Projecting a list of lists efficiently in F#

Question

I have to do projection of a list of lists which returns all combinations with each element from each list. For example:

projection([[1]; [2; 3]]) = [[1; 2]         


        

Answer 1

let crossProduct listA listB listC listD listE = 
  listA |> Seq.collect (fun a -> 
  listB |> Seq.collect (fun b -> 
  listC |> Seq.collect (fun c -> 
  listD |> Seq.collect (fun d -> 
  listE |> Seq.map (fun e -> a,b,c,d,e))

Answer 2

This function is Haskell's sequence (although sequence is more generic). Translating to F#:

let sequence lss =
    let k l ls = [ for x in l do for xs in ls -> x::xs ]
    List.foldBack k lss [[]]

in interactive:

> test projection 10;;
Real: 00:00:12.240, CPU: 00:00:12.807, GC gen0: 163, gen1: 155, gen2: 4
val it : int = 3628800
> test sequence 10;;
Real: 00:00:06.038, CPU: 00:00:06.021, GC gen0: 75, gen1: 74, gen2: 0
val it : int = 3628800

General idea: avoid explicit recursion in favor to standard combinators (fold, map etc.)

Answer 3

You implementation is slow because of the @ (i.e List concat) operation, which is a slow operation and it is being done many a times in recursive way. The reason for @ being slow is that List are Linked list in functional programming and to concat 2 list you have to first go till the end of the list (one by one traversing through elements) and then append another list .

Please look at the suggested references in comments. I hope those will help you out.

Answer 4

Here's a tail-recursive version. It's not as fast as some of the other solutions (only 25% faster than your original function), but memory usage is constant, so it works for extremely large result sets.

let cartesian l = 
  let rec aux f = function
    | [] -> f (Seq.singleton [])
    | h::t -> aux (fun acc -> f (Seq.collect (fun x -> (Seq.map (fun y -> y::x) h)) acc)) t
  aux id l

Answer 5

First of all, try to avoid list concatenation (@) whenever possible, since it's O(N) instead of O(1) prepend.

I'd start with a (relatively) easy to follow plan of how to compute the cartesian outer product of lists.

• Prepend each element of the first list to each sublist in the cartesian product of the remaining lists.
• Take care of the base case.

First version:

let rec cartesian = function
  | [] -> [[]]
  | L::Ls -> [for C in cartesian Ls do yield! [for x in L do yield x::C]]

This is the direct translation of the sentences above to code.

Now speed this up: instead of list comprehensions, use list concatenations and maps:

let rec cartesian2 = function
  | [] -> [[]]
  | L::Ls -> cartesian2 Ls |> List.collect (fun C -> L |> List.map (fun x->x::C))

This can be made faster still by computing the lists on demand via a sequence:

let rec cartesian3 = function
  | [] -> Seq.singleton []
  | L::Ls -> cartesian3 Ls |> Seq.collect (fun C -> L |> Seq.map (fun x->x::C))

This last form is what I use myself, since I most often just need to iterate over the results instead of having them all at once.

Some benchmarks on my machine: Test code:

let test f N = 
  let fss0 = List.init N (fun i -> List.init (i+1) (fun j -> j+i*i+i))
  f fss0 |> Seq.length

Results in FSI:

> test projection 10;;
Real: 00:00:18.066, CPU: 00:00:18.062, GC gen0: 168, gen1: 157, gen2: 7
val it : int = 3628800
> test cartesian 10;;
Real: 00:00:19.822, CPU: 00:00:19.828, GC gen0: 244, gen1: 121, gen2: 3
val it : int = 3628800
> test cartesian2 10;;
Real: 00:00:09.247, CPU: 00:00:09.250, GC gen0: 94, gen1: 52, gen2: 2
val it : int = 3628800
> test cartesian3 10;;
Real: 00:00:04.254, CPU: 00:00:04.250, GC gen0: 359, gen1: 1, gen2: 0
val it : int = 3628800