How to combine elements of multiple lists in scala

Question

I am new to Scala. I have three List.

List("XX", None,None,None)

List( None,"YY",None,None)

List(None,None,None, "ZZ")

I need to merge these list to create a single list which should look like

List("XX","YY",None,"ZZ")

Is there any way in scala to achieve this result? Thanks

Does the order matter? Does the types are always String or None? — roterl
– roterl, Commented Sep 29, 2015 at 18:16

Nyavro · Accepted Answer · 2015-09-29 11:29:57Z

5

Maybe you need this?

val list: List[List[Option[String]]] = List(
  List(Some("XX"), None, None, None),
  List(None, Some("YY"), None, None),
  List(None, None, None, Some("ZZ"))
)
list.tail.foldLeft(list.head) {
  case (acc, item) => acc.zip(item).map {
    case (Some(x), _) => Some(x)
    case (_, y) => y
  }
}

answered Sep 29, 2015 at 11:29

Nyavro

8,8762 gold badges29 silver badges35 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

som-snytt Over a year ago

The fold function can be { case (a, x) => (a, x).zipped map { case (y, z) => y orElse z } } where zipped avoids an intermediate collection.

Seth Tisue · Accepted Answer · 2015-09-29 13:37:28Z

3

Your lists are List[AnyRef]. That's a type that should never appear in ordinary Scala code.

I would suggest representing your data using List[Option[String]]:

scala> List(List(Some("XX"), None, None, None),
     |      List(None, Some("YY"), None, None),
     |      List(None, None, None, Some("ZZ")))
res2: List[List[Option[String]]] = List(...

Now your problem is easy to solve:

scala> res2.transpose.map(_.flatten.headOption)
res6: List[Option[String]] = List(Some(XX), Some(YY), None, Some(ZZ))

I've assumed here that if there are multiple Somes in the same position, you want to take the first one.

answered Sep 29, 2015 at 13:37

Seth Tisue

30.6k12 gold badges88 silver badges152 bronze badges

Comments

Shyamendra Solanki · Accepted Answer · 2015-09-29 11:53:43Z

2

val l1 = List("XX", None, None, None)
val l2 = List(None, "YY", None, None)
val l3 = List(None, None, None, "ZZ")

val listOfList = List(l1, l2, l3) 

// Assuming all lists are of same length.

val lT = listOfList.transpose  // swap column and rows
val result = lT.map{ _.reduce{ (a, b) => 
  if (a.toString == "None") b else a
}}
// List[Serializable] = List(XX, YY, None, ZZ)

answered Sep 29, 2015 at 11:53

Shyamendra Solanki

8,8512 gold badges33 silver badges25 bronze badges

2 Comments

som-snytt Over a year ago

Testing for None that way is especially stringly (i.e., bad). Bad as in awful.

Shyamendra Solanki Over a year ago

@som-snytt point accepted. but starting from given data, its a way to achieve the result. others have made the point that starting place should be List of options. I agree.

elm · Accepted Answer · 2015-09-29 12:08:13Z

2

For

val xs = List(list1,list2,list2)

consider

xs.flatten.distinct

answered Sep 29, 2015 at 12:08

elm

20.5k15 gold badges72 silver badges116 bronze badges

Collectives™ on Stack Overflow

How to combine elements of multiple lists in scala

4 Answers 4

1 Comment

Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related