Menu
Is there any way to efficiently do this, perhaps through toBuffer or to methods? My real problem is I'm building a List off a parser, as follows:
lazy val nodes: Parser[List[Node]] = phrase(( nodeA | nodeB | nodeC).*) But after building it, I want it to be a buffer instead - I'm just not sure how to build a buffer straight from the parser.
862
If you want to convert ListBuffer into a List , use . toList . I mention this because that particular conversion is performed in constant time. Note, though, that any further use of the ListBuffer will result in its contents being copied first.
Advertisements. Scala provides a data structure, the ListBuffer, which is more efficient than List while adding/removing elements in a list. It provides methods to prepend, append elements to a list.
Lists are immutable whereas arrays are mutable in Scala.
to indeed does the trick, and it is pretty trivial to use:
scala> val l = List(1,2,3) l: List[Int] = List(1, 2, 3) scala> l.to[ListBuffer] res1: scala.collection.mutable.ListBuffer[Int] = ListBuffer(1, 2, 3) Works in scala 2.10.x
For scala 2.9.x, you can do:
scala> ListBuffer.empty ++= l res1: scala.collection.mutable.ListBuffer[Int] = ListBuffer(1, 2, 3) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
