Scala List Comprehensions

Question

I'm trying to generate a list in scala according to the formula:

for n > 1 f(n) = 4*n^2 - 6*n + 6 and for n == 1 f(n) = 1

currently I have:

def lGen(end: Int): List[Int] = {
    for { n <- List.range(3 , end + 1 , 2) } yields { 4*n*n - 6*n - 6 }
}

For end = 5 this would give the list:

List(24 , 76)

Right now I'm stuck on trying to find a gracefull way to make this function give

List(1 , 24 , 74)

Any suggestions would be greatly appreciated.

-Lee

Luigi Plinge · Answer 1 · 2012-02-18T04:59:42.260

13

I'd separate out the "formula" from the list generation:

val f : Int => Int = {
  case 1 => 1
  case x if x > 1 => 4*x*x - 6*x + 6
}

def lGen(end: Int) = (1 to end by 2 map f).toList

or

def lGen(end: Int) = List.range(1, end + 1, 2) map f

edited Feb 18 '12 at 04:59

answered Jan 26 '12 at 06:33

Luigi Plinge

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I like how this approach keeps the definition intact. – LeeG Jan 26 '12 at 06:39
is it change the result if I write instead of `case x if x > 1`, `case x > 1`? – Slow Harry Feb 14 '14 at 19:35

score 5 · Accepted Answer · answered Jan 26 '12 at 06:17

5

How about this:

scala> def lGen(end: Int): List[Int] =
         1 :: List.range(3, end+1, 2).map(n => 4*n*n - 6*n + 6)

scala> lGen(5)
res0: List[Int] = List(1, 24, 76)

answered Jan 26 '12 at 06:17

fotNelton

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What is the :: operator doing in this case? – LeeG Jan 26 '12 at 06:29
1

It's a "cons" operator, i.e. it prepends the `1` to the list. – fotNelton Jan 26 '12 at 06:55

Scala List Comprehensions

2 Answers2