变换元组序列的列表在斯卡拉

Question

比方说，我有一个看起来像这样3元组序列：

Seq("m1" -> 1, "m2" -> 2) 
Seq("m3" -> 3, "m4" -> 4) 
Seq("m5" -> 5, "m2" -> 6) 

我要地图对这些并返回看起来像下面的3个新的记录：

Seq("m1" -> Some(1), "m2" -> Some(2), "m3" -> None, "m4" -> None, "m5" -> None) 
Seq("m1" -> None, "m2" -> None, "m3" -> Some(3), "m4" -> Some(4), "m5" -> None) 
Seq("m1" -> None, "m2" -> Some(6), "m3" -> None, "m4" -> None, "m5" -> Some(5)) 

，我正在寻找新的集合包含了不同的一组基础上，相应的原始序列是否不包含元组从最初的名单中Some(v)或None键和值的关键。

我设法从原来的列表中拔出钥匙：

case class SiteReading(val site: String, val measures: Seq[(String, Double)]) 
val xs = getSomeSiteReadingsFromSomewhere() 
val keys = xs.flatMap(_.measures.map(t => t._1)).distinct 

我现在正计划通过观察两个电流值和一套独特的再次经历整个列表，产生一个新的列表键。我想知道在这个集合框架中是否有一些漂亮的东西可以让它变得更清洁，更容易处理？也许是免费的？

Answer 1

val s1 = Seq("m1" -> 1, "m2" -> 2) 
val s2 = Seq("m3" -> 3, "m4" -> 4) 
val s3 = Seq("m5" -> 5, "m2" -> 6) 
val ss = Seq(s1, s2, s3) 
def foo(xss: Seq[Seq[(String,Int)]]): Seq[Seq[(String,Option[Int])]] = { 
    val keys = xss.flatMap(_.map(_._1)).toSet 
    xss.map{ xs => 
    val found = xs.map{ case (s,i) => (s, Some(i)) } 
    val missing = (keys diff xs.map(_._1).toSet).map(x => (x, None)).toSeq 
    (found ++ missing).sortBy(_._1) 
    } 
} 

scala> foo(ss).foreach(println) 
List((m1,Some(1)), (m2,Some(2)), (m3,None), (m4,None), (m5,None)) 
List((m1,None), (m2,None), (m3,Some(3)), (m4,Some(4)), (m5,None)) 
List((m1,None), (m2,Some(6)), (m3,None), (m4,None), (m5,Some(5))) 

Answer 2

这里是一个办法。在键上映射并查询每个映射以查看它是否包含键。

让一组键来遍历。

scala> val ms = (1 to 5).map(i => "m" + i) 
ms: scala.collection.immutable.IndexedSeq[String] = Vector(m1, m2, m3, m4, m5) 

三元组序列

scala> val s1 = Seq("m1" -> 1, "m2" -> 2).toMap 
s1: scala.collection.immutable.Map[String,Int] = Map(m1 -> 1, m2 -> 2) 

scala> val s2 = Seq("m3" -> 3, "m4" -> 4).toMap 
s2: scala.collection.immutable.Map[String,Int] = Map(m3 -> 3, m4 -> 4) 

scala> val s3 = Seq("m5" -> 5, "m2" -> 6).toMap 
s3: s3: scala.collection.immutable.Map[String,Int] = Map(m5 -> 5, m2 -> 6) 

map键Seq在每个Set，并尝试得到钥匙。

scala> ms.map(m => m -> s1.get(m)) 
res19: scala.collection.immutable.IndexedSeq[(String, Option[Int])] = 
Vector((m1,Some(1)), (m2,Some(2)), (m3,None), (m4,None), (m5,None)) 

scala> ms.map(m => m -> s2.get(m)) 
res20: scala.collection.immutable.IndexedSeq[(String, Option[Int])] = 
Vector((m1,None), (m2,None), (m3,Some(3)), (m4,Some(4)), (m5,None)) 

scala> ms.map(m => m -> s3.get(m)) 
res21: scala.collection.immutable.IndexedSeq[(String, Option[Int])] = 
Vector((m1,None), (m2,Some(6)), (m3,None), (m4,None), (m5,Some(5))) 

Answer 3

这里是我的解决方案：

val s1 = Seq("m1" -> 1, "m2" -> 2) 
val s2 = Seq("m3" -> 3, "m4" -> 4) 
val s3 = Seq("m5" -> 5, "m2" -> 6) 

def process(ss: Seq[(String, Int)]*): Seq[Seq[(String, Option[Int])]] = { 
    val asMap = ss map (_.toMap) 
    val keys = asMap.flatMap(_.keys).sorted 
    for(m <- asMap) yield keys.map(k => k -> m.get(k)) 
} 

val Seq(r1, r2, r3) = process(s1, s2, s3) 

结果：

r1: Seq[(String, Option[Int])] = ArrayBuffer((m1,Some(1)), (m2,Some(2)), (m2,Some(2)), (m3,None), (m4,None), (m5,None)) 
r2: Seq[(String, Option[Int])] = ArrayBuffer((m1,None), (m2,None), (m2,None), (m3,Some(3)), (m4,Some(4)), (m5,None)) 
r3: Seq[(String, Option[Int])] = ArrayBuffer((m1,None), (m2,Some(6)), (m2,Some(6)), (m3,None), (m4,None), (m5,Some(5))) 

Answer 4

我喜欢雷克斯·科尔的答案。我评论说，这个解决方案也运行良好，可能更清晰简明。

def denormalize(xss: Seq[Seq[(String, Double)]]): Seq[Map[String, Option[Double]]] = { 
    val keys = xss.flatMap(_.map(_._1)).distinct.sorted 
    val base = keys.map(_ -> None).toMap[String, Option[Double]] 
    xss.map(base ++ _.map(t => t._1 -> Option(t._2))) 
} 

它会同样工作和一套。我不确定哪个表现更好。我可能会测试两者。