My task is to remove the duplicates from a list. To do that I have to first sort the list.
I have written the function that sorts the list and the one that remove the
duplicates(once they are sorted) but I don't know how to combine them.
Example:
input: [4;5;2;2;1;3;3]
output: [1;2;3;4;5]
let rec setify = function
| [] -> []
| x :: l -> insert x (setify l)
and insert elem = function
| [] -> [elem]
| x :: l -> if elem < x then elem :: x :: l
else x :: insert elem l;;
let rec rem =function
|[] -> []
| x :: []-> x :: []
| x :: y :: rest -> if x = y then rem (y :: rest)
else x :: rem (y :: rest) ;;
You want to make the function that takes a list, creates the sorted list, and deduplicates that. In other words, you want:
let task list =
let sorted_list = setify list in
rem sorted_list
It is possible to do this in arbitrarily more complicated ways, but the above is one straightforward, one-action-per-line version. Since the phrasing of the title of your question invites it, here is one of the more sophisticated ways:
(* it's possible to write a generic combinator of functions, that takes two functions f and g *)
let combine f g =
(* and returns a function *)
fun x ->
(* that maps x to f(g(x)) *)
f (g x)
(* this function is typed as:
val combine : ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b = <fun>
*)
(* the task can then be expressed as the combination of setify and rem: *)
let task = combine rem setify
Don't use this style unless something is actually gained from it. Most of
the times it only makes programs less readable and slower with no corresponding benefit. *)
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]
If I have an input of a tuple containing two lists of integers of the same length, and I want my output to be a list of these two lists zipped, after extracting these two lists from the tuple how do I zip each individual element into one list? For example, if my input is twolists= ([1;2;3], [4;5;6]), then I want my output to be [(1,4); (2,5); (3,6)]. How do I zip each element and add it to my output?
The function name and type is as follows:
let rec pairlists twolists = ...
val pairlists : 'a list * 'b list -> ('a * 'b) list = fun
So far I have:
let rec pairlists twolists =
let (l1, l2) = twolists in
let rec zip (l1,l2) =
match l1 with
|[] -> l2
|x :: xs -> x :: zip(l2, xs) in
twolists ;;
but this is clearly not doing what I want.
Are you looking for List.combine ?
val combine : 'a list -> 'b list -> ('a * 'b) list
Transform a pair of lists into a list of pairs: combine [a1; ...; an] [b1; ...; bn] is [(a1,b1); ...; (an,bn)].
Raises Invalid_argument if the two lists have different lengths. Not tail-recursive.
If your result list should contain elements that consist of the elements of both sublists, then you obviously have to destructure each sublist on each iteration.
If the lists are guaranteed to have the same lengths, the solution can be as simple as:
let rec zip paired_lists =
match paired_lists with
| [], [] -> []
| h1::t1, h2::t2 -> (h1, h2)::(zip (t1, t2))
| _, _ -> failwith "oops, the lists seems to have different lengths"
;;
zip ([1;2;3], [4;5;6]);;
- : (int * int) list = [(1, 4); (2, 5); (3, 6)]
But this one is not tail-recursive, which is obviously not good. The second sub-optimal thing is this reconstruction of tuple of lists on each iteration (I'm a newbie in OCaml, so chances are compiler is smart enough to avoid the unnecessary allocations, but still...). Fixing both flaws is trivial too:
let zip_tr paired_lists =
let list1, list2 = paired_lists in
let rec aux l1 l2 acc =
match l1, l2 with
| [], [] -> List.rev acc
| h1::t1, h2::t2 -> aux t1 t2 (h1, h2)::acc
| _, _ -> failwith "oops, the lists seems to have different lengths"
in aux list1 list2 []
;;
zip_tr ([1;2;3], [4;5;6]);;
- : (int * int) list = [(1, 4); (2, 5); (3, 6)]
The signature of your code does not match the expected signature :
line 2, characters 11-13:
Warning 26: unused variable l2.
Line 2, characters 7-9:
Warning 26: unused variable l1.
val pairlists : 'a list * 'a list -> 'a list = <fun>
Indeed, both possible matches return either a 'a list (this is l2) or x::zip... which is also a list of 'a type.
There should be sth like (x,y)::list in your code.
In addition, pairlists is not recursive and does not need to be declared as such, only zip is recursive.
The end of your function shall be like this (otherwise zip has no effect) :
....
let rec zip (l1,l2) =
match l1 with
|[] -> l2
|x :: xs -> x :: zip(l2, xs) in
zip twolists ;;
In addition to the other solutions mentioned, ocaml-4.08 onwards enables you to provide let+ and and+ operators which will zip a list sum-wise, where you might otherwise think of using applicatives. Whether it is an improvement on them is in the eye of the beholder:
let (let+) list f = List.map f list
let (and+) a b =
let rec loop first second =
match first,second with
first_hd::first_tl,second_hd::second_tl ->
(first_hd,second_hd)::(loop first_tl second_tl)
| _ -> []
in
loop a b
let pairlists = function
first,second ->
let+ elt1 = first
and+ elt2 = second in
[elt1 ; elt2]
(* example *)
let () =
let res = pairlists ([1;2;3], [4;5;6]) in
List.iter
(fun list -> List.iter (fun i -> Printf.printf "%d " i) list ;
print_endline "")
res
Here by way of comparison is the more traditional approach if you are using applicatives
let pure x = [x]
let (<*>) aps args =
List.concat (List.map (fun f -> List.map (fun x -> f x) args) aps)
let (<|>) aps args =
let rec loop args_rest aps_rest =
match args_rest,aps_rest with
args_hd::args_tl,aps_hd::aps_tl ->
(aps_hd args_hd)::(loop args_tl aps_tl)
| _ -> []
in
loop args aps
let pairlists = function
first,second ->
let two_list a b = a :: [b] in
pure two_list <*> first <|> second
(* example *)
let () =
let res = pairlists ([1;2;3], [4;5;6]) in
List.iter
(fun list -> List.iter (fun i -> Printf.printf "%d " i) list ;
print_endline "")
res
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> *)
Say we have such a lazy list:
type 'a lazy_list_t = Cons of 'a * (unit -> 'a lazy_list_t)
Does it make sense to have a function like the iter in regular list:
val iter : ('a -> unit) -> 'a list -> unit
List.iter f [a1; ...; an] applies function f in turn to a1; ...; an. It is equivalent to begin f a1; f a2; ...; f an; () end.
Or is it possible to produce iter_lazy like
val iter_lazy: ('a -> unit) -> 'a lazy_list -> unit
No, it does not make much sense.
First, and you probably noticed it, all your list are infinite (you do not have an empty element). So, only examples of inhabitant of your type are somehow using a recursive function, eg. :
let omega =
let rec f n = Cons (n, fun () -> f (n + 1)) in
f 0
This implements the infinite stream [ 0, 1, 2, 3, ...
If you WANT a diverging program you could implement :
let rec iter f (Cons (n, g)) = f n; iter f (g ())
but if you do iter print_int omega it will result output all integers which will take some time.
So itering is not an option. What would work is "mapping", you can implement the function :
val map: ('a -> 'b) -> 'a lazy_list_t -> 'b lazy_list
let rec map f (Cons (x, g)) = Cons (f x, fun () -> map f (g ()))
Notice how the recursive call to map is "protected" by the "fun () ->" so it will not trigger "right away" but only each time the tail of your lazy list is forced.
You can use this to lazily compute on infinite streams, eg :
let evens = map ((*) 2) omega
computes the stream [0; 2; 4; 6; 8; ...
Note, that you could use it to implement some sort of "iter" by mapping a function that does a side_effect eg.
let units = map print_int evens
will output right away the number "0" and outputs the stream [(); (); (); ... and each time you force one of the "tail" of this stream it will output the corresponding number (it can happen multiple times). Example:
(* Force the tail *)
val tl : 'a lazy_list_t -> 'a lazy_list_t
let tl (Cons (_, g)) = g ()
let () = begin
tl units; (* ouputs "2" *)
tl (tl units); (* outputs "24" *)
tl units; (* outputs "2" *)
end
(I haven't tried the code so there may be some typos).
Suppose I have some code like this:
List.map (fun e -> if (e <> 1) then e + 1 else (*add nothing to the list*))
Is there a way to do this? If so, how?
I want to both manipulate the item if it matches some criteria and ignore it if it does not. Thus List.filter wouldn't seem to be the solution.
SML has a function mapPartial which does exactly this. Sadly this function does not exist in OCaml. However you can easily define it yourself like this:
let map_partial f xs =
let prepend_option x xs = match x with
| None -> xs
| Some x -> x :: xs in
List.rev (List.fold_left (fun acc x -> prepend_option (f x) acc) [] xs)
Usage:
map_partial (fun x -> if x <> 1 then Some (x+1) else None) [0;1;2;3]
will return [1;3;4].
Or you can use filter_map from extlib as ygrek pointed out.
Both Batteries and Extlib provide an equivalent of mapPartial: their extended List module sprovide a filter_map function of the type ('a -> 'b option) -> 'a list -> 'b list, allowing the map function to select items as well.
Another solution would be to use directly a foldl :
let f e l = if (e <> 1)
then (e + 1)::l
else l
in List.fold_left f [] list
But my preference is filter_map as Michael Ekstrand provided
Alternatively you can filter your list then apply the map on the resulted list as follows :
let map_bis predicate map_function lst =
List.map map_function (List.filter predicate lst);;
# val map_bis : ('a -> bool) -> ('a -> 'b) -> 'a list -> 'b list = <fun>
Usage :
# map_bis (fun e -> e<>1) (fun e -> e+1) [0;1;2;3];;
- : int list = [1; 3; 4]
You can also map values to singleton lists if you want to keep them or empty lists if you don't, and then concat the results.
List.concat (List.map (fun e -> if (e <> 1) then [e + 1] else []) my_list)
use
let rec process = function
| 1 :: t -> process t
| h :: t -> (h + 1) :: (process t)
| [] -> []
or tail recursive
let process =
let rec f acc = function
| 1 :: t -> f acc t
| h :: t -> f ((h + 1) :: acc) t
| [] -> List.rev acc in
f []
or with a composition of standard functions
let process l =
l |> List.filter ((<>)1)
|> List.map ((+)1)
The OCaml standard library has had List.filter_map since 4.08. This can therefore now be written as:
List.filter_map (fun e -> if e <> 1 then Some (e + 1) else None)