I'm used to JaneStreet's Core library. Its List module has a neat init function:
List.init;;
- : int -> f:(int -> 'a) -> 'a list = <fun>
It allows you to create a list with using a custom function to initialize elements:
List.init 5 ~f:(Fn.id);;
- : int list = [0; 1; 2; 3; 4]
List.init 5 ~f:(Int.to_string);;
- : string list = ["0"; "1"; "2"; "3"; "4"]
However, this function doesn't seem to exist in Pervasives, which is sad. Am I missing something, or do I have to implement it myself? And if I do need to write it, how do I achieve this?
EDIT:
I have written an imperative version of init, but it doesn't feel right to have to resort to OCaml's imperative features in such a case. :(
let init n ~f =
let i = ref 0 in
let l = ref [] in
while !i < n do
l := (f !i) :: !l;
incr i;
done;
List.rev !l
;;
EDIT 2:
I've opened a pull request on OCaml's GitHub to have this feature included.
EDIT 3:
The feature was released in OCaml 4.06.
A recursive implementation is fairly straightforward. However, it is not tail-recursive, which means that you'll risk a stack overflow for large lists:
let init_list n ~f =
let rec init_list' i n f =
if i >= n then []
else (f i) :: (init_list' (i+1) n f)
in init_list' 0 n f
We can transform it into a tail-recursive version using the usual techniques:
let init_list n ~f =
let rec init_list' acc i n f =
if i >= n then acc
else init_list' ((f i) :: acc) (i+1) n f
in List.rev (init_list' [] 0 n f)
This uses an accumulator and also needs to reverse the intermediate result, as the list is constructed in reverse. Note that we could also use f (n-i-1) instead of f i to avoid reversing the list, but this may lead to unexpected behavior if f has side-effects.
An alternative and shorter solution is to simply use Array.init as a starting point:
let init_list n ~f = Array.(init n f |> to_list)
You can copy the code from JaneStreet and use it.
The code look's like (but not exactly the same) :
let init n ~f =
if n < 0 then raise (Invalid_argument "init");
let rec loop i accum =
if i = 0 then accum
else loop (i-1) (f (i-1) :: accum)
in
loop n []
;;
You can find the original code inside core_list0.ml from the package core_kernel.
Related
How do you efficiently split a list in 2, preserving the order of the elements?
Here's an example of input and expected output
[] should produce ([],[])
[1;] can produce ([1;], []) or ([], [1;])
[1;2;3;4;] should produce ([1; 2;], [3; 4;])
[1;2;3;4;5;] can produce ([1;2;3;], [4;5;]) or ([1;2;], [3;4;5;])
I tried a few things but I'm unsure which is the most efficient... Maybe there is a solution out there that I'm missing completely(calls to C code don't count).
My first attempt was to use List's partition function with a ref to 1/2 the length of the list. This works but you walk through the whole list when you only need to cover half.
let split_list2 l =
let len = ref ((List.length l) / 2) in
List.partition (fun _ -> if !len = 0 then false else (len := !len - 1; true)) l
My next attempt was to use a accumulator and then reverse it. This only walks through half the list but I call reverse to correct the order of the accumulator.
let split_list4 l =
let len = List.length l in
let rec split_list4_aux ln acc lst =
if ln < 1
then
(List.rev acc, lst)
else
match lst with
| [] -> failwith "Invalid split"
| hd::tl ->
split_list4_aux (ln - 1) (hd::acc) tl in
split_list4_aux (len / 2) [] l
My final attempt used function closures for the accumulator and it works but I have no idea how efficient closures are.
let split_list3 l =
let len = List.length l in
let rec split_list3_aux ln func lst =
if ln < 1
then
(func [], lst)
else
match lst with
| hd::tl -> split_list3_aux (ln - 1) (fun t -> func (hd::t)) tl
| _ -> failwith "Invalid split" in
split_list3_aux (len / 2) (fun t -> t) l
So is there a standard way to split a list in OCaml(preserving element order) that's most efficient?
You need to traverse the whole list for all of your solutions. The List.length function traverses the whole list. But it's true that your later solutions re-use the tail of the original list rather than constructing a new list.
It is difficult to say how fast any given bit of code is going to be just by inspection. Generally it's good enough to think in aysmptotic O(f(n)) terms, then work on slow functions in detail through timing tests (of realistic data).
All of your answers look to be O(n), which is the best you can do since you clearly need to know the length of the list to get the answer.
Your split_list2 and split_list3 solutions look pretty complicated to me, so I would expect (intuitively) them to be slower. A closure is a fairly complicated data structure containing a function and the environment of accessible variables. So it's problaby not all that fast to construct one.
Your split_list4 solution is what I would code up myself.
If you really care about timings you should time your solutions on some long lists. Keep in mind that you might get different timings on different systems.
Couldn't give up this question. I had to find a way that I could walk through this list one time to create a split with order preserved..
How about this?
let split lst =
let cnt = ref 0 in
let acc = ref ([], []) in
let rec split_aux c l =
match l with
| [] -> cnt := (c / 2)
| hd::tl ->
(
split_aux (c + 1) tl;
let (f, s) = (!acc) in
if c < (!cnt)
then
acc := ((hd::f), s)
else
acc := (f, hd::s)
)
in
split_aux 0 lst; !acc
This code is in Haskell. How can i do the same thing in OCAML?
perfect n = [x | x<-[1..n], sum(f x) == x]
f x = [i | i<-[1..x-1], x `mod` i ==0]
While Jeffrey's answer is correct, using appropriate libraries (in this case, sequence), you can get something that is similar in terseness and semantics to the Haskell style:
module S = Sequence
let sum = S.fold (+) 0
let f x = S.filter (fun i -> x mod i = 0) S.(1 -- (x-1))
let perfect n = S.filter (fun x -> sum (f x) = x) S.(1 -- n)
You're using many (really nice) features of Haskell that don't exist in OCaml.
For list comprehensions, you can use List.filter.
For the notation [x .. y] you can use this range function:
let range a b =
let rec go accum i =
if i > b then List.rev accum else go (i :: accum) (i + 1)
in
go [] a
For sum you can use this:
let sum = List.fold_left (+) 0
I'm having a problem with understanding how F# works. I come from C# and I think that I'm trying to make F# work like C#. My biggest problem is returning values in the correct format.
Example:
Let's say I have function that takes a list of integers and an integer.
Function should print a list of indexes where values from list match passed integer.
My code:
let indeks myList n = myList |> List.mapi (fun i x -> if x=n then i else 0);;
indeks [0..4] 3;;
However it returns:
val it : int list = [0; 0; 0; 3; 0]
instead of just [3] as I cannot ommit else in that statement.
Also I have targeted signature of -> int list -> int -> int list and I get something else.
Same goes for problem no. 2 where I want to provide an integer and print every number from 0 to this integer n times (where n is the iterated value):
example:
MultiplyValues 3;;
output: [1;2;2;3;3;3]
Best I could do was to create list of lists.
What am I missing when returning elements?
How do I add nothing to the return
example: if x=n then n else AddNothingToTheReturn
Use List.choose:
let indeks lst n =
lst
|> List.mapi (fun i s -> if s = n then Some i else None)
|> List.choose id
Sorry, I didn't notice that you had a second problem too. For that you can use List.collect:
let f (n : int) : list<int> =
[1 .. n]
|> List.collect (fun s -> List.init s (fun t -> s))
printfn "%A" (f 3) // [1; 2; 2; 3; 3; 3]
Please read the documentation for List.collect for more information.
EDIT
Following s952163's lead, here is another version of the first solution without the Option type:
let indeks (lst : list<int>) (n : int) : list<int> =
lst
|> List.fold (fun (s, t) u -> s + 1, (if u = n then (s :: t) else t)) (0, [])
|> (snd >> List.rev)
This one traverses the original list once, and the (potentially much shorter) newly formed list once.
The previous answer is quite idiomatic. Here's one solution that avoids the use of Option types and id:
let indeks2 lst n =
lst
|> List.mapi (fun i x -> (i,x))
|> List.filter (fun x -> (fst x) % n = 0 )
|> List.map snd
You can modify the filter function to match your needs.
If you plan to generate lots of sequences it might be a good idea to explore Sequence (list) comprehensions:
[for i in 1..10 do
yield! List.replicate i i]
If statements are an expression in F# and they return a value. In this case both the IF and ELSE branch must return the same type of value. Using Some/None (Option type) gets around this. There are some cases where you can get away with just using If.
I want to write a function that does builds a list between two ints, inclusive
rec myFunc x y would build a list with all the ints between x and y, including x and y
For the logic right now I have something like this:
let rec buildList i n = let x = i+1 in if i <= n then i::(buildList x n)
But this gives me an error "Expression has type 'a list but but an expression was expected of type unit.
I thought buildList is returning a list of ints, and i as an int, so the cons operator would be valid, but its saying it should be void?
Why does this happen, and how do I fix it?
If the condition is true, you return the list i::(buildList x n). If it's not true, what do you return ?
Add else [] to your function to return the empty list when the condition is not met.
When you don't have any else, the compiler supposes it is else () (hence the error message).
Your if is missing an else condition
I suggest that you use a tail recursive function:
let buildList x y =
let (x,y) = if x<y then (x,y) else (y,x) in
let rec aux cpt acc =
if cpt < x then acc
else aux (cpt-1) (cpt::acc)
in aux y []
First, make sure that you ordered your boundaries correctly (idiot-proof), and then construct the list thank to a local recursive function which takes an accumulator.
Two alternatives relying on batteries' package,
Using unfold, which purpose is to build list,
let range ~from:f ~until:u =
BatList.unfold f (function | n when n <= u -> Some (n, succ n) | _ -> None)
Using Enum, allowing to work with lazy datastructure,
# BatList.of_enum ## BatEnum.(1--9);;
- : int list = [1; 2; 3; 4; 5; 6; 7; 8; 9]
My suggestion, this respects the ordering of the arguments.
let rec iota n m =
let oper = if n < m then succ else pred in
if n = m then [n] else n :: iota (oper n) m
Edit:
The operator selection is inside the recursive part, it should better be outside like this:
let iota n m =
let oper = if n < m then succ else pred in
let rec f1 n m = if n = m then [n] else n :: f1 (oper n) m in
f1 n m
At more than 200000 elements I get a stack overflow (so here we are)
# iota 0 250000;;
Stack overflow during evaluation (looping recursion?).
Todo: tail recursion
let buildList i n =
let rec aux acc i =
if i <= n then
aux (i::acc) (i+1)
else (List.rev acc)
in
aux [] i
Test:
# buildList 1 3;;
- : int list = [1; 2; 3]
# buildList 2 1;;
- : int list = []
# buildList 0 250000;;
- : int list =
[0; 1; 2; 3; .... 296; 297; 298; ...]
i'm trying to learn ocaml right now and wanted to start with a little program, generating all bit-combinations:
["0","0","0"]
["0","0","1"]
["0","1","0"]
... and so on
My idea is the following code:
let rec bitstr length list =
if length = 0 then
list
else begin
bitstr (length-1)("0"::list);
bitstr (length-1)("1"::list);
end;;
But i get the following error:
Warning S: this expression should have type unit.
val bitstr : int -> string list -> string list = <fun>
# bitstr 3 [];;
- : string list = ["1"; "1"; "1"]
I did not understand what to change, can you help me?
Best regards
Philipp
begin foo; bar end executes foo and throws the result away, then it executes bar. Since this makes only sense if foo has side-effects and no meaningful return value ocaml emits a warning if foo has a return value other than unit, since everything else is likely to be a programmer error (i.e. the programmer does not actually intend for the result to be discarded) - as is the case here.
In this case it really does make no sense to calculate the list with "0" and then throw it away. Presumably you want to concatenate the two lists instead. You can do this using the # operator:
let rec bitstr length list =
if length = 0 then
[list]
else
bitstr (length-1)("0"::list) # bitstr (length-1)("1"::list);;
Note that I also made the length = 0 case return [list] instead of just list so the result is a list of lists instead of a flat list.
Although sepp2k's answer is spot on, I would like to add the following alternative (which doesn't match the signature you proposed, but actually does what you want) :
let rec bitstr = function
0 -> [[]]
| n -> let f e = List.map (fun x -> e :: x) and l = bitstr (n-1) in
(f "0" l)#(f "1" l);;
The first difference is that you do not need to pass an empty list to call the function bitsr 2 returns [["0"; "0"]; ["0"; "1"]; ["1"; "0"]; ["1"; "1"]]. Second, it returns a list of ordered binary values. But more importantly, in my opinion, it is closer to the spirit of ocaml.
I like to get other ideas!
So here it is...
let rec gen_x acc e1 e2 n = match n with
| 0 -> acc
| n -> (
let l = List.map (fun x -> e1 :: x) acc in
let r = List.map (fun x -> e2 :: x) acc in
gen_x (l # r) e1 e2 (n - 1)
);;
let rec gen_string = gen_x [[]] "0" "1"
let rec gen_int = gen_x [[]] 0 1
gen_string 2
gen_int 2
Result:
[["0"; "0"]; ["0"; "1"]; ["1"; "0"]; ["1"; "1"]]
[[0; 0]; [0; 1]; [1; 0]; [1; 1]]