Scala:合法的方式来合并每个键的最大值的地图列表？

Question

我有一个Map [Int,Int]列表,它们都有相同的键(从1到20),我想将它们的内容合并到一个Map [Int,Int]中.

我已经阅读了关于合并使用|+|scalaz库的地图的堆栈溢出的另一篇文章.

我想出了以下解决方案,但对我来说似乎很笨拙.

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)
}

说明:val factors创建一个地图列表,每个地图代表2-20个数字的素因子; 然后将这18张地图折叠成一张包含每个键最大值的地图.

UPDATE

使用@folone的建议,我最终得到以下代码(对我的原始版本有一个明显的改进,我不必将地图更改为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)
  }
}

Answer 1

此解决方案不适用于一般Maps,但如果您使用immutable.HashMaps,您可以考虑以下merged方法:

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

创建一个新映射,它是this和参数哈希映射的合并.

如果两个键相同,则使用指定的冲突解决功能.冲突解决函数将始终从此哈希映射中获取第一个参数,并从中获取第二个参数.

合并的方法平均比执行遍历和从头开始重构新的不可变哈希映射(或++)更高效.

使用案例:

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)))
}

Answer 2

正如您的标签所示，您可能对 scalaz 解决方案感兴趣。开始：

> 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._

在最大运算下，存在 Ints 的 Semigroup 实例：

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

让我们使用它：

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)

如果您对这种“标记”（事物）的工作原理感兴趣@@，这里有一个很好的解释：http://etorreborre.blogspot.de/2011/11/practical-uses-for-unboxed-tagged-types.html