在scala中划分一个元组序列

Question

I often write the same code to partition a sequence of tuples:

def groupBy[A, B](s: Seq[(A, B)]) : Map[A, Seq[B]] = s.groupBy (_._1) map { case (k, values) => (k, values.map(_._2))}

Is there a better way ?

Answer 1

Scalaz foldMap is clearer, although possibly less efficient:

import scalaz._, Scalaz._

def groupBy[A, B](l: List[(A,B)]) = l foldMap {t => Map(t._1 → List(t._2))}

Since Scalaz only offers typeclass instances for concrete types like List or Vector you can only use those, not Seq . If you want a generic version you can use the typeclass explicitly:

def groupBy[S[_]: Foldable, A, B](s: S[(A, B)]) =
   s foldMap {t => Map(t._1 → List(t._2))}

This will work for List , Vector , or generally any S[_]: Foldable - but still not Seq where you don't know the concrete type.