" -> List("1","2","3") "b" -> List("1") "c" -> List("1","3") "j" -> List("2") "k" -> List("2") 我试过这个: m.map(_.swap) 但它返回一个Map [List [String],String]: List("a","c") -> "1" List("a","k") -> "2" List("a","c") -> "3" 解决方法
地图反演有点复杂.
val m = Map("1" -> List("a","c"),"2" -> List("a","k"),"3" -> List("a","c")) m flatten {case(k,vs) => vs.map((_,k))} groupBy (_._1) mapValues {_.map(_._2)} //res0: Map[String,Iterable[String]] = Map(j -> List(2),a -> List(1,2,3),b -> List(1),c -> List(1,k -> List(2)) 将地图展平为一组元组. groupBy将使用旧值作为新键创建一个新Map.然后通过删除键(先前值)元素来取消元组值. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |