如何将Map [String，Seq [String]]转换为Map [String，String]

Question

我有一个Map [String，Seq [String]]并希望基本上将它转换为Map [String，String]，因为我知道序列只有一个值。

Answer 1

Someone else already mentioned mapValues , but if I were you I would do it like this:

scala> val m = Map(1 -> Seq(1), 2 -> Seq(2))
m: scala.collection.immutable.Map[Int,Seq[Int]] = Map(1 -> List(1), 2 -> List(2))

scala> m.map { case (k,Seq(v)) => (k,v) }
res0: scala.collection.immutable.Map[Int,Int] = Map(1 -> 1, 2 -> 2)

Two reasons:

1. The mapValues method produces a view of the result Map, meaning that the function will be recomputed every time you access an element. Unless you plan on accessing each element exactly once, or you only plan on accessing a very small percentage of them, you don't want that recomputation to take place.

2. Using a case with (k,Seq(v)) ensures that an exception will be thrown if the function ever sees a Seq that doesn't contain exactly one element. Using _(0) or _.head will throw an exception if there are zero elements, but will not complain if you had more than one, which will likely result in mysterious bugs later on when things go missing without errors.

Answer 2

You can use mapValues() .

scala> Map("a" -> Seq("aaa"), "b" -> Seq("bbb"))
res0: scala.collection.immutable.Map[java.lang.String,Seq[java.lang.String]] = M
ap(a -> List(aaa), b -> List(bbb))

scala> res0.mapValues(_(0))
res1: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(a
-> aaa, b -> bbb)

Answer 3

我想通过以下方式得到了它：

mymap.flatMap(x => Map(x._1 -> x._2.head))

Answer 4

Yet another suggestion:

m mapValues { _.mkString }

This one's agnostic to whether the Seq has multiple elements -- it'll just concatenate all the strings together. If you're concerned about the recomputation of each value, you can make it happen up-front:

(m mapValues { _.mkString }).view.force