I'm trying to write following code without recursion:
let rec traverse lst =
match lst with
| a::b::t ->
(* Something that return None*)
traverse (b::t)
| _ -> ()
How to do it in imperative way ?
In an imperative way:
let traverse li =
let state = ref li in
while !state <> [] do
let x = List.hd !state in
state := List.tl !state;
(* do whatever you want *)
done
If you need to access the second element of the list, just use the appropriate List.hd call. But you may need to check that the list isn't empty first.
I see no reason to do that this way, which is heavier, less efficient and less flexible than a recursive loop.
Related
I am new to F# and am having trouble with my code. Its a simple problem to define a function, search, with that take a boolean function and a list and return an index. So for example:
> search (fun x -> x > 10) [ 2; 12; 3; 23; 62; 8; 2 ];;
val it : int = 1
> search (fun s -> s < "horse") [ "pig"; "lion"; "horse"; "cow"; "turkey" ];;
val it : int = 3
What I have as of right now finds the right match but what I cant figure out is how to return a number instead of the rest of the list. I know I'm getting the list instead of a value back because I wrote "if f head then list". What I don't know is what I should put there instead or if what I have is not going to get the result I want.
Below is the code I have written.
let rec search f list =
match list with
| [] -> [-1]
| head::tail ->
if f head then list
else search f tail
Returning a number is easy, you just... return it. Your problem is that you don't have a number to return, because you can't derive it directly from the current state. You have to keep track of the number yourself, using some internal state variable.
When using recursion you change state by calling your function recursively with "modified" arguments. You're already doing that with the list here. To keep internal state in a recursive function you have to introduce another argument, but not expose it outside. You can solve that by using an internal recursive helper function. Here's one that keeps track of the previous item and returns that when it encounters a match:
let search f list =
let rec loop list prev =
match list with
| [] -> None
| head::tail ->
if f head then prev
else loop tail (Some head)
in
loop list None
That's a silly example, but I don't want to just solve your homework for you, because then you wouldn't learn anything. Using this you should be able to figure out how to keep a counter of which position the current item is in, and return that when it matches. Good luck!
You typically define an inner recursive function to help you carry state as you loop, and then call the inner function with an initial state.
let search predicate list =
let rec loop list index =
match list with
| [] -> -1
| head::tail ->
if predicate head then index
else loop tail (index + 1)
loop list 0
i'm studying the programming language Standard ML and i am wondering how i can iterate a list with a check condition.
In other languages we have for loops like :
var input;
for(var i = 0; i < arr.length; i++) {
if(arr[i] == input) {
//you have arrived at your condition...
} else {
//other case
}
}
f.ex
i want to iterate through a list and check if the input variable matches a existing element in the list.
i = 5
xs = [1,5,2,3,6] --> the element matches after one iteration.
fun check i nil = []
| check i (x::xs) = if i=x
then //dowork
else //iterate;
I've gone through many documentations on how to implement this without success.
It would be really helpful if someone could give me some explaining regarding how i can use let val A in B end; inside or outside of if conditions for this kind of work.
how i can iterate a list with a check condition
fun check i nil = []
| check i (x::xs) = if i=x
then //dowork
else //iterate;
i want to iterate through a list and check if the input variable matches a existing element in the list.
I would call this a predicate combinator. It already exists in the standard library and is called List.exists. But you can also make it yourself:
fun exists p [] = false
| exists p (x::xs) = p x orelse exists p xs
This is a simplification of the if-then-else you're attempting, which would look like:
fun exists p [] = false
| exists p (x::xs) = if p x then true else exists p xs
If-then-else isn't really necessary when the result type is a boolean, since orelse, andalso and not are short-circuiting (will not evaluate their second operand if the result can be determined with the first).
Using this List.exists function to check if a list contains a specific element, you have to construct a p that compares the list element with some given value, e.g.:
fun check y xs = List.exists (fn x => ...) xs
This may seem a bit more complicated than simply writing check recursively from scratch,
fun check y [] = false
| check y (x::xs) = ... orelse check y xs
but a solution using higher-order functions is preferred for several reasons.
One is that a seasoned reader will quickly detect what you're doing when seeing List.exists: Ah, you're scanning a list for an element given a predicate. Whereas if your function is explicitly recursive, the reader will have to read the entire recursion scheme: OK, the function doesn't do anything funky, which I'd have known if I'd seen e.g. List.exists.
I am actually sitting over a hour on a problem and don´t find a solution for it.
I have this data type:
type 'a tree = Empty | Node of 'a * 'a tree * 'a tree
And i have to find a function which converts a given tree in a ordered list. There is also no invariant like that the left child has to be less then the right. I already found a "normal" recursion solution but not a tail recursive solution. I already thought about to build a unordered list and sort it with List.sort, but this uses a merge sort which is not tail recursive. Maybe someone has a good advice.
Thank you!
If you want to traverse the tree in order and return a list, that means our function inorder must have the type 'a tree -> 'a list.
let rec inorder t =
match t with
| Empty -> []
| Node (v, l, r) -> List.append (inorder l) (v :: (inorder r)) (* ! *)
However List.append is in tail position, not inorder. Another problem is we have two calls to inorder. If we put inorder l in tail position, inorder r could not possibly be in tail position - and vice versa.
A neat way to work around this problem is continuation passing style. We take our function above and convert it into a helper function with an extra parameter for our continuation, return
(* convert to helper function, add an extra parameter *)
let rec loop t return =
match t with
| Empty -> ...
| Node (v, l, r) -> ...
The continuation represents "what to do next", so instead of sending values directly out of our function, we must hand them to the continuation instead. That means for the Empty case, we'll return [] - instead of simply []
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) -> ...
For the Node (v, l, r) case, now that we have an extra parameter we can write our own continuation that informs loop what to do next. So to construct our sorted list, we will need to loop l, then loop r (or vice versa), then we can append them. We'll write our program just like this.
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) ->
loop l ... (* build the left_result *)
loop r ... (* build the right_result *)
return (List.append left_result (v :: right_result))
In this next step, we'll fill in the actual lambda syntax for the continuations.
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) ->
loop l (fun left ->
loop r (fun right ->
return (List.append left (v :: right))))
Last, we define inorder which is a call to loop with the default continuation, identity.
let identity x =
x
let inorder t =
let rec loop t return =
match t with
| Empty -> return []
| Node (v, l, r) ->
loop r (fun right ->
loop l (fun left ->
return (List.append left (v :: right))))
in
loop t identity
As you can see loop r (fun right -> ...) is in tail position for the Node branch. loop l (fun left -> ...) is in tail position of the first continuation. And List.append ... is in tail position of the second continuation. Provided List.append is a tail-recursive procedure, inorder will not grow the stack.
Note using List.append could be a costly choice for big trees. Our function calls it once per Node. Can you think of a way to avoid it? This exercise is left for the reader.
I'm currently interested in F# as it is different to everything I have used before. I need to access the first element of each list contained within a large list.
If I assume that the main list contains 'x' amount of lists that themselves contain 5 elements, what would be the easiest way to access each of the first elements.
let listOfLists = [ [1;2;3;4;5]; [6;7;8;9;10]; [11;12;13;14;15] ]
My desired output would be a new list containing [1;6;11]
Currently I have
let rec firstElements list =
match list with
| list[head::tail] ->
match head with
| [head::tail] -> 1st # head then firstElements tail
To then expand on that, how would I then get all of the second elements? Would it be best to create new lists without the first elements (by removing them using a similar function) and then reusing this same function?
You can use map to extract the head element of each child list:
let firstElements li =
match li with [] -> None | h::_ -> Some h
let myfirstElements = List.map firstElements listOfLists
I'm using Ocaml's speak with little lookup on F# so this may be inaccurate, but the idea applies.
EDIT: You can also use List.head which makes it more concise and will return a int list instead of int option list. However, it throws an exception if you hits an empty list. Most of the time, I'd avoid using List.head or List.tail in this case.
The easisest way to access the first element in a list is List.head. As you have a list of lists, you just need to List.map this function:
let listOfLists = [ [1;2;3;4;5]; [6;7;8;9;10]; [11;12;13;14;15] ]
listOfLists
|> List.map List.head
//val it : int list = [1; 6; 11]
Now if you need to access other elements, you can use List.item or just index into your list with xs.[1]. But please keep in mind, that for large lists this will be inefficient and if you want fast look up use an array.
listOfLists
|> List.map (List.item 1)
//val it : int list = [2; 7; 12]
With indexing:
listOfLists
|> List.map (fun x -> x.[1])
This is my function
let rec helper inputList = function
| [] -> []
| a :: b :: hd ->
if a = b then helper ([b::hd])
else a :: helper (b::hd)
It's not complete, however I can't see why I keep getting the error in the title at helper ([b::hd]). I've tried helper (b::hd) or helper (b::hd::[]) however all come up with errors. How do I make it so that it works?
When you use function you are supplying a pattern for the parameter of the function. But you already have a parameter named inputList. So this function helper is expecting two parameters (but it ignores the first).
You can fix this by removing inputList.
You also have a problem in your first recursive call to helper. Your expression [b :: hd] is a list of lists. I suspect that you want something more like just b :: hd here.
There is at least one other problem, but I hope this helps get you started.
There are multiple errors here. One is that the keyword function means we have an implicit parameter over which we are working. So the pattern matching happens on that "invisible" parameter. But here you defined probably the explicit one: inputList. So we can remove that one:
let rec helper = function
| [] -> []
| a :: b :: hd -> if a = b then helper ([b::hd]) else a :: helper (b:: hd)
Next there is a problem with the types: in the recursion, you use:
helper ([b::hd]); and
a :: helper (b:: hd)
But you put these on the same line, and that makes no sense, since the first one passes a list of lists of elements, and the second a list of elements. So the result of the first one would be a list of list of elements, and the second one a list of elements. It does not make sense to merge these.
If I understood correctly that you want to ensure that no two consecutive elements should occur that are equal, then we should rewrite it to:
let rec helper = function
| [] -> []
| a :: b :: hd -> if a = b then helper (b::hd) else a :: helper (b:: hd)
You have defined two patterns here:
one for the empty list; and
one for a list with at least two elements.
The second one will perform recursion on the tail of the list b :: hd. So that means that eventually when we pass it a list with n elements, it will recursively work on a list with n-1 elements, n-2 elements, etc. But eventually it will have one element. And there is no case for that. So we need to add a case for the one element pattern:
let rec helper = function
| [] -> []
| h :: [] -> h :: []
| a :: b :: hd -> if a = b then helper (b::hd) else a :: helper (b:: hd)