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.
Related
When declaring a function, I've 3 different ways:
let f x = ...
let f = (fun x -> ...)
let f = function
| ... -> (pattern matching)
It's this last one that I don't fully understand how it works.
I was doing a function that, considering a list (we'll assume it has integers in it but could be anything), reverses it, pretty basic, but with a complexity of O(n). After struggling for an hour at least I check the answer, and it is written like this:
let reverse lst =
let rec aux acc = function
| [] -> acc
| hd :: tl -> aux (hd :: acc) tl
in
aux [] lst
I thought that using the key word function was just another way of doing patter matching, but when I do this:
let reverse lst =
let rec aux acc =
match aux with
| [] -> acc
| hd :: tl -> aux (hd :: acc) tl
in
aux [] lst
It doesn't work, and idk why. On top of that, why can we add tl at the end of the first function? Isn't aux a single argument function?
There are a few problems with this question. First, the code you give as the solution for reverse is not valid OCaml. It matches aux (which is a function) against list patterns. Most likely aux was supposed to be acc. But even so it doesn't seem right because it should have two arguments (the accumulated result and the input that still needs to be processed).
Second, your two code examples are the same. You seem to be saying that one works and one doesn't work. That doesn't make sense since they're the same.
IMHO you need to rewrite the question if you want to get a helpful answer.
Ocaml uses currying, which means that a two-argument function is the same thing that a function whose return value is a function.
To define a two-argument function, you can combine all the ways you know of creating one-argument functions:
let f x y = x + y
let f x = (fun y -> x + y)
let f x = function
| y -> x + y
let f = (fun x -> (fun y -> x + y))
let f = function
| x -> function
| y -> x + y
let f x = (let g y = x + y in g)
etc, etc.
All these definitions for f lead to the same result:
val f : int -> int -> int = <fun>
# f 3 4;;
- : int = 7
Note that the signature of f is:
val f : int -> int -> int = <fun>
If we added parentheses to better understand this signature, it would be this:
val f : int -> (int -> int) = <fun>
Meaning that f is a one-argument function whose return value is a one-argument function whose return value is an int.
Indeed, if we partially apply f:
# f 3;;
- : int -> int = <fun>
# let add_three = f 3;;
val add_three : int -> int = <fun>
# add_three 4;;
- : int = 7
The code you give at the end of your question is wrong. It's most likely intended to be this:
let reverse lst =
let rec aux acc l =
match l with
| [] -> acc
| hd :: tl -> aux (hd :: acc) tl
in
aux [] lst;;
val reverse : 'a list -> 'a list = <fun>
# reverse [1;2;3;4;5];;
- : int list = [5; 4; 3; 2; 1]
I want to implement the List.assoc function using List.find, this is what I have tried:
let rec assoc lista x = match lista with
| [] -> raise Not_found
| (a,b)::l -> try (List.find (fun x -> a = x) lista)
b
with Not_found -> assoc l x;;
but it gives me this error:
This expression has type ('a * 'b) list but an expression was expected of type 'a list
The type variable 'a occurs inside 'a * 'b
I don't know if this is something expected to happen or if I'm doing something wrong. I also tried this as an alternative:
let assoc lista x = match lista with
| [] -> raise Not_found
| (a,b)::l -> match List.split lista with
| (l1,l2) -> let ind = find l1 (List.find (fun s -> compare a x = 0))
in List.nth l2 ind;;
where find is a function that returns the index of the element requested:
let rec find lst x =
match lst with
| [] -> raise Not_found
| h :: t -> if x = h then 0 else 1 + find t x;;
with this code the problem is that the function should have type ('a * 'b) list -> 'a -> 'b, but instead it's (('a list -> 'a) * 'b) list -> ('a list -> 'a) -> 'b, so when I try
assoc [(1,a);(2,b);(3,c)] 2;;
I get:
This expression has type int but an expression was expected of type
'a list -> 'a (refering to the first element of the pair inside the list)
I don't understand why I don't get the expected function type.
First off, a quick suggestion on making your assoc function more idiomatic OCaml: have it take the list as the last argument.
Secondly, why are you attempting to implement this in terms of find? It's much easier without.
let rec assoc x lista =
match lista with
| [] -> raise Not_found
| (a, b) :: xs -> if a = x then b else assoc x xs
Something like this is simpler and substantially more efficient with the way lists work in OCaml.
Having the list as the last argument, even means we can write this more tersely.
let rec assoc x =
function
| [] -> raise Not_found
| (a, b) :: xs -> if a = x then b else assoc x xs
As to your question, OCaml infers the types of functions from how they're used.
find l1 (List.find (fun s -> compare a x = 0))
We know l1 is an int list. So we must be trying to find it in an int list list. So:
List.find (fun s -> compare a x = 0)
Must return an int list list. It's a mess. Try rethinking your function and you'll end up with something much easier to reason about.
I have a list of (string * int) list elements and I need to find the biggest int element and return the corresponding(string * int) element.
I have something like this atm, but problem is, I think my approach is more of "typical programming"
let it = [] in
for x = 0 to length LIST - 1 do
let str = ((List.nth LIST x).string) in
let num = ((List.nth LIST x).int) in
let it = it # [num, str] in
let (str, num) = List.hd(List.rev it) in
[str, num]
What I tried to do is to loop through the list and add the string and int value in another list, then sort them, reverse it and then take the head, which should be the max int, then I need to return the pair in (string * int)
Your code is not a well-formed OCaml code. It highlights, however, some number of issues with your understanding of OCaml.
First of all, by default, values in OCaml are immutable. For example,
let x = 0 in
for i = 0 to 10 do
let x = x + 1 in
print_int x;
done
You will get 11111111111 as the output. This is because, during the loop, you are just computing every time the x+1 expression, where x is always 0 and you will always get 1 as the result. This is because, let x = <expr> in <body> is not changing the existing variable x but is creating a new variable x (shadowing any previous definitions) and make it available in the scope of the <body> expression.
Concerning your problem in general, it should be solved as a recursive function greatest_element, which has the following definition,
for an empty list [] it is undefined;
for a list of one element [x] is it is x;
otherwise, for a list of x::xs it is max x (greatest_element xs),
where max x y is x if it is greater or equal to y.
Finally, it looks like you have missed the first steps in OCaml and before solving this task you have to move back and to learn the basics. In particular, you have to learn how to call functions, bind variables, and in general what are the lexical conventions and syntax of the language. If you need pointers, feel free to ask.
First of all, it doesn't seem that you did any kind of sorting in
the code that you provided.
Assuming that your list is of type
(string * int) list then a possible to find the element with the
maximum integer using recursion:
let max_in_list list =
let rec auxiliary max_str max_int = function
| []
-> (max_str, max_int)
| (crt_str, crt_int)::tail when crt_int > max_int
-> auxiliary crt_str crt_int tail
| _::tail
-> auxiliary max_str max_int tail
in
match list with
| []
-> None
| (fst_str, fst_int)::tail
-> Some (auxiliary fst_str fst_int tail)
let check = max_in_list [("some", 1); ("string", 3); ("values", 2)]
You could write a generic maxBy function. This allows you to get the max of any list -
let rec maxBy f = function
| [] -> None
| [ x ] -> Some x
| x :: xs ->
match (maxBy f xs) with
| Some y when (f y) > (f x) -> Some y
| _ -> Some x
(* val maxBy : ('a -> 'b) -> 'a list -> 'a option = <fun> *)
let data = [("a", 3); ("b", 2); ("c", 6); ("d", 1)]
(* val data : (string * int) list = [("a", 3); ("b", 2); ("c", 6); ("d", 1)]*)
maxBy (fun (_, num) -> num) data
(* - : (string * int) option = Some ("c", 6) *)
maxBy (fun (str, _) -> str) data
(* - : (string * int) option = Some ("d", 1) *)
maxBy (fun x -> x) [3; 2; 6; 1]
(* - : int option = Some 6 *)
maxBy (fun x -> x) ["c"; "d"; "b"; "a"]
(* - : string option = Some "d" *)
maxBy (fun x -> x) []
(* - : 'a option = None *)
It can be fun to rewrite the same function in various ways. Here's another encoding -
let maxBy f list =
let rec loop r = function
| [] -> r
| x::xs when (f x) > (f r) -> loop x xs
| _::xs -> loop r xs
in
match list with
| [] -> None
| x::xs -> Some (loop x xs)
(* val maxBy : ('a -> 'b) -> 'a list -> 'a option = <fun> *)
As I am going through the website:
http://www.cs.princeton.edu/courses/archive/fall14/cos326/sec/03/precept03_sol.ml
I have got a question according to the Ocaml structure. To be more specific, I have questions according to the code:
let rec reduce (f:'a -> 'b -> 'b) (u:'b) (xs:'a list) : 'b =
match xs with
| [] -> u
| hd::tl -> f hd (reduce f u tl);;
What does the f hd do at the very last line? (I understand that reduce f u tl is calling the function itself again.)
My second question is how to use a function to implement another function in Ocaml. For the code:
let times_x (x: int) (lst: int list) : int list =
map (fun y -> y*x) lst
What does fun y -> y*x do? what does lst do at the end of the code?
Thank you for the help!
The code that has been provided is a reduce function that takes three parameters - a function that maps inputs of type 'a and 'b to an output of type 'b, a value of type 'b, and as list of elements of type 'a.
For example, the length example from the lecture:
let length (lst: int list) : int =
reduce (fun _ len -> len + 1) 0 lst
The first parameter to reduce is a function that, when given two parameters, discards the first one and returns the second parameter incremented by one. The second is a value (0) to be used as an accumulator. The third is a list to find the length of.
The behavior of this recursive reduce function is to return the second parameter (an accumulator as used in the length example) once the provided list is empty, and otherwise run the provided function using the head of the list and the recursed value.
Once again going to the length example, say we give it a list with a single element [1].
Our call to length becomes reduce (fun _ len -> len + 1) 0 [1]
Recall reduce:
let rec reduce (f:'a -> 'b -> 'b) (u:'b) (xs:'a list) : 'b =
match xs with
| [] -> u
| hd::tl -> f hd (reduce f u tl);;
First, we match [1] against [], which fails. Since it is a non-empty list, we run f hd (reduce f u tl)
Recall that f is the parameter that length provided: fun _ len -> len + 1
Therefore, we effectively run the following:
(fun _ len -> len + 1) 1 (reduce (fun _ len -> len + 1) 0 [])
In this case, the length function discards the first parameter since the values in the list are not necessary to know the length of the list.
The recursive portion will match against [] and return the value of u at the time, which is 0.
Therefore, one level up, (fun _ len -> len + 1) 1 (reduce (fun _ len -> len + 1) 0 []) becomes (fun _ len -> len + 1) 1 0 and returns 0 + 1, simplifying to our expected value 1, which represents the length of the list.
Now, to your second question, in regards to times_x. This performs a mapping. For example, we can map [1;2;3;4;5] to [3;6;9;12;15] with a mapping fun x -> x * 3.
Here times_x is defined as follows:
let times_x (x: int) (lst: int list) : int list =
map (fun y -> y*x) lst
times_x takes an integer and a list. Using the above example, we could call it with times_x 3 [1;2;3;4;5] to get [3;6;9;12;15].
Beyond this I recommend looking into how map and reduce functions work in general.
I hope this answer was adequate at addressing your question.
let rec add_tail l e = match l with
| [] -> [e]
| (h::t) -> h::(add_tail t e)
let rec fill_help l x n = match n = 0 with
true -> l
| false -> add_tail(l, x); fill_help(l, x, n-1)
let fill x n =
let l = [] in
fill_help(l, x, n)
and I'm getting the error in the interpreter
# #use "prac.ml";;
val prod : int list -> int = <fun>
val add_tail : 'a list -> 'a -> 'a list = <fun>
File "prac.ml", line 13, characters 21-27:
Error: This expression has type 'a * 'b
but an expression was expected of type 'c list
line 13 would be
| false -> add_tail(l, x); fill_help(l, x, n-1)
First of all you call fill_help with a tuple as an argument ((l, x, n-1)) even though it's not defined to take one. You should call fill_help as fill_help l x (n-1) instead. Same for add_tail.
Secondly you call a side-effect-free function (add_tail) and throw away its return value. This is almost always an error. It looks like you expect l to be different after the call to add_tail. It won't be. You probably want fill_help (add_tail l x) x (n-1).