How to specify a 'next' function in Scala for loop

Question

I have a code snippet like this:

val step = Step.Montyly  //this could be Yearly, daily, hourly, etc.
val (lower, upper) = (****, ****) //unix timestamps, which represent a time range 
val timeArray = Array[Long](0)

var time = lower
while (time <= upper) {
    timeArray +: = time
    time = step.next(time) // eg, Step.Hourly.next(time) = time + 3600, Step.Monthly.next(time) = new DateTime(time).addMonths(1).toTimeStamp()
}

return timeArray

Though this is written in Scala, it is a non-functional way. I'm new to Scala, wondering if this can be rewritten in functional way. I know things like the following:

for(i <- 1 to 10 by step) yield i

But step is a fixed value here, how to make 'i' can be generated by a 'next function' rather than a fixed value?

Reactormonk · Accepted Answer · 2016-11-02 08:18:21Z

5

You'll have to change the flow a bit to make it functional (without mutable Array and without var). Stream.iterate works with an initial start value, and applies a function repeatedly to generate the next elements.

Stream.iterate(lower)(step.next).takeWhile(_ < upper)

edited Nov 2, 2016 at 8:18

answered Oct 30, 2016 at 8:40

Reactormonk

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3 Comments

stefanobaghino Over a year ago

I like this! Furthermore, the same function is available on List, if one wants a strictly-evaluated solution.

Aries Over a year ago

@Reactormonk, thanks! I'm now able to clear non-functional codes in my project. I like this elegant and expressive way :)

Reactormonk Over a year ago

@stefanobaghino yeah, but it won't terminate.

The Archetypal Paul · Accepted Answer · 2016-10-30 14:45:15Z

1

Reactormonk's answer is probably what I would use in practice, but here's an alternative fully-functional tail-recursive one (taking any step function you like)

  def next(n: Long) = n + 3 // simple example for illustration

  def steps(current: Long, upper: Long, f: Long => Long, result: List[Long]): Array[Long] =
  {
    if (current >= upper)
      result.reverse.toArray
    else 
      steps(f(current), upper, f, current :: result)
  } 
  steps(1, 20, next, List(0))
  //> res0: Array[Long] = Array(0, 1, 4, 7, 10, 13, 16, 19)

edited Oct 30, 2016 at 14:45

answered Oct 30, 2016 at 8:59

The Archetypal Paul

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