Scala generate a collection of elements based on a number of required elements and a known first element

Question

Looking for an elegant Scala way to generate a collection of elements based on a number of required elements and a known first element.

For example:

def generateListOfData(numberOfElements: Int = 1): Seq[Data] = {
if(numberOfElements == 1)
   Seq(aData(id = id)) //aData is some builder
else 
   // generate a `Seq` of elements with `id = randomInt`
}

What would be the most elegant approach to do it?

Answer 1

You can generate range from 1 to numberOfElements - 1 map it into Data and prepend with default element:

def generateListOfData(numberOfElements: Int = 1): Seq[Data] = {
    if (numberOfElements > 0) {
      val rnd = new scala.util.Random
      aData(id = id) +: (1 to numberOfElements - 1)
        .map(_ => aData(id = rnd.nextInt()))
    }
    Seq[Data]()
}

Or using foldLeft and pattern matching with guards:

def generateListOfData(numberOfElements: Int = 1): Seq[Data] = {
    val rnd = new scala.util.Random
    numberOfElements match {
      case _ if numberOfElements > 0 => (1 to numberOfElements - 1)
        .foldLeft(Seq(aData(id = id))) { (acc, _) => aData(id = rnd.nextInt()) +: acc }
        .reverse
      case _ => Seq()
    }
}