CODEWITHSUNDEEP
listscala
Ska
Ska

Reputation: 113

How to replace a given item in a list?

This describes the problem pretty well:

scala> var l2 = List(1,2,3)
l2: List[Int] = List(1, 2, 3)

scala> l2(2) = 55
<console>:10: error: value update is not a member of List[Int]
              l2(2) = 55
              ^

Upvotes: 11

Views: 28850

Answers (5)

Mika&#235;l Mayer
Mika&#235;l Mayer

Reputation: 10701

But you can wrap the assignment in an implicit if you don't like updated

 implicit class RichList[A](l: List[A]) {
      def update(which: Int, what: A): List[A] = {
         l.updated(which, what)
      }
 }

 val l = List(1, 2, 3)
 val l2 = l(2) = 55

 List(1, 2, 55)

Upvotes: 3

Nishadi Wickramanayaka
Nishadi Wickramanayaka

Reputation: 94

Lists are immutable in scala. But if you want to replace an element in a List you can use "updated" method like this

val num:List[Int]=List(1,5,4,7,8,2,10)
num=num.updated(0,22)

Since Lists are immutable,this creates a copy of the first List called num(as we assigned the new list to 'num') by replacing 0th element to 22.So the genaral updated method is 

listname.updated(index,value)

Upvotes: 1

Calum
Calum

Reputation: 5926

scala.List is immutable, meaning you cannot update it in place. If you want to create a copy of your List which contains the updated mapping, you can do the following:

val updated = l2.updated( 2, 55 )

There are mutable ordered sequence types as well, in scala.collection.mutable, such as Buffer types which seem more like what you want. If you try the following you should have more success:

scala> import scala.collection._
import scala.collection._
scala> val b = mutable.Buffer(1,2,3)
b: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1, 2, 3)
scala> b(2) = 55
scala> b
res1: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1, 2, 55)

Edit: Just to note that some other answers have mentioned that you should use a "mutable List type" - this is true, but "List" in Scala just refers to the single-linked list, whereas in Java it's generally used for any ordered, random-access collection. There is also a DoubleLinkedList, which is more like a Java LinkedList, and a MutableList, which is a type used for the internals of some other types.

Generally speaking what you probably want in Scala is a Buffer for this job; especially since the default implementation is an ArrayBuffer, which is pretty close to being the same as ArrayList, most peoples' default, in Java.

If you ever want to find out what the closest "mapping" of a Java collections interface to the Scala world is, though, the easiest thing to do is probably just check what JavaConversions does. In this case you can see the mapping is to Buffer:

scala.collection.mutable.Buffer <=> java.util.List

Upvotes: 32

Tim Blair
Tim Blair

Reputation: 29

Your problem is that Lists in scala are immutable, you need to use a mutable list instead. Import scala.collection.mutable.Queue and use that instead. Queue is a MutableList, so it should do what you want.

Upvotes: 2

Janx
Janx

Reputation: 3333

The List you are creating is immutable.

scala> val l = List(1,2,3)
l: List[Int] = List(1, 2, 3)
scala> l.getClass
res3: java.lang.Class[_] = class scala.collection.immutable.$colon$colon

Use a mutable List instead and you should be fine.

Upvotes: 3

Related Questions