F# Create list of multiples of x?

Question

I want to create a list that is the multiples of a number. For example [2; 4; 6; 8; 10] would be the multiples of 2 between 0 and 10.

How would I dynamically create such a list of the multiples of x? Is it possible to do it without setting an upper bound?

One way to do it would be to create a list between 0 and some crazy large number and then filter it using the mod function. Trying to test this, creating a list of 0 to some crazy large number caused an out of memory exception (after a 30 second or so wait).

I feel like F# has some super simple and awesome way to build such a list but I'm too much of a newb to know what it is, yet. Help?

Answer 1

This produces an infinite sequence of multiples:

let multiples n = Seq.unfold (fun i -> Some(i, i + n)) n

multiples 3 |> Seq.take 3 //seq [3; 6; 9]

This is a bit more code, but faster:

let multiples n =
  let rec loop i =
    seq {
      yield i
      yield! loop (i + n)
    }
  loop n

It's basically equivalent to the following C#:

static IEnumerable<int> Multiples(int n) {
    int i = n;
    while (true) {
        yield return i;
        i += n;
    }
}

Answer 2

Sequences (IEnumerables) give the laziness you want here:

let multiplesOfN n =
    seq {
        for i in 1 .. 1000000 do
            yield i * n
    }

let first6multsof3 = 
    multiplesOfN 3 |> Seq.take 6

printfn "%A" (first6multsof3 |> Seq.toList)

or with your filter-mod strategy:

seq { 1 .. 1000000} |> Seq.filter (fun x -> x%3=0) |> Seq.take 6 |> Seq.toList 

Answer 3

[ firstValue..Step..endValue]

[ 2..2..10] => [2; 4; 6; 8; 10]

other way

Seq.initInfinite id |> Seq.map (((+) 1) >> ((*) 2))

Answer 4

List.init 10 ((*) 3)
val it : int list = [0; 3; 6; 9; 12; 15; 18; 21; 24; 27]

You can play with the arguments and Seq.skip to get whatever you need.

For example, for [2; 4; 6; 8; 10]:

List.init 6 ((*) 2)
|> List.tail

Or:

List.init 6 ((*) 2)
|> Seq.skip 1
|> List.ofSeq