I have a map of type:
Map[Int, Seq[Option[User]]]
I want to transform this to:
Map[Int, Seq[User]]
I just want to remove any Option[User] from the Seq.
Asked
Active
Viewed 2,464 times
0
-
1And what do you want to do with items that are `None` that may exist in any of these sequences? – Tzach Zohar Aug 10 '16 at 15:13
-
@TzachZohar I want to remove the key from the map. – cool breeze Aug 10 '16 at 15:16
-
1If there exists a single element in the sequence that is None you want to remove the key? – Yuval Itzchakov Aug 10 '16 at 15:18
-
@YuvalItzchakov sorry, I want to remove the item from the Seq, not remove the key. – cool breeze Aug 10 '16 at 15:20
2 Answers
7
Simple Seq#flatten does the magic:
scala> Map(1 -> Seq(Option(1),None, Option(3),None), 2 -> Seq(None))
res1: Map[Int,Seq[Option[Int]]] =
Map(1 -> List(Some(1), None, Some(3), None), 2 -> List(None))
scala> res1.mapValues(_.flatten)
res2: Map[Int,Seq[Int]] = Map(1 -> List(1, 3), 2 -> List())
pedrofurla
- 12,763
- 1
- 38
- 49
-
2Just remember that mapValues works counterintuitvely, by facading the orignal map rather than creating a new one. If this is a problem, you'd need: res1.map {case(k,v) -> k, v.flatten} – Iadams Aug 10 '16 at 15:56
-
what? "Counterintuitively" why? "by facading the orignal map rather than creating a new one" What number 2. So give me an example where this is observable. – pedrofurla Aug 10 '16 at 17:51
-
See http://stackoverflow.com/questions/14882642/scala-why-mapvalues-produces-a-view-and-is-there-any-stable-alternatives – Iadams Aug 10 '16 at 17:53
0
yourMap.map({ case (key, seq) => {
(key, seq.filter(uOpt => !uOpt.isEmpty).map(_.get))
} })
Or,
val newMap = for {
(key, seq) <- yourMap
newSeq = for {
userOpt <- seq
if !userOpt.isEmpty
} yield userOpt.get
} yield (key -> newSeq)
sarveshseri
- 13,738
- 28
- 47
-
1
-
Yup... that works in this case, but some times (specially for beginners) knowing why your code works is more important. – sarveshseri Aug 10 '16 at 15:26
-