如何合并两个元组列表？

Question

I have two lists in Scala, how to merge them such that the tuples are grouped together?

Is there an existing Scala list API which can do this or need I do it by myself?

Input:

 List((a,4), (b,1), (c,1), (d,1))
 List((a,1), (b,1), (c,1))

Expected output:

List((a,5),(b,2),(c,2),(d,1))

Answer 1

You can try the following one-line:

scala> ( l1 ++ l2 ).groupBy( _._1 ).map( kv => (kv._1, kv._2.map( _._2).sum ) ).toList
res6: List[(Symbol, Int)] = List(('a,5), ('c,2), ('b,2), ('d,1))

Where l1 and l2 are the lists of tuples you want merge.

Now, the breakdown:

• (l1 ++ l2) you just concatenate both lists
• .groupBy( _._1) you group all tuples by their first element. You will receive a Map with the first element as key and lists of tuples starting with this element as values.
• .map( kv => (kv._1, kv._2.map( _._2).sum ) ) you make a new map, with similar keys, but the values are the sum of all second elements.
• .toList you convert the result back to a list.

Alternatively, you can use pattern matching to access the tuple elements.

( l1 ++ l2 ).groupBy( _._1 ).map{
  case (key,tuples) => (key, tuples.map( _._2).sum ) 
}.toList

Answer 2

Alternatively you can also use mapValues to shorten the code.

mapValues , as you can probably guess, allows you to re-map just the value for each (key, value) pair in the Map created by groupBy .

In this case the function passed to mapValues reduces each (Char, Int) tuple to just the Int then sums the resulting List of Ints.

(l1 ::: l2).groupBy(_._1).mapValues(_.map(_._2).sum).toList

If the order of the output list needs to follow your example, just add sorted which relies on an Ordering[(Char, Int)] implicit instance.

(l1 ::: l2).groupBy(_._1).mapValues(_.map(_._2).sum).toList.sorted

Answer 3

If you can assume that both List[(A,B)] are ordered according to Ordering[A] , you could write something like:

def mergeLists[A,B](one:List[(A,B)], two:List[(A,B)])(op:(B,B)=>B)(implicit ord:Ordering[A]): List[(A,B)] = (one,two) match {
    case (xs, Nil) => xs
    case (Nil, ys) => ys
    case((a,b)::xs,(aa,bb)::ys) =>
      if (a == aa) (a, op(b,bb)) :: mergeLists(xs,ys)(op)(ord)
      else if (ord.lt(a,aa)) (a, b) :: mergeLists(xs, (aa,bb)::ys)(op)(ord)
      else (aa, bb) :: mergeLists((a,b) :: xs, ys)(op)(ord)
}

Unfortunately this isn't tail recursive.