How do you write a f# recursive function that accepts a positive integer n and a list xs as input, and returns a list containing only the first n elements in xs.
let rec something n xs =
..
something 3 [1..10] = [1;2;3]
The short answer is: Don't, just use Seq.take.
A simple version would be something like:
let rec take n list =
match n with
| 0 -> []
| _ -> List.head list :: take (n - 1) (List.tail list)
A tail-recursive could look like:
let rec take n list =
let rec innertake m innerlist acc =
match m with
| 0 -> List.rev acc
| _ -> innertake (m - 1) (List.tail innerlist) ((List.head innerlist) :: acc)
innertake n list []
Note that neither of these does anything to handle the case that the input list is shorter than the requested number of items.
Related
So consider getting a list of [1; 2; 3; 4; 5; 6; 7; 8; 9] and reshape it into [[1; 2; 3]; [4; 5; 6]; [7; 8; 9]]. How would you do that in OCaml? I want a simple function or something from the standard library.
Turns out, it can be easily done with 3 lines of code, considering that the length of the list is divisible by 3.
let rec re_shape = function
| x :: xs :: xz :: xt -> [x; xs; xz] :: re_shape xt
| _ -> []
How this works is that for each iteration it cons a list of 3 to the rest of the function, till it reaches the end. The last line is added for safety.
As you have shown an effort to solve this, for your consideration, see a strategy below for generalizing this to allow for any length.
The partition function will allow us to get the first n elements from the list and the remainder, raising Invalid_argument if there aren't n elements in the list.
The chunks function applies this recursively to the remainder to build a list of lists.
let partition n lst =
let rec partition' n (first, rest) =
match n, rest with
| 0, _ -> (List.rev first, rest)
| _, [] -> raise (Invalid_argument "List not long enough")
| _, x::xs -> partition' (n-1) (x :: first, xs)
in
partition' n ([], lst)
let rec chunks n lst =
match partition n lst with
| first, [] -> [first]
| first, rest -> first :: chunks n rest
| exception (Invalid_argument _) ->
raise (Invalid_argument (Format.sprintf "List length not evenly divisible by %d" n))
This second function is not tail-recursive, though that can readily be addressed in OCaml 4.14 and later with:
let[#tail_mod_cons] rec chunks n lst =
...
How do you write a F# recursive function that accepts a positive integer n and a list xs as input, and returns a list except first n elements in xs?
let rec something n xs = .. something 7 [1..10] = [8; 9; 10]
I don't think that recursion is the most efficient way to solve this problem, but you can do it like this:
let rec something n xs =
if n > List.length xs || n < 0 then failwith "incorrect parameter n - out of range"
else if n = 0 then xs
else something (n-1) (xs |> List.tail)
let res = something 7 [1..10]
open System
Console.WriteLine(res)
//something 7 [1..10] = [8; 9; 10]
The simple answer is to use List.skip ... i.e. [0..10] |> List.skip 5
To reimplement List.skip you'd be looking at something like:
let rec listSkip n list =
match (n, list) with
| 0, list -> list
| _, [] -> failwith "The index is outside the legal range"
| n, _ when n < 0 -> failwith "The index cannot be negative"
| n, _ :: tl -> listSkip (n - 1) tl
As this is recursion is eligible for tail-call optimization, performance should be similar to an explicit loop.
I've avoided an explicit guard checking List.length against n because List.length requires iteration of the entire list ( which we'd have to check each round of the recursion ). Thus it's cheaper just to try and remove n items and fail if we run into an empty list before n reaches 0.
With a list of integers such as:
[1;2;3;4;5;6;7;8;9]
How can I create a list of list of ints from the above, with all new lists the same specified length?
For example, I need to go from:
[1;2;3;4;5;6;7;8;9] to [[1;2;3];[4;5;6];[7;8;9]]
with the number to split being 3?
Thanks for your time.
So what you actually want is a function of type
val split : int list -> int -> int list list
that takes a list of integers and a sub-list-size. How about one that is even more general?
val split : 'a list -> int -> 'a list list
Here comes the implementation:
let split xs size =
let (_, r, rs) =
(* fold over the list, keeping track of how many elements are still
missing in the current list (csize), the current list (ys) and
the result list (zss) *)
List.fold_left (fun (csize, ys, zss) elt ->
(* if target size is 0, add the current list to the target list and
start a new empty current list of target-size size *)
if csize = 0 then (size - 1, [elt], zss # [ys])
(* otherwise decrement the target size and append the current element
elt to the current list ys *)
else (csize - 1, ys # [elt], zss))
(* start the accumulator with target-size=size, an empty current list and
an empty target-list *)
(size, [], []) xs
in
(* add the "left-overs" to the back of the target-list *)
rs # [r]
Please let me know if you get extra points for this! ;)
The code you give is a way to remove a given number of elements from the front of a list. One way to proceed might be to leave this function as it is (maybe clean it up a little) and use an outer function to process the whole list. For this to work easily, your function might also want to return the remainder of the list (so the outer function can easily tell what still needs to be segmented).
It seems, though, that you want to solve the problem with a single function. If so, the main thing I see that's missing is an accumulator for the pieces you've already snipped off. And you also can't quit when you reach your count, you have to remember the piece you just snipped off, and then process the rest of the list the same way.
If I were solving this myself, I'd try to generalize the problem so that the recursive call could help out in all cases. Something that might work is to allow the first piece to be shorter than the rest. That way you can write it as a single function, with no accumulators
(just recursive calls).
I would probably do it this way:
let split lst n =
let rec parti n acc xs =
match xs with
| [] -> (List.rev acc, [])
| _::_ when n = 0 -> (List.rev acc, xs)
| x::xs -> parti (pred n) (x::acc) xs
in let rec concat acc = function
| [] -> List.rev acc
| xs -> let (part, rest) = parti n [] xs in concat (part::acc) rest
in concat [] lst
Note that we are being lenient if n doesn't divide List.length lst evenly.
Example:
split [1;2;3;4;5] 2 gives [[1;2];[3;4];[5]]
Final note: the code is very verbose because the OCaml standard lib is very bare bones :/ With a different lib I'm sure this could be made much more concise.
let rec split n xs =
let rec take k xs ys = match k, xs with
| 0, _ -> List.rev ys :: split n xs
| _, [] -> if ys = [] then [] else [ys]
| _, x::xs' -> take (k - 1) xs' (x::ys)
in take n xs []
I'm wondering how can I write a function that divides a given list to sublists in a given point, swaps these sublists and returns a resulting list.
For example:
swap([1;3;5;6],2) => [5;6;1;3]
I suppose that the code that I developed is correct?
let rec swap (l,n) =
let rec loop t (count,laux) =
match t with
| h::t when count < n -> loop t (count+1, h::laux)
| h::t -> h::t# List.rev laux
| []->[]
in
loop l (0,[])
;;
You're almost there. The problem is your function handles the case when length of l is greater or equals to n incorrectly.
The pattern [] doesn't mean input list is empty; it means we come to the end of the list. What you should do at that point is returning the accumulator acc in the reverse order.
I rearrange patterns a little bit so base cases come first:
let rec swap (l, n) =
let rec loop xs count acc =
match xs with
| _ when count = n -> xs # List.rev acc
| [] -> List.rev acc
| h::t -> loop t (count+1) (h::acc)
in loop l 0 []
What I'm hoping to make this function do is:
Generate a list of random integers of length specified by count
Generate another random number to replace first element of list
Sort the list
Split list in half, discarding second half
Discard first element of list
Repeat 2-5 unless list is empty
What I have so far (but not working) is below. What is the matter with it?
let go count =
let rec cut l =
if List.length l = 0 then l
printfn "%A" l
let list = System.Random().Next(100)::List.tail l
let cut list =
let firstHalf= list |> Seq.take (List.length list / 2) |> Seq.toList
firstHalf
let listSorted = List.sort list
cut (List.tail listSorted)
let r = System.Random()
let list1 = List.init count (fun numbers -> r.Next(100))
printfn "List = %A" list1
cut list1
A few tips:
Don't test if a list is empty by List.length L = 0. Each test will take as long as the amount of elements in the list. Test with pattern matching instead, that's (almost) instantanteous:
Don't instantiate a new instance of a random number generator each time your cut function is called: let list = System.Random().... Doing that means that you're likely to get the same numbers (each instantiaion seeds the generator with the current system time). Just move your declaration r = System.Random() up a bit, and use that generator throughout your code.
example:
let rec cut l =
match l with
| [] -> // the list is empty, end the recursion here
| head::tail -> // the list consists of the head element and the rest
// you can refer to head and tail in your code here
let newlist = r.next(100) :: tail
You're declaring a function called 'cut' inside your recursive 'cut' function, which means that the last call to 'cut' in your recursive function actually calls the non-recursive one you defined inside. Use different names there.
You've written 'if List.length l = 0 then l', which (apart from not using a pattern match) also presents a problem: an 'if' in F# is an expression, like the ? operator in C#. In C# that would mean something like
(l.Count == 0) ? l : //other case missing! error! danger!
Another tip: once your list is sorted, you don't need to sort again each time you add a new random element. You can write code that inserts a new element in a sorted list that would be more efficient than adding an element and sorting afterwards. I'll leave the insert-into-sorted-list as an excercise.
I hope these tips are useful.
Here it is as simple as making functions for each of your statements.
let rnd = new System.Random()
let genList n =
[for i = 0 to n-1 do yield rnd.Next()]
let replaceHead v lst = match lst with
| [] -> []
| (x::xs) -> (v::xs)
let splitInHalf lst =
let len = (lst |> List.length) / 2
let rec loop n lst =
match (n,lst) with
| 0,_ -> []
| _,[] -> []
| _,(x::xs) -> x :: (loop (n-1) xs)
loop len lst
let start n =
let lst = genList n
let rec loop l =
match l with
| [] -> []
| ls -> match ls |> replaceHead (rnd.Next())
|> List.sort
|> splitInHalf with
| [] -> []
| xs -> xs |> List.tail |> loop
loop lst
start 1
here is my try...
let go count =
System.Random() |> fun rnd -> // With ranomizer ... (we will need it)
let rec repeat = function // So we got recursion
| x::xs when xs.Length <> 1 -> // while we have head and tail
printfn "%A" xs
rnd .Next(100) :: (List.tail xs) // Add random value
|> Seq.sort // Sort
|> Seq.take( abs(xs.Length /2) ) // Make a half
|> Seq.skip 1 // Remove first (just skip)
|> List.ofSeq // Make the list
|> repeat // So and repeat
| x::xs -> printfn "%A" xs
| _ -> () // If we have no head and tail
repeat <| List.init count (fun _ -> rnd.Next(100)) // do it with our random list
It does look like homework :)
But here is my take on it:
#light
// Create random integer sequence
let random_integers_of_length l =
(l, new System.Random())
|> Seq.unfold (fun (c, rnd) -> if c = 0 then None else Some (rnd.Next(), (c-1, rnd)))
|> Seq.cache
let rec mutate numbers =
printfn "%A" (List.ofSeq numbers); // pretty print the list
match numbers with
| _ when (Seq.length numbers) <= 1 -> printfn "Done.." // if length is 1 or 0 we can stop.
| _ ->
numbers
|> Seq.skip 1 // discard first element
|> Seq.append (random_integers_of_length 1) // append random number at the start
|> Seq.sort // sort
|> Seq.take ((Seq.length numbers) / 2) // take the first half, ignore the rest
|> Seq.skip 1 // discard first element
|> mutate // do it again.