Convert Seq[Seq[String,String]] to Map[String,Seq[String]]

Question

I need to convert this structure

val seq = Seq(Seq("a","aa"), Seq("b","bb"), Seq("a", "a2"), Seq("b","b2") )

to this Map:

val map2 = Map ( "a" -> Seq("aa","a2"), "b" -> Seq("bb","b2") )

cannot use toMap because it only works with Tuple2 as input. Any ideas how to approach this?

Xavier Guihot · Accepted Answer · 2018-05-06 15:59:03Z

2

You can first group by the first item of each sub-seq and then map the resulting grouped values to only keep the second element of subsequences:

Seq(Seq("a","aa"), Seq("b","bb"), Seq("a", "a2"), Seq("b","b2") )
  .groupBy(_(0)) // Map(b -> List(List(b, bb), List(b, b2)), a -> List(List(a, aa), List(a, a2)))
  .mapValues(_.map(_(1))) // Map(b -> List(bb, b2), a -> List(aa, a2))

which returns:

Map(b -> List(bb, b2), a -> List(aa, a2))

Similar: instead of using _(0) and _(1) you could also use .groupBy(_.head).mapValues(_.map(_.last))

The mapValues part can be made a bit more explicit this way:

.mapValues{
  case valueLists => // List(List(b, bb), List(b, b2))
    valueLists.map{
      case List(k, v) => v // List(b, bb) => bb
    }
}

edited May 6, 2018 at 15:59

answered May 6, 2018 at 15:47

Xavier Guihot

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3 Comments

ps0604 Over a year ago

Thanks for the response. If I use .groupBy(_.head).mapValues(_.map(_.last)) will the order of the keys be preserved? It seems that it doesn't.

Xavier Guihot Over a year ago

As we're producing a Map, the order of keys is not conserved as there isn't a notion of order with Map.

Xavier Guihot Over a year ago

Based on this answer, if instead of a Map you want List((a,List(aa, a2)), (b,List(bb, b2))) (whose keys are ordered as they appeared first), you can add this 2nd step to the produced unorderedMap: Seq(Seq("a","aa"), Seq("b","bb"), Seq("a", "a2"), Seq("b","b2") ).map(_(0)).distinct.map(k => (k, unorderedMap(k))). A foldLeft type of solution might also work.

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