Question

假设我有3个元组序列，看起来像这样：

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

我想映射这些并返回3条新的记录，如下所示：

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))

我正在寻找的新集合包含来自原始列表的不同密钥集的密钥以及Some(v)或None的值，具体取决于相应的原始序列是否包含元组。

我设法从原始列表中取出密钥：

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

我现在计划再次浏览整个列表，通过查看当前值和唯一键集来生成新列表。我想知道在集合框架中是否有一些漂亮的东西使它更清洁，更容易处理？也许没有点？

Answer 1

这是一种方法。映射键并查询每个映射以查看它是否包含密钥。

制作一组密钥以进行迭代。

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

三个元组序列

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每个Seq上的Set个密钥，并尝试获取密钥。

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 2

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 3

我喜欢Rex Kerr的回答。我评论说这个解决方案也运行良好，可能更清晰简洁。

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)))
}

它同样适用于一套。我不确定哪个表现更好。我可以测试两者。

Answer 4

这是我的解决方案：

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)))

转换Scala中的元组序列列表

4 个答案: