How do I filter a list without using List.filter in OCaml?

Question

I have to write a function that, given two lists, it returns a list of the elements of the first one whose square is present in the second one (sry for my english). I can't do it recursively and i can't use List.filter. this is what i did:

let lst1= [1;2;3;4;5];;
let lst2= [9;25;10;4];;

let filquadi lst1 lst2 =
  let aux = [] in 
  List.map(fun x -> if List.mem (x*x) lst2 then x::aux else []) lst1;;

It works but it also prints [] when the number doesn't satisfy the if statement:

filquadi lst1 lst2 ;;
- : int list list = [[]; [2]; [3]; []; [5]]

how can I return a list of numbers instead of a list of a list of numbers?

- : int list = [2;3;5]

Answer 1

You can use List.concat to put things together at the end:

List.concat (List.map ...)

As a side comment, aux isn't doing anything useful in your code. It's just a name for the empty list (since OCaml variables are immutable). It would probably be clearer just to use [x] instead of x:: aux .

As another side comment, this is a strange sounding assignment. Normally the reason to forbid use of functions from the List module is to encourage you to write your own recursive solution (which indeed is educational). I can't see offhand a reason to forbid the use of recursion, but it's interesting to combine functions from List in different ways.

Answer 2

Your criteria don't say you can't use List.fold_left or List.rev , so...

let filter lst1 lst2 =
  List.fold_left 
    (fun init x -> 
       if List.mem x lst2 then x::init 
       else init) 
    [] lst1 
  |> List.rev

If you're not supposed to use recursion, this is technically cheating, because List.fold_left works recursively, but then so does basically anything working with lists.