How to sort a list of scala objects by sort order of other list?

Question

I am having following 2 lists in scala.

    case class Parents(name: String, savings: Double)
    case class Children(parentName: String, debt: Double)

    val parentList:List[Parents] = List(Parents("Halls",1007D), Parents("Atticus",8000D), Parents("Aurilius",900D))

    val childrenList:List[Children] = List(Children("Halls",9379.40D), Children("Atticus",9.48D), Children("Aurilius",1100.75D))

    val sortedParentList:List[Parents] = parentList.sortBy(_.savings).reverse

// sortedParentList = List(Parents(Atticus,8000.0), Parents(Halls,1007.0), Parents(Aurilius,900.0))

now my parenList is Sorted By savings in decreasing order, I want my childrenList to be sorted in the way that it follows parentList Order. i.e. expected order will be following

// sortedParentList = List(Children(Atticus,9.48D), Children(Halls,9379.40D), Children(Aurilius,1100.75D))

I would suggest to have a Map for Children instead of trying to propagate external sort order to another List — Sergey
– Sergey, Commented Jul 11, 2017 at 9:27
Or better, have a List[(Parents, Children)] and sort everything at once - you'll be able to call .unzip on such a list to get separate lists — Sergey
– Sergey, Commented Jul 11, 2017 at 9:27
I agree with @Sergey. Rather have a link from parent to child(ren). The fact that you have parents and children does indicate that there should be some association between them. — sheunis
– sheunis, Commented Jul 11, 2017 at 9:30

Dima · Accepted Answer · 2017-07-11 16:09:11Z

6

Well, if you know both lists are initially in the same order (you can always ensure that by sorting both by name), you can just sort them both in one go:

 val (sortedParentList, sortedChildrenList) = (parents zip children)
   .sortBy(-_._1.savings)
   .unzip

Or you can define the ordering ahead of time, and use it to sort both lists:

val order = parentList.map(p => p.name -> -p.savings).toMap
val sortedParentList = parentList.sortBy(order(_.name))
val sortedChildrenList = childrenList.sortBy(order(_.parentName))

Or you can sort parents first (maybe, they are already sorted), and then define the order:

val order = sortedParentList.zipWithIndex.map { case(p, idx) => p.name -> idx }.toMap
val sortedChildrenList = childrenList.sortBy(c => order(c.parentName))

edited Jul 11, 2017 at 16:09

answered Jul 11, 2017 at 10:44

Dima

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3 Comments

axnet Over a year ago

I am trying last one, its giving me error error: missing parameter type for expanded function ((x$1) => x$1.parentName)

axnet Over a year ago

I am using last one because, I do not want to disturb the order of the parentList. and I want to use that order at many subsequent places

Dima Over a year ago

Ok, it didn't like my wildcard ... I fixed the snippet above. Try it now.

sheunis · Accepted Answer · 2017-07-11 09:46:46Z

1

case class Parents(name: String, savings: Double)
case class Children(parentName: String, debt: Double)

val familiesList: List[(Parents, Children)] = List(
  Parents("Halls",1007D) -> Children("Halls",9379.40D),
  Parents("Atticus",8000D) -> Children("Atticus",9.48D),
  Parents("Aurilius",900D) -> Children("Aurilius",1100.75D))

val (sortedParents, sortedChildren) = familiesList.sortBy {
  case (parents, _) => -parents.savings
}.unzip

edited Jul 11, 2017 at 9:46

sheunis

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answered Jul 11, 2017 at 9:30

Sergey

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