This is an excerpt from the Scala Cookbook (partially modified for the internet). This is Recipe 11.2, “How to Create a Mutable List in Scala (ListBuffer)”
You want to use a mutable list — a
LinearSeq, as opposed to an
IndexedSeq — but a Scala
List isn’t mutable.
Use the Scala ListBuffer class, and convert the
ListBuffer to a
List when needed. The following examples demonstrate how to create a
ListBuffer, and then add and remove elements, and then convert it to a
List when finished:
import scala.collection.mutable.ListBuffer var fruits = new ListBuffer[String]() // add one element at a time to the ListBuffer fruits += "Apple" fruits += "Banana" fruits += "Orange" // add multiple elements fruits += ("Strawberry", "Kiwi", "Pineapple") // remove one element fruits -= "Apple" // remove multiple elements fruits -= ("Banana", "Orange") // remove multiple elements specified by another sequence fruits --= Seq("Kiwi", "Pineapple") // convert the ListBuffer to a List when you need to val fruitsList = fruits.toList
List is immutable, if you need to create a list that is constantly changing, the preferred approach is to use a
ListBuffer while the list is being modified, then convert it to a
List when a
List is needed.
ListBuffer Scaladoc states that a
ListBuffer is “a Buffer implementation backed by a list. It provides constant time prepend and append. Most other operations are linear.” So, don’t use
ListBuffer if you want to access elements arbitrarily, such as accessing items by index (like
ArrayBuffer instead. See Recipe 10.4, “Understanding the Performance of Collections” for more information.
Although you can’t modify the elements in a
List, you can create a new
List from an existing one, typically prepending items to the original list with the
scala> val x = List(2) x: List[Int] = List(2) scala> val y = 1 :: x y: List[Int] = List(1, 2) scala> val z = 0 :: y z: List[Int] = List(0, 1, 2)
This is discussed more in Recipe 11.3, “Adding Elements to a List”.
help support my writing
& keep this website running