Scala: idiomatic way to merge list of maps with the greatest value of each key?

Question

I have a List of Map[Int, Int], that all have the same keys (from 1 to 20) and I'd like to merge their contents into a single Map[Int, Int].

I've read another post on stack overflow about merging maps that uses |+| from the scalaz library.

I've come up with the following solution, but it seems clunky to me.

val defaultMap = (2 to ceiling).map((_,0)).toMap
val factors: Map[Int, Int] = (2 to ceiling). map(primeFactors(_)).
        foldRight(defaultMap)(mergeMaps(_, _))

def mergeMaps(xm: Map[Int, Int], ym: Map[Int, Int]): Map[Int,Int] = {
    def iter(acc: Map[Int,Int], other: Map[Int,Int], i: Int): Map[Int,Int] = {
      if (other.isEmpty) acc
      else iter(acc - i + (i -> math.max(acc(i), other(i))), other - i, i + 1)
    }
    iter(xm, ym, 2)
  }

def primeFactors(number: Int): Map[Int, Int] = {
  def iter(factors: Map[Int,Int], rem: Int, i: Int): Map[Int,Int] = {
    if (i > number) factors
    else if (rem % i == 0) iter(factors - i + (i -> (factors(i)+1)), rem / i, i)
    else iter(factors, rem, i + 1)
  }
  iter((2 to ceiling).map((_,0)).toMap, number, 2)
}

Explanation: val factors creates a list of maps that each represent the prime factors for the numbers from 2-20; then these 18 maps are folded into a single map containing the greatest value for each key.

UPDATE

Using the suggestion of @folone, I end up with the following code (a definite improvement over my original version, and I don't have to change the Maps to HashMaps):

import scalaz._
import Scalaz._
import Tags._

/**
 * Smallest Multiple
 *
 * 2520 is the smallest number that can be divided by each of the numbers 
 * from 1 to 10 without any remainder. What is the smallest positive number 
 * that is evenly divisible by all of the numbers from 1 to 20?
 *
 * User: Alexandros Bantis
 * Date: 1/29/13
 * Time: 8:07 PM
 */
object Problem005 {

  def findSmallestMultiple(ceiling: Int): Int = {
    val factors = (2 to ceiling).map(primeFactors(_).mapValues(MaxVal)).reduce(_ |+| _)
    (1 /: factors.map(m => intPow(m._1, m._2)))(_ * _)
  }

  private def primeFactors(number: Int): Map[Int, Int] = {
    def iter(factors: Map[Int,Int], rem: Int, i: Int): Map[Int,Int] = {
      if (i > number) factors.filter(_._2 > 0).mapValues(MaxVal)
      else if (rem % i == 0) iter(factors - i + (i -> (factors(i)+1)), rem / i, i)
      else iter(factors, rem, i + 1)
    }
    iter((2 to number).map((_,0)).toMap, number, 2)
  }

  private def intPow(x: Int, y: Int): Int = {
    def iter(acc: Int, rem: Int): Int = {
      if (rem == 0) acc
      else iter(acc * x, rem -1)
    }
    if (y == 0) 1 else iter(1, y)
  }
}

axel22 · Answer 1 · 2013-02-27T19:44:12.367

This solution does not work for general Maps, but if you are using immutable.HashMaps you may consider the merged method:

def merged[B1 >: B](that: HashMap[A, B1])(mergef: ((A, B1), (A, B1)) ⇒ (A, B1)): HashMap[A, B1]

Creates a new map which is the merge of this and the argument hash map.

Uses the specified collision resolution function if two keys are the same. The collision resolution function will always take the first argument from this hash map and the second from that.

The merged method is on average more performant than doing a traversal and reconstructing a new immutable hash map from scratch, or ++.

Use case:

val m1 = immutable.HashMap[Int, Int](1 -> 2, 2 -> 3)
val m2 = immutable.HashMap[Int, Int](1 -> 3, 4 -> 5)
m1.merged(m2) {
  case ((k1, v1), (k2, v2)) => ((k1, math.max(v1, v2)))
}

George · Accepted Answer · 2013-03-01T09:38:11.727

As your tags suggest, you might be interested in a scalaz solution. Here goes:

> console
[info] Starting scala interpreter...
[info] 
Welcome to Scala version 2.10.0 (OpenJDK 64-Bit Server VM, Java 1.7.0_15).
Type in expressions to have them evaluated.
Type :help for more information.

scala> import scalaz._, Scalaz._, Tags._
import scalaz._
import Scalaz._
import Tags._

There exists a Semigroup instance for Ints under a maximum operation:

scala> Semigroup[Int @@ MaxVal]
res0: scalaz.Semigroup[scalaz.@@[Int,scalaz.Tags.MaxVal]] = scalaz.Semigroup$$anon$9@15a9a9c6

Let's just use it:

scala> val m1 = Map(1 -> 2, 2 -> 3) mapValues MaxVal
m1: scala.collection.immutable.Map[Int,scalaz.@@[Int,scalaz.Tags.MaxVal]] = Map(1 -> 2, 2 -> 3)

scala> val m2 = Map(1 -> 3, 4 -> 5) mapValues MaxVal
m2: scala.collection.immutable.Map[Int,scalaz.@@[Int,scalaz.Tags.MaxVal]] = Map(1 -> 3, 4 -> 5)

scala> m1 |+| m2
res1: scala.collection.immutable.Map[Int,scalaz.@@[Int,scalaz.Tags.MaxVal]] = Map(1 -> 3, 4 -> 5, 2 -> 3)

If you're interested in how this "tagging" (the @@ thing) works, here's a good explanation: http://etorreborre.blogspot.de/2011/11/practical-uses-for-unboxed-tagged-types.html

Xavier Guihot · Answer 3 · 2019-03-31T18:00:29.460

Starting Scala 2.13, another solution only based on the standard library consists in merging the Maps as sequences before applying a groupMapReduce which (as its name suggests) is an equivalent of a groupBy followed by a mapping and a reduce step on values:

// val map1 = Map(1 -> 2, 2 -> 3)
// val map2 = Map(1 -> 3, 4 -> 5)
(map1.toSeq ++ map2).groupMapReduce(_._1)(_._2)(_ max _)
// Map[Int,Int] = Map(2 -> 3, 4 -> 5, 1 -> 3)

This:

concatenates the two maps as a sequence of tuples (List((1,2), (2,3), (1,3), (4,5))). For conciseness, map2 is implicitly converted to Seq to adopt the type of map1.toSeq - but you could choose to make it explicit by using map2.toSeq.
groups elements based on their first tuple part (group part of groupMapReduce)
maps grouped values to their second tuple part (map part of groupMapReduce)
reduces mapped values (_ max _) by taking their max (reduce part of groupMapReduce)

Scala: idiomatic way to merge list of maps with the greatest value of each key?

UPDATE

3 Answers3