Scala: grouped from right? - scala

Question

In Scala, grouped works from left to right.
val list = List(1,2,3,4,5)
list.grouped(2).toList
=> List[List[Int]] = List(List(1, 2), List(3, 4), List(5))
But what if I want:
=> List[List[Int]] = List(List(1), List(2, 3), List(4, 5))
?
Well I know this works:
list.reverse.grouped(2).map(_.reverse).toList.reverse
It seems not efficient, however.

Answer 1

Then you could implement it by yourself:
def rgrouped[T](xs: List[T], n: Int) = {
val diff = xs.length % n
if (diff == 0) xs.grouped(n).toList else {
val (head, toGroup) = xs.splitAt(diff)
List(head, toGroup.grouped(n).toList.head)
}
}
Quite ugly, but should work.

Answer 2

Here is my attempt:
def rightGrouped[T](ls:List[T], s:Int) = {
val a = ls.length%s match {
case 0 => ls.grouped(s)
case x => List(ls.take(x)) ++ ls.takeRight(ls.length-x).grouped(s)
}
a.toList
}
Usage:
scala> rightGrouped(List(1,2,3,4,5),3)
res6: List[List[Int]] = List(List(1, 2), List(3, 4, 5))
I initially tried without pattern matching, but it was wrong when the list was "even"
val ls = List(1,2,3,4,5,6)
val s = 3
val x = ls.length % s
List(ls.take(x)) ++ ls.takeRight(ls.length-x).grouped(s)
produced:
List(List(), List(1, 2, 3), List(4, 5, 6))

Answer 3

val l =List(list.head)::(list.tail grouped(2) toList)
EDIT:
After #gzm0 pointed out my mistake I have fixed the solution, though it works only for n=2
def group2[T](list: List[T]) ={
(list.size % 2 == 0) match {
case true => list.grouped(2).toList
case false => List(list.head) :: (list.tail grouped(2) toList)
}
}
println(group2(List()))
println(group2(List(1,2,3,4,5)))
println(group2(List(1,2,3,4,5,6)))
List()
List(List(1), List(2, 3), List(4, 5))
List(List(1, 2), List(3, 4), List(5, 6))

Answer 4

Staying consistent with idiomatic use of the Scala Collections Library such that it also works on things like String, here's an implementation.
def groupedRight[T](seqT: Seq[T], width: Int): Iterator[Seq[T]] =
if (width > 0) {
val remainder = seqT.length % width
if (remainder == 0)
seqT.grouped(width)
else
(seqT.take(remainder) :: seqT.drop(remainder).grouped(width).toList).iterator
}
else
throw new IllegalArgumentException(s"width [$width] must be greater than 0")
val x = groupedRight(List(1,2,3,4,5,6,7), 3).toList
// => val x: List[Seq[Int]] = List(List(1), List(2, 3, 4), List(5, 6, 7))
val sx = groupedRight("12345", 3).toList
// => val sx: List[Seq[Char]] = List(12, 345)
val sx = groupedRight("12345", 3).toList.map(_.mkString)
// => val sx: List[String] = List(12, 345)