根据条件压缩非顺序列表答案

【问题标题】：Zip non sequential list based on condition根据条件压缩非顺序列表
【发布时间】：2018-01-23 09:34:30
【问题描述】：

我有一个场景，我根据条件在 scala 中压缩两个列表。它们可能不按顺序排列。最好的方法是什么？

我想将具有相同 requestId 的 DirectRetailCM 和 DirectRetailCM 分组为一个元组。

object Main extends App {
   case class SalesDoc(val id: Int, val name: String, val requestId: String) {}
   val list = List(
     SalesDoc(1, "ILLEGAL", "1"),
     SalesDoc(2, "DirectRetailCM", "1"),

     SalesDoc(3, "DirectRetailOffsetInvoice", "2"),
     SalesDoc(4, "DirectRetailCM", "2"),
     SalesDoc(5, "OTHER", "2"),

     SalesDoc(5, "DirectRetailCM", "LEFTOUT"),
     SalesDoc(6, "ILLEGAL2", "4"),

     SalesDoc(5, "OTHER", "3"),
     SalesDoc(7, "DirectRetailOffsetInvoice", "4"),
     SalesDoc(8, "DirectRetailCM", "4")
  )

 // I expect zip results of drOffsetInvoice and drCms as
List(
  (SalesDoc(3, "DirectRetailOffsetInvoice", "2"), SalesDoc(4, "DirectRetailCM", "2")),
  (SalesDoc(7, "DirectRetailOffsetInvoice", "4"), SalesDoc(8, "DirectRetailCM", "4"))
  )
}

我能想到的最初方法是

组 directRetailCM - list.filter(e => e.name == "DirectRetailCM")
组 DirectRetailOffsetInvoice - list.filter(e => e.name == "DirectRetailOffsetInvoice")
同时压缩 - 但可能不按顺序进行
可能存在没有对应的行

您能否建议我需要考虑的任何其他方法？

【问题讨论】：

标签： scala scala-collections scalaz

【解决方案1】：

// You don't need the val keyword for a case class
case class SalesDoc(id: Int, name: String, requestId: String)

val list = List(
  SalesDoc(1, "ILLEGAL", "1"),
  SalesDoc(2, "DirectRetailCM", "1"),

  SalesDoc(3, "DirectRetailOffsetInvoice", "2"),
  SalesDoc(4, "DirectRetailCM", "2"),
  SalesDoc(5, "OTHER", "2"),

  SalesDoc(5, "DirectRetailCM", "LEFTOUT"),
  SalesDoc(6, "ILLEGAL2", "4"),

  SalesDoc(5, "OTHER", "3"),
  SalesDoc(7, "DirectRetailOffsetInvoice", "4"),
  SalesDoc(8, "DirectRetailCM", "4")
)

// Find all of the DirectRetailOffsetInvoice items
val offsets = list.filter(_.name == "DirectRetailOffsetInvoice")

// Map over all of the DirectRetailOffsetInvoice items and see if there is matching DirectRetailCM item
val maybeMatched = offsets.map(offset => {
  val maybeCm = list.find(i => i.requestId == offset.requestId && i.name == "DirectRetailCM")

  // Return a tuple of type (SalesDoc, Option[SalesDoc])
  (offset, maybeCm)
})

// Map over the tuples and only take the ones where there was a match, and extract it from the Option to create a tuple of (SalesDoc, SalesDoc)
val output = maybeMatched.collect { case (s1, Some(s2)) => (s1, s2) }

output.foreach(println)
// (SalesDoc(3,DirectRetailOffsetInvoice,2),SalesDoc(4,DirectRetailCM,2))
// (SalesDoc(7,DirectRetailOffsetInvoice,4),SalesDoc(8,DirectRetailCM,4))

【讨论】：

谢谢 Tyler，我可能会在 DirectRetailCM 上再使用一个过滤器来避免每个 DirectRetailOffsetInvoice 循环。
如果您的行数比您的示例建议的多得多，您可能希望构建一个地图，以便您可以更快地查找。

【解决方案2】：

您可以使用标准的 Scala 组合器实现此目的

list
  .filter(sd => sd.name == "DirectRetailCM" || sd.name == "DirectRetailOffsetInvoice")
  .groupBy(_.requestId)
  .flatMap {
     case (_, List(a,b)) => List(a->b)
     case _ => List.empty
  }

这给了你：

 res3: scala.collection.immutable.Map[SalesDoc,SalesDoc] = 
       Map(
         SalesDoc(3,DirectRetailOffsetInvoice,2) -> SalesDoc(4,DirectRetailCM,2), 
         SalesDoc(7,DirectRetailOffsetInvoice,4) -> SalesDoc(8,DirectRetailCM,4))

如果输入序列未排序且DirectRetailOffsetInvoice 在DirectRetailCM 之前，则需要对其进行处理。

【讨论】：

【解决方案3】：

list.filter(s => s.name == "DirectRetailCM" || s.name == "DirectRetailOffsetInvoice")
    .groupBy(_.requestId)
    .collect { case (_, List(a, b)) => (a, b) }
    .toList

// List[(SalesDoc, SalesDoc)]

【讨论】：