Can you simplify this, please?

groupBy

scala> case class A(a: Int, b: Int)
defined class A

scala> val as = Seq(A(1, 100), A(2, 100), A(3, 101))
as: Seq[A] = List(A(1,100), A(2,100), A(3,101))

scala> as.groupBy(_.b)
res3: scala.collection.immutable.Map[Int,Seq[A]] = Map(101 -> List(A(3,101)), 100 -> List(A(1,100), A(2,100)))

scala> res3.map{case (b, as) => b -> as.map(_.a)}
res5: scala.collection.immutable.Map[Int,Seq[Int]] = Map(101 -> List(3), 100 -> List(1, 2))

You could use a mutable MultiMap:

import collection.mutable.{HashMap, MultiMap, Set}
val mm = new HashMap[Int, Set[Int]] with MultiMap[Int, Int]
as.foldLeft(mm)((mm, a) => mm.addBinding(a.b, a.a))

See the MultiMap scaladocs for more details.

// using mapValues simplifies it a tiny bit
as.groupBy(_.b).mapValues(_.map(_.a))

// scalaz allows you to sum any foldable which has a monoid. Map has a monoid when its value-type has a monoid, which in this case is a List (which has a monoid)
import scalaz._
import std.map._ // monoid for Map
import std.list._ // monoid for List
import syntax.foldable._ // suml 

as.toList.map{ case A(a,b) => Map(b -> List(a))}.suml