Reputation: 73
I have a F# list and I'm taking two elements of that list.
If the list has 10 elements in it :
let rnd = new Random()
let elem1 = list.Item(rnd.Next(0,9))
let elem2 = list.Item(rnd.Next(0,9))
There is a chance elem1 and elem2 are equal.
I have checked some workarounds and most of them work using a do while, but I don't want to implement a function that may never end in F#.
Is there a way to create a restriction in the random function?
First random : 0 <= x <= 9
Second random : 0 <= y <= 9 <> x
Upvotes: 2
Views: 115
Reputation: 26184
A simple solution:
let rnd = new Random()
let ndx1 = rnd.Next(9)
let ndx2 =
let x = rnd.Next(8)
if x < ndx1 then x else x + 1
let elem1, elem2 = list.[ndx1], list.[ndx2]
Another way, using maths and calling the random function once:
let r = Random().Next(9 * 8)
let x = 1 + r + r / 9
let elem1, elem2 = list.[x / 9], list.[x % 9]
which may be generalised to:
let getTwoElements lst =
let c = List.length lst
let x, y = Math.DivRem(Random().Next(c * (c-1)) * (c+1) / c + 1, c)
lst.[x], lst.[y]
Upvotes: 3
Reputation: 19897
There are lots of way to achieve this. A simple one would be something like this:
open System
open System.Linq
let rnd = new Random()
let elem1 = list.Item(rnd.Next(0,9))
let elem2 = list.Where(fun x->x <> elem1).ElementAt(rnd.Next(0,8))
Upvotes: 0
Reputation: 13577
A more declarative approach, taking into account your comment about points in the image:
let rnd = System.Random()
/// this will build you a list of 10 pairs of indices where a <> b.
let indices =
Seq.initInfinite (fun _ -> rnd.Next(0,10), rnd.Next(0,10))
|> Seq.filter (fun (a,b) -> a <> b)
|> Seq.take 10
|> List.ofSeq
/// map indices into actual points.
let elems =
let points = list |> Array.ofList
List.map (fun (a, b) -> points.[a], points.[b]) indices
As a side note, do not use random access on lists. They're not made for that and performance of that is poor. Convert them to an array first.
Upvotes: 3