在scala中将贴图列表转换为单个贴图

可能重复：

我有以下需要转换为单个贴图的贴图列表。最好的方法是什么

地图列表的类型为：

Iterable[Map[String, Map[String,Float]]

例如：

val list1 = List(Map(1216 -> Map(key1 -> 144.0)), 
     Map(1253 -> Map(key1 -> 144.0)), 
     Map(1359 -> Map(key1 -> 144.0))

val list2 = List(Map(1216 -> Map(key2 -> 148.0)), 
     Map(1200 -> Map(key2 -> 144.0)), 
     Map(1359 -> Map(key2 -> 144.0))

我想要一张结果图

val map3 = Map(
        1216 -> Map(key1 -> 1440, key2 -> 148.0), 
        1359 -> Map(key1 -> 1440, key2 -> 144.0))

---

谢谢

首先，找到新地图中应该存在的所有键，这是两个地图列表中都存在的键（我假设）：

然后选择所有这些关键点的值：

val alllist = list1 ++ list2
val map3 = keys.map(k => k -> alllist.flatMap(_ get k).reduceLeft(_ ++ _))

---

虽然效率不是很高，但它能完成任务。

如果我做对了，像这样的方法就行了：

scala> import scalaz._
import scalaz._

scala> import Scalaz._
import Scalaz._

scala> list1 ++ list2
res2: List[scala.collection.immutable.Map[Int,scala.collection.immutable.Map[java.lang.String,Double]]] = List(Map(1216 -> Map(key1 -> 144.0)), Map(1253 -> Map(key1 -> 144.0)), Map(1359 -> Map(key1 -> 144.0)), Map(1216 -> Map(key2 -> 148.0)), Map(1200 -> Map(key2 -> 144.0)), Map(1359 -> Map(key2 -> 144.0)))


scala> .foldLeft(Map[Int,Seq[Map[String,Double]]]()) { case (acc, v) => 
     | v.mapValues(Seq(_)) |+| acc
     | }
res8: scala.collection.immutable.Map[Int,Seq[Map[String,Double]]] = Map(1216 -> List(Map(key2 -> 148.0), Map(key1 -> 144.0)), 1253 -> List(Map(key1 -> 144.0)), 1359 -> List(Map(key2 -> 144.0), Map(key1 -> 144.0)), 1200 -> List(Map(key2 -> 144.0)))


scala> .map { case(i, m) => (i, m reduce (_ ++ _)) }
res11: scala.collection.immutable.Map[Int,Map[String,Double]] = Map(1216 -> Map(key2 -> 148.0, key1 -> 144.0), 1253 -> Map(key1 -> 144.0), 1359 -> Map(key2 -> 144.0, key1 -> 144.0), 1200 -> Map(key2 -> 144.0))y2 -> 148.0)), Map(1200 -> Map(key2 -> 144.0)), Map(1359 -> Map(key2 -> 144.0)))

scala> import scalaz._
import scalaz._

scala> import Scalaz._
import Scalaz._

scala> list1 ++ list2
res2: List[scala.collection.immutable.Map[Int,scala.collection.immutable.Map[java.lang.String,Double]]] = List(Map(1216 -> Map(key1 -> 144.0)), Map(1253 -> Map(key1 -> 144.0)), Map(1359 -> Map(key1 -> 144.0)), Map(1216 -> Map(key2 -> 148.0)), Map(1200 -> Map(key2 -> 144.0)), Map(1359 -> Map(key2 -> 144.0)))


scala> .foldLeft(Map[Int,Seq[Map[String,Double]]]()) { case (acc, v) => 
     | v.mapValues(Seq(_)) |+| acc
     | }
res8: scala.collection.immutable.Map[Int,Seq[Map[String,Double]]] = Map(1216 -> List(Map(key2 -> 148.0), Map(key1 -> 144.0)), 1253 -> List(Map(key1 -> 144.0)), 1359 -> List(Map(key2 -> 144.0), Map(key1 -> 144.0)), 1200 -> List(Map(key2 -> 144.0)))


scala> .map { case(i, m) => (i, m reduce (_ ++ _)) }
res11: scala.collection.immutable.Map[Int,Map[String,Double]] = Map(1216 -> Map(key2 -> 148.0, key1 -> 144.0), 1253 -> Map(key1 -> 144.0), 1359 -> Map(key2 -> 144.0, key1 -> 144.0), 1200 -> Map(key2 -> 144.0))y2 -> 148.0)), Map(1200 -> Map(key2 -> 144.0)), Map(1359 -> Map(key2 -> 144.0)))