Scala: split card size n into a list (max card size 3)

val a = Map("k1" -> "v1", "k2" -> "v2", "k3" -> "v3", "k4" -> "v4", "k5" -> "v5", "k6" -> "v6", "k7" -> "v7", "k8" -> "v8", "k9" -> "v9", "k0" -> "v0")
a.grouped(3).toList

This gives you:

res2: List[scala.collection.immutable.Map[String,String]] = List(Map(k2 -> v2, k0 -> v0, k5 -> v5), Map(k9 -> v9, k6 -> v6, k7 -> v7), Map(k1 -> v1, k4 -> v4, k3 -> v3), Map(k8 -> v8))

The only thing he did not order

To maintain order, you can do something like this:

a.toList.sortBy(_._1).grouped(3).toList.map(_.toMap)

What gives you:

res6: List[scala.collection.immutable.Map[String,String]] = List(Map(k0 -> v0, k1 -> v1, k2 -> v2), Map(k3 -> v3, k4 -> v4, k5 -> v5), Map(k6 -> v6, k7 -> v7, k8 -> v8), Map(k9 -> v9))

Note that your start is Mapnot sorted correctly (the last element is "k0", but it must be the first). But if you want to keep the insertion order and have a list of cards grouped by 3, this should work:

val b = scala.collection.mutable.LinkedHashMap("k1" -> "v1", "k2" -> "v2", "k3" -> "v3", "k4" -> "v4", "k5" -> "v5", "k6" -> "v6", "k7" -> "v7", "k8" -> "v8", "k9" -> "v9", "k0" -> "v0")
b.toList.grouped(3).toList.map(_.toMap)

It leads to:

res8: List[scala.collection.immutable.Map[String,String]] = List(Map(k1 -> v1, k2 -> v2, k3 -> v3), Map(k4 -> v4, k5 -> v5, k6 -> v6), Map(k7 -> v7, k8 -> v8, k9 -> v9), Map(k0 -> v0))