Combining two lists in Scala

问题:

From 2 lists of the form List[(Int, String):

l1 = List((1,"a"),(3,"b"))
l2 = List((3,"a"),(4,"c"))

how can I combine the Integers where the Strings are the same to get this third list:

l3 = List((4,"a"),(3,"b"),(4,"c"))

Right now I'm traversing both of the lists and adding if the strings are the same, but I think there should be a simple solution with pattern matching.

回答1:

val l = l1 ::: l2
val m = Map[String, Int]()
(m /: l) {
  case (map, (i, s)) => { map.updated(s, i + (map.get(s) getOrElse 0))}
}.toList // Note: Tuples are reversed.

But I suppose there is a more elegant way to do the updated part.

回答2:

How about,

(l1 ++ l2).groupBy(_._2).mapValues(_.unzip._1.sum).toList.map(_.swap)

Unpacking this a little on the REPL helps to show what's going on,

scala> l1 ++ l2
res0: List[(Int, java.lang.String)] = List((1,a), (3,b), (3,a), (4,c))

scala> res0.groupBy(_._2)
res1: ... = Map(c -> List((4,c)), a -> List((1,a), (3,a)), b -> List((3,b)))

scala> res1.mapValues(_.unzip)
res2: ... = Map(c -> (List(4),List(c)), a -> (List(1, 3),List(a, a)), b -> (List(3),List(b)))                         

scala> res1.mapValues(_.unzip._1)                                                                                                                                                                      
res3: ... = Map(c -> List(4), a -> List(1, 3), b -> List(3))                                                                                    

scala> res1.mapValues(_.unzip._1.sum)
res4: ... = Map(c -> 4, a -> 4, b -> 3)                                                                                                               

scala> res4.toList                                                                                                                                                                                     
res5: List[(java.lang.String, Int)] = List((c,4), (a,4), (b,3))                                                                                                                                        

scala> res5.map(_.swap)
res6: List[(Int, java.lang.String)] = List((4,c), (4,a), (3,b))

回答3:

With Scalaz, this is a snap.

import scalaz._
import Scalaz._

val l3 = (l1.map(_.swap).toMap |+| l2.map(_.swap).toMap) toList

The |+| method is exposed on all types T for which there exists an implementation of Semigroup[T]. And it just so happens that the semigroup for Map[String, Int] is exactly what you want.

回答4:

for ( (k,v) <- (l1++l2).groupBy(_._2).toList ) yield ( v.map(_._1).sum, k )

回答5:

Note that with this solution, the lists are traversed twice.

val l3 = (l1 zip l2).foldRight(List[(Int, String)]()) {
  case ((firstPair @ (firstNumber, firstWord),
        secondPair @ (secondNumber, secondWord)),
        result) =>
    if (firstWord == secondWord)
      ((firstNumber + secondNumber), firstWord) :: result
    else
      firstPair :: secondPair :: result
}

回答6:

Another opaque onetwo-liner of questionable efficiency yet indubitable efficacy:

val lst = l1 ++ l2
lst.map(_._2).distinct.map(i => (lst.filter(_._2 == i).map(_._1).sum, i))

回答7:

val a = List(1,1,1,0,0,2)
val b = List(1,0,3,2)

scala> List.concat(a,b)
res31: List[Int] = List(1, 1, 1, 0, 0, 2, 1, 0, 3, 2)

(or) 

scala> a.:::(b)
res32: List[Int] = List(1, 0, 3, 2, 1, 1, 1, 0, 0, 2)

(or) 

scala> a ::: b
res28: List[Int] = List(1, 1, 1, 0, 0, 2, 1, 0, 3, 2)

回答8:

An alternative to Miles Sabin's answer using Scala 2.13's new groupMapReduce method which is (as its name suggests) an equivalent (more efficient) of a groupBy followed by mapValues and a reduce step:

(l1 ::: l2).groupMapReduce(_._2)(_._1)(_ + _).toList.map(_.swap)
// List[(Int, String)] = List((3,b), (4,a), (4,c))

This:

• prepends l1 to l2

• groups elements based on their second tuple part (group part of groupMapReduce)

• maps grouped values to their first tuple part (map part of groupMapReduce)

• reduces values (_ + _) by summing them (reduce part of groupMapReduce)

• and finally swaps tuples' parts.

This is an equivalent version performed in one pass (for the group/map/reduce part) through the List of:

(l1 ::: l2).groupBy(_._2).mapValues(_.map(_._1).reduce(_ + _)).toList.map(_.swap)