I'm trying to rewrite the List.length function without using recursion. Here's my code:
(* given *)
type 'a list =
| []
| (::) of 'a * 'a list
let nil : 'a list = []
let cons (hd : 'a) (tl : 'a list): 'a list = hd :: tl
let length (ls : 'a list): int =
let i = fold_left(fun x y -> Fun.const 1 :: y) [] ls in
fold_left(fun x y -> x + y) 0 i
OCaml gave me an error on the last line fold_left(fun x y -> x + y) 0 i and saying my i here has type ('a -> int) list but an expression was expected of type int list, is there any way I can fix this? Thank you!
It's not entirely clear to me what you are trying to achieve with Fun.const, but you can actually achieve length with a single fold_left:
let length l =
fold_left (fun acc _ -> acc+1) 0 l
There are a few issues with your attempt.
As Fun.const creates in this case a function which always returns 1, and your expression parses as: (Fun.const 1) :: y your first use of List.fold_left is generating a list of functions which all return one. But there are not the value 1.
Thus when you use fold_left (fun x y -> x + y) 0 i you are trying to add a function to an int. Clearly this doesn't work. You need to apply some argument to that function.
let length (ls : 'a list): int =
let i = List.fold_left(fun x y -> Fun.const 1 :: y) [] ls in
List.fold_left(fun x y -> x + y ()) 0 i
This compiles. But... the type doesn't look quite right.
val length : (unit -> int) list list -> int = <fun>
This results from having the argument order to the function List.fold_left takes wrong. The first argument is the initial value. The second is the first element from the list.
If we change that:
let length (ls : 'a list): int =
let i = List.fold_left(fun y _ -> Fun.const 1 :: y) [] ls in
List.fold_left(fun x y -> x + y ()) 0 i
The type of this function is:
val length : 'a list -> int = <fun>
But as Blackbeans has noted, there is a much simpler way to write this using a single call to List.fold_left.
If i put this code to utop line by line, then it works, but if I compile it as programm, then I see error.
module List = Core.Std.List;;
let () =
let shifts = [ (-1, -1); (0, -1) ] in
let first = List.nth shifts 0 in
let (a, b) = first in
Printf.printf "%d %d\n" a b;
;;
And error message:
Error: This expression has type (int * int) option
but an expression was expected of type 'a * 'b
What wrong with types and why it works in utop?
It looks to me like Core.Std.List.nth has the type 'a list -> int -> 'a option in order to handle the case when the int is out of range. The standard library List.nth has the type 'a list -> int -> 'a. It raises an exception for an out-of-range int.
If you're using Core.Std.List in one case but the standard List in the other, this would explain the difference.
I'm trying to build a type LazyList but hide the definition of LazyList.t. I've got the following files:
LazyList.ml
type 'a t =
| Cell of ('a * 'a t) option Lazy.t
;;
let rec from_list l = ...;;
let rec from_string s = ...;;
let rec from_in_channel c = ...;;
let rec to_list l = ...;;
LazyList.mli
type 'a t;;
val from_list : 'a list -> 'a t;
val from_string : string -> char t;
val from_in_channel : in_channel -> char t;
val to_list : 'a t -> 'a list;
When I run ocamlc LazyList.mli I get the following error:
File "LazyList.mli", line 1, characters 9-10:
Error: Syntax error
What gives? Do I need to expose the constructor if I'm going to be writing all of the methods I need in LazyList.ml?
This is only a syntax error. Semicolons cannot appear in mli file. Also you really do not need to use this ugly double semicolons in ml file, although it is not an error.
I've got a basic function which checks a list for duplicates and returns true if they are found, false otherwise.
# let rec check_dup l = match l with
[] -> false
| (h::t) ->
let x = (List.filter h t) in
if (x == []) then
check_dup t
else
true
;;
Yet when I try to use this code I get the error
Characters 92-93:
let x = (List.filter h t) in
^
Error: This expression has type ('a -> bool) list
but an expression was expected of type 'a list
I don't really understand why this is happening, where is the a->bool list type coming from?
The type ('a -> bool) list is coming from the type of filter and from the pattern match h::t in combination. You're asking to use a single element of the list, h, as a predicate to be applied to every element of the list t. The ML type system cannot express this situation. filter expects two arguments, one of some type 'a -> bool where 'a is unknown, and a second argument of type 'a list, where 'a is the same unknown type as in the first argument. So h must have type 'a -> bool and t must have type 'a list.
But you have also written h::t, which means that there is another unknown type 'b such that h has type 'b and t has type 'b list. Put this together and you get this set of equations:
'a -> bool == 'b
'a list == 'b list
The type checker looks at this and decides maybe 'a == 'b, yielding the simpler problem
'a -> bool == 'a
and it can't find any solution, so it bleats.
Neither the simpler form nor the original equation has a solution.
You are probably looking for List.filter (fun x -> x = h) t, and you would probably be even better off using List.exists.
For complete this answer I post the final function for search duplicate value in array:
let lstOne = [1;5;4;3;10;9;5;5;4];;
let lstTwo = [1;5;4;3;10];;
let rec check_dup l = match l with
[] -> false
| (h::t) ->
let x = (List.filter (fun x -> x = h) t) in
if (x == []) then
check_dup t
else
true;;
and when the function run:
# check_dup lstOne
- : bool = true
# check_dup lstTwo
- : bool = false
#
I'm learning to deal with Lists and Tuples in F# and a problem came up. I have two lists: one of names and one with names,ages.
let namesToFind = [ "john", "andrea" ]
let namesAndAges = [ ("john", 10); ("andrea", 15) ]
I'm trying to create a function that will return the first age found in namesAndAges given namesToFind. Just the first.
So far I have the following code which returns the entire tuple ("john", 10).
let findInList source target =
let itemFound = seq { for n in source do
yield target |> List.filter (fun (x,y) -> x = n) }
|> Seq.head
itemFound
I tried using fst() in the returning statement but it does not compile and gives me "This expression was expected to have type 'a * 'b but here has type ('c * 'd) list"
Thanks for any help!
There are lots of functions in the Collections.List module that can be used. Since there are no break or a real return statement in F#, it is often better to use some search function, or write a recursive loop-function. Here is an example:
let namesToFind = [ "john"; "andrea" ]
let namesAndAges = [ "john", 10; "andrea", 15 ]
let findInList source target =
List.pick (fun x -> List.tryFind (fun (y,_) -> x = y) target) source
findInList namesToFind namesAndAges
The findInList function is composed of two functions from the Collections.List module.
First we have the List.tryFind predicate list function, which returns the first item for which the given predicate function returns true.
The result is in the form of an option type, which can take two values: None and Some(x). It is used for functions that sometimes give no useful result.
The signature is: tryFind : ('T -> bool) -> 'T list -> 'T option, where 'T is the item type, and ('T -> bool) is the predicate function type.
In this case it will search trough the target list, looking for tuples where the first element (y) equals the variable x from the outer function.
Then we have the List.pick mapper list function, which applies the mapper-function to each one, until the first result that is not None, which is returned.
This function will not return an option value, but will instead throw an exception if no item is found. There is also an option-variant of this function named List.tryPick.
The signature is: pick : ('T -> 'U option) -> 'T list -> 'U, where 'T is the item type, 'U is the result type, and ('T -> 'U option) is the mapping function type.
In this case it will go through the source-list, looking for matches in the target array (via List.tryFind) for each one, and will stop at the first match.
If you want to write the loops explicitly, here is how it could look:
let findInList source target =
let rec loop names =
match names with
| (name1::xs) -> // Look at the current item in the
// source list, and see if there are
// any matches in the target list.
let rec loop2 tuples =
match tuples with
| ((name2,age)::ys) -> // Look at the current tuple in
// the target list, and see if
// it matches the current item.
if name1 = name2 then
Some (name2, age) // Found a match!
else
loop2 ys // Nothing yet; Continue looking.
| [] -> None // No more items, return "nothing"
match loop2 target with // Start the loop
| Some (name, age) -> (name, age) // Found a match!
| None -> loop rest // Nothing yet; Continue looking.
| [] -> failwith "No name found" // No more items.
// Start the loop
loop source
(xs and ys are common ways of writing lists or sequences of items)
First let's look at your code and annotate all the types:
let findInList source target =
let itemFound =
seq {
for n in source do
yield target |> List.filter (fun (x,y) -> x = n) }
|> Seq.head
itemFound
The statement yield List.Filter ... means you're creating a sequence of lists: seq<list<'a * 'b>>.
The statement Seq.head takes the first element from your sequence of lists: list<'a * 'b>.
So the whole function returns a list<'a * 'b>, which is obviously not the right type for your function. I think you intended to write something like this:
let findInList source target =
let itemFound =
target // list<'a * 'b>
|> List.filter (fun (x,y) -> x = n) // list<'a * 'b>
|> Seq.head // 'a * 'b
itemFound // function returns 'a * 'b
There are lots of ways you can get the results you want. Your code is already half way there. In place of filtering by hand, I recommend using the built in val Seq.find : (a' -> bool) -> seq<'a> -> 'a method:
let findAge l name = l |> Seq.find (fun (a, b) -> a = name) |> snd
Or you can try using a different data structure like a Map<'key, 'value>:
> let namesAndAges = [ ("john", 10); ("andrea", 15) ] |> Map.ofList;;
val namesAndAges : Map<string,int> = map [("andrea", 15); ("john", 10)]
> namesAndAges.["john"];;
val it : int = 10
If you want to write it by hand, then try this with your seq expression:
let findInList source target =
seq {
for (x, y) in source do
if x = target then
yield y}
|> Seq.head
Like fst use this(below) . This way you can access all the values.
This is from F# interactive
let a = ((1,2), (3,4));
let b = snd (fst a);;
//interactive output below.
val a : (int * int) * (int * int) = ((1, 2), (3, 4))
val b : int = 2