Scala如何展平嵌套地图[String，Any]

Question

I have nested map as below:

val x: Map[String, Any] = 
  Map("a" -> "apple", "b" -> "ball", "c" -> Map("x" -> "cat", "y" -> 12))

and I want to convert it into:

Map("a" -> "apple", "b" -> "ball", "x" -> "cat", "y" -> 12)

However, if I try to invoke flatten to x then I get exception.

x.flatten
Error:(40, 14) No implicit view available from (String, Any) => scala.collection.GenTraversableOnce[B].
println(mx.flatten)
Error:(40, 14) not enough arguments for method flatten: (implicit asTraversable: ((String, Any)) => scala.collection.GenTraversableOnce[B])scala.collection.immutable.Iterable[B].
Unspecified value parameter asTraversable.
println(x.flatten)

So, how can I provide implicit view in order to flatten the above map?

Answer 1

It's a bit strange you want to loose the information about 'c'.
But anyway, the compiler is complaining because it does not know how to convert the (String,Any), your Key -> Value pair into a traversable and that is logical.
You could provide the compiler with a hint that in case the 'Any' is a Map, it should only use the values and loose the key.

For example:

x.flatten {
  case ((key, map : Map[String, Any])) => map
  case ((key, value)) => Map(key -> value)
}.toMap

This returns

Map(a -> apple, b -> ball, x -> cat, y -> 12)

Note: the 'toMap' is needed because the 'flatten' returns a List[(String,Any)].