What's the best way to create a dynamically growing array in Scala?

Question

I wanted to add items dynamically into an array. But it seems Scala arrays and lists don't provide any methods for adding items dynamically due to the immutable nature.

So I decided to use List data type to make use of this :: method to achieve this. My code look like this

var outList = List(Nil)
val strArray = Array("ram","sam","bam")

for (str<-strArray)
     outList = str :: outList

Though it works in some way, the problem is the new strings are pre-appended into the list. But the ideal requirement is order of the data. Yeah, I know what you are thinking, you can reverse the final result list to get the original order. But the problem is it's a huge array. And I believe it's not a solution though it solves the problem. I believe there should be a simple way to solve this...

And my reason for hacking Scala is to learn the functional way of coding. Having var (mutable type) and populating the list on the fly seems to me is not a functional way of solving things.

How can I do it?

Ideally, I want to achieve something like this in Scala (below the C# code)

List<int> ls = new List<int>();    
for (int i = 0; i < 100; i++)
    ls.Add(i);

Answer 1

But it seems Scala Arrays & Lists doesn't provide any methods for adding items dynamically due to the immutable nature.

Well, no. Scala Arrays are just Java arrays, so they are mutable:

val arr = Array(1,2)

arr(0) = 3 // arr == Array(3, 2)

But just as Java (and C/C++/C#/etc.) arrays, you can't change the size of an array.

So you need another collection, which is backed by an array, but does allow resizing. A suitable collection in Scala is scala.collection.mutable.ArrayBuffer, java.util.ArrayList in Java, etc.

If you want to get a List instead of an Array in the end, use scala.collection.mutable.ListBuffer instead.

Answer 2

If you want to work with immutable structures, you can use the ++ method:

scala> val orgList = List(1,2,3)
orgList: List[Int] = List(1, 2, 3)

scala> val list2Add = List(4,5,6)
list2Add: List[Int] = List(4, 5, 6)

scala> val newList = orgList ++ list2Add
newList: List[Int] = List(1, 2, 3, 4, 5, 6)

If you want to do more work on the elements than just adding them you can use higher order functions:

val newList = orgList ++ list2Add.map(_ * 2)
newList: List[Int] = List(1, 2, 3, 8, 10, 12)

Or with a for loop:

val newList = orgList ++ {for(x <- list2Add) yield 2*x}

Or you could create some recursive loop:

def addAll(toList: List[Int], fromList: List[Int]): List[Int] =
  fromList match {
    case x :: tail => addAll(2*x :: toList, tail)
    case Nil => toList
  }

val newList = addAll(orgList, list2Add )

but in this case the ordering of the added elements will be in reversed:

List(12, 10, 8, 1, 2, 3)

If you want performance when working with lists, it is better to reverse the result than try to add new elements in the end. Adding elements in the end to a list is nooot good :-)

Answer 3

Going after retronym's answer:

If you still want to use list there are a few ways to prepend an item to the list. What you can do is (yes the top part is still wrong):

scala> var outList : List[String] = Nil
outList: List[String] = List()

scala> val strArray = Array("a","b","c")
strArray: Array[java.lang.String] = Array(a, b, c)

scala> for(s <- strArray)      
     | outList = outList :+ s

scala> outList
res2: List[String] = List(a, b, c)

Note the :+ operator. If you rather append, you'd use s +: outList.

Now who says programming in Scala isn't fun? ;)

P.S. Maybe the reason why you'd want to make them immutable is the speed. Handling large data will be more efficient with immutable data types. Am I right?

Answer 4

If you want to use a mutable Buffer, as retronym mentioned. It looks like this:

scala> var outList = scala.collection.mutable.Buffer[String]()
outList: scala.collection.mutable.Buffer[String] = ArrayBuffer()

scala> for(str<-strArray) outList += str                          

scala> outList
res10: scala.collection.mutable.ListBuffer[String] = ListBuffer(ram, sam, bam)

Anyway, perhaps it is better to directly do the things you want to do with the strArray. E.g:

strArray map(_.toUpperCase) foreach(println)

Answer 5

Okay, there are a few things to clear up.

This is wrong, you're make a one element list, containing an empty list:

scala> var outList = List(Nil) 
outList: List[object Nil] = List(List())

Nil is the empty list:

scala> var outList: List[String] = Nil
outList: List[String] = List()

Or, if you prefer:

scala> var outList = List[String]() 
outList: List[String] = List()

Without more context, it's hard to know what you mean by 'dynamically'. Your example code would be better written as:

scala> val strArray = Array("ram","sam","bam")
strArray: Array[java.lang.String] = Array(ram, sam, bam)

scala> strArray toList
res0: List[java.lang.String] = List(ram, sam, bam)

If you want a mutable collection that can grow and efficiently handle prepend, append, and insert operations, you could use scala.mutable.Buffer.

Answer 6

If you are looking to create a new collection, you can use the yield key word:

val outlist = for(i <- 0 to 100) yield i

Or:

val arrList = "Some" :: "Input" :: "List" :: Nil
val outlist = for ( i <- arrList ) yield i

Technically, outlist is a Seq in both of the above examples so you may need to call the toList method on it if you need some of List's methods.

Answer 7

We can use ArrayBuffer as suggested. However, i have created a simple program which takes no of parameters for an Array of int and it will let you enter the numbers and finally we are displaying them.

val arraySize = scala.io.StdIn.readLine().toInt
  val arr = new ArrayBuffer[Int]() ++ (1 to arraySize).map{
    i =>
      scala.io.StdIn.readLine().toInt
  }
  println(arr.mkString(","))