Scala: Remove none elements from map and flatten - scala

Question

I have a map:
Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3"))
I want to remove all None elements and flatten the map. What is the easiest way to accomplish that? I only found this way:
Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3")).filter(_._2.nonEmpty).map(item => (item._1 -> item._2.getOrElse(Nil)))
The result is:
Map(key1 -> value1, key3 -> value3)
Do you know a better way?

Answer 1

My take using pattern matching is:
Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3")).collect {
case (key, Some(value)) => key -> value
}
// Map(key1 -> value1, key3 -> value3)
Collect acts like combined map + filter

Answer 2

You can use for-comprehension + pattern-matching:
for((k, Some(v)) <- yourMap) yield k -> v

Answer 3

Using partition over the map, like this,
val (flattened,_) = map.partition(_._2.isDefined)

Answer 4

My take using for comprehensions:
val m = Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3"))
for( (key,value) <- m if(value.isDefined)) yield (key,value.get)

Answer 5

You could define the following helpers too, which allow for more compact syntax
implicit class RichPairedOptionIterableOps[A, B, Repr[_]](
iterable: IterableOps[(A, Option[B]), Repr, Repr[(A, Option[B])]]
) {
def collectWithSome: Repr[(A, B)] = iterable.collect { case (a, Some(b)) => a -> b }
def collectWithNone: Repr[A] = iterable.collect { case (a, None) => a }
}
on your example:
Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3")).collectWithSome