My task is to create a histogram that output the number of times that an element it is in a list.
Input:[2;2;2;3;4;4;1]
Output[(2, 3); (2, 2); (2, 1); (3, 1); (4, 2); (4, 1); (1, 1)]
Expected output : [(2, 3); (3, 1); (4, 2); (1, 1)]
My code:
let rec count a ls = match ls with
|[] -> 0
|x::xs when x=a -> 1 + count a xs
|_::xs -> count a xs
let rec count a = function
|[] -> 0
|x::xs when x=a -> 1 + count a xs
|_::xs -> count a xs
let rec histo l = match l with
|[] -> []
|x :: xs -> [(x, count x l)] # histo xs ;;
What have i done wrong?
The issue is that xs contains potentially elements that are equal to x. This is what you see in your ouput : (2,3) means that there is 3 times 2 in the list; xs is then equals to [2;2;3;4;4;1]... and so on.
Also (not impacting the conclusion): you have 2 definitions of count, but they are identical.
To implement an histogram, use Hashtbl :
let h = Hashtbl.create 1000;;
List.iter (fun x -> let c = try Hashtbl.find h x with Not_found -> 0 in Hashtbl.replace h x (c+1)) your_list;;
Hashtbl.fold (fun x y acc -> (x,y)::acc) h [];;
Related
How to create a tuple list from one single list, like so:
[1; 2; 4; 6] -> [(1, 2); (4, 6)]
I want to do it using function List.fold_left since I'm trying to learn that currently but don't know how... Is there a way? Or should I leave it like that?
This is a working code that doesn't use List.fold_left:
let rec create_tuple acc l = match l with
| [] -> acc
| x :: y :: l' -> create_tuple (acc # [(x, y)]) l'
| _ -> acc
List.fold_left reads elements one by one. There is no direct way to make it read elements two by two.
It really is pointless complication (great for teaching, though), but if you absolutely want to use List.fold_left here, your accumulator needs to somehow record the state of the traversal:
either you have read an even number of elements so far,
or you have read an odd number and then you have to record what was the last element you read, so that, upon reading the following one, you can pair them.
Here is a way to do it. I use an algebraic datatype to represent the state.
(* This is the type that we’ll use for the accumulator;
the option component is the state of the traversal.
(None, acc) means that we have read an even number of elements so far;
(Some x, acc) means that we have read an odd number of elements so far,
the last of which being x. *)
type 'a accumulator = 'a option * ('a * 'a) list
let folder (state, acc) x =
match state with
| None -> (Some x, acc)
| Some y -> (None, (y,x)::acc)
let create_pairs l =
let (_, acc) = List.fold_left folder (None, []) l in
List.rev acc
Also notice how I avoid the complexity bug that I outlined in a comment: I add elements in reverse order (i.e. at the head of the accumulating list), and at the very end I reverse that list.
#Maëlan's answer is beautiful, but what if we want to get triples rather than pairs? Is there a way we can use List.fold_left to handle this more generically?
let chunks n lst =
let (_, _, acc) = List.fold_left
(fun (counter, chunk, lst') x ->
if counter = n - 1 then
(0, [], List.rev (x :: chunk) :: lst')
else
(counter + 1, x :: chunk, lst'))
(0, [], [])
lst
in
List.rev acc
Using this, chunks 2 [1; 2; 4; 6] returns [[1; 2]; [4; 6]]. We can map this to the result you're looking for with a very simple function that takes a list with two elements and creates a tuple with two elements.
chunks 2 [1; 2; 4; 6] |> List.map (fun [x; y] -> (x, y))
And we get:
[(1, 2), (4, 6)]
This could be used to implement a triples function.
let create_triples lst =
chunks 3 lst |> List.map (fun [x; y; z] -> (x, y, z));;
And now create_triples [1; 2; 3; 4; 5; 6; 7; 8; 9] returns [(1, 2, 3); (4, 5, 6); (7, 8, 9)].
I tried this question(using List.fold_left) and this is the best I could come up with:
type 'a node = First of 'a | Second of ('a * 'a)
let ans =
List.fold_left
(
fun a e ->
match a with
| [] -> (First e)::a
| (First f)::tl -> Second(f, e)::tl
| (Second n)::tl -> (First e)::(Second n)::tl
)
[]
[1; 2; 3; 4; 5; 6; ]
let () =
List.iter
(
fun e ->
match e with
| First f ->
print_endline(string_of_int f)
| Second (f, s) ->
Printf.printf "(%d, %d)" f s
)
(List.rev ans)
Just to make my answer all there...
type 'a node = One of 'a | Two of ('a * 'a)
let ans =
(List.map
(
fun e ->
match e with
| One _ -> failwith "Should only be Two's"
| Two (f, s) -> (f, s)
)
(List.filter
(
fun e ->
match e with
| One _ -> false
| Two _ -> true
)
(List.rev
(List.fold_left
(
fun a e ->
match a with
| [] -> (One e)::[]
| (One o)::tl -> (Two (o, e))::tl
| (Two t)::tl -> (One e)::(Two t)::tl
)
[]
(List.init 10 (fun x -> x + 1))
)
)
)
)
let () =
List.iter
(fun (f, s) -> Printf.printf "(%d, %d) " f s)
ans
I’m trying to create a function that takes an int list as an argument and returns the sum of the product between an int and its position in the list. To put in an example this : multSum [5; 11; 15] should return (5 * 1 + 11 * 2 + 15 * 3) = 72.
It should be written recursively and I’m trying while avoiding List.map or List.filter or any other prefabricated functions.
By dividing and reigning the query above, I have so far started by trying the following :
let rec tir f acc l =
match l with
|[] -> acc
|h::t -> tir f (f acc h) t ;;
val tir : ('a -> 'b -> 'a) -> 'a -> 'b list -> 'a = <fun>
then I moved to this :
let rec carto f a b =
match (a,b) with
|([],[])->([])
|(h1::t1,h2::t2)->(f h1 h2):: (carto f t1 t2)
|_->invalid_arg "carto";;
val carto : ('a -> 'b -> 'c) -> 'a list -> 'b list -> 'c list = <fun>
with the final idea to be able to do that :
let prod arg1 arg2 =
tir (+) 1 (carto ( * ) arg1 arg2);;
val prod : int list -> int list -> int = <fun>
But I am stuck now and I’m not sure of my orientation from here forward. I thought of trying to search for the index in a "l" and replace each index int in the acc, in order to make it work but I'm afraid I'm rather complicating things... Any help please ?
Edit 1 :
let rec multSum l =
let rec indices n xs = match xs with
| [] -> []
| h::t -> n::(indices (n+1) t)in
let rec tir f acc l =
match l with
|[] -> acc
|h::t -> tir f (f acc h) t in
let rec carto f a b =
match (a,b) with
|([],[])->([])
|(h1::t1,h2::t2)->(f h1 h2):: (carto f t1 t2)
|_->invalid_arg "carto" in
let prod arg1 arg2 =
tir (+) 0 (carto ( * ) arg1 arg2) in
prod l (indices 1 l);;
val multSum : int list -> int = <fun>
Building on your replies, surely these are 'fold' and 'map' rewritten. At least, I'm sure now that I was on the right track. I have come to put together the whole code as signaled above in Edit 1.
It seems to be working well... I know that I want a recursive function and here it is. But, do you think it could be done even shorter recursively of course?
#coredump is quite right about this looking like an ideal scenario for a fold, but the extra functions aren't really that necessary. We can just use a tuple to pass the index and sum information around, then when we're done, discard the index information from the tuple.
let sum_list_prod lst =
let (_, result) = List.fold_left
(fun (i, sum) x -> (i + 1, sum + i * x))
(1, 0)
lst
in
result
Edit: A simple implementation of a left fold to demonstrate the recursion going on here.
let rec foldl f init lst =
match lst with
| [] -> init
| first :: rest -> foldl f (f init first) rest
So working through a simple example with sum_list_prod:
sum_list_prod [2; 3; 4]
Calls the fold like so:
List.fold_left (fun (i, sum) x -> (i + 1, sum + i * x)) (1, 0) [2; 3; 4]
And as that evaluates:
List.fold_left (fun (i, sum) x -> (i + 1, sum + i * x)) (1, 0) [2; 3; 4]
List.fold_left (fun (i, sum) x -> (i + 1, sum + i * x)) (2, 2) [3; 4]
List.fold_left (fun (i, sum) x -> (i + 1, sum + i * x)) (3, 8) [4]
List.fold_left (fun (i, sum) x -> (i + 1, sum + i * x)) (4, 20) []
(4, 20)
And then we throw away the 4 because we don't need it anymore and are just left with 20.
Your tir functions looks like a fold; in fact has the exact same type as List.fold_left:
# List.fold_left;;
- : ('a -> 'b -> 'a) -> 'a -> 'b list -> 'a = <fun>
In the following snippets the prod function looks like a map2
# List.map2;;
- : ('a -> 'b -> 'c) -> 'a list -> 'b list -> 'c list = <fun>
You can use a fold and a map to compute the function you want, but you also need first to build a list of indices from the list of values. You could do this as follows:
let rec indices n xs = match xs with
| [] -> []
| h::t -> n::(indices (n+1) t);;
For example:
# indices 1 [5;1;3];;
- : int list = [1; 2; 3]
This is not recursive terminal, if you first computed the length of the list, how would you build the list in a recursive terminal way?
Then you should be able to call prod on a list xs and on a secondary list indices 1 xs. It is a bit wasteful because you need to build an auxiliary list, but it looks quite simple to me to understand, higher-order functions like map or fold do work on whole lists so there are fewer corner cases to consider.
But, it might be better to first write a direct recursive function for your particular problem before going the more abstract route.
The direct recursive function also requires no additional memory allocation. If you write a recursive terminal function you'll carry additional accumulator values:
the current position in the list, initially 1
the current sum of products, initially 0
Then, your function has the following skeleton:
let rec f xs index product = match xs with
| [] -> ...
| h::t -> ...
You can wrap it in a main function g:
let g xs = f xs 1 0;;
I am attempting to create a function in OCaml that gives the "k-average" of consecutive elements in a list. For example:
average 4 [1; 2; 3; 4; 5; 6] = [2; 3; 4]
since the average of 1, 2, 3, 4 is 2, of 2, 3, 4, 5 is 3, and of 3, 4, 5, 6 is 4.
I have created a function that averages the list, but with every 2 elements:
let rec average2 xs = match xs with
| [] -> []
| x :: [] -> [x]
| x :: x' :: xs -> if xs = [] then [(x + x') / 2] else [(x + x') / 2] #
(average2 (x'::xs))
How can I modify this to allow me to average k-elements?
What you should do is just verify that the list has the proper length and then two recursive functions will do it easily :
let average n l =
if List.length l < n then failwith "List is too small"
else
(* this function computes one k-average and returns the result *)
let rec aux2 acc i = function
| hd :: tl when i < n -> aux2 (acc + hd) (i + 1) tl
| _ -> acc / n
in
let rec aux acc l = match l with
(* the resulting list is reversed *)
| [] -> List.rev acc
| _ :: tl ->
(* Get the k-average of the k first elements of the list *)
let avgn = aux2 0 0 l in
(* if the rest of the list is too small, we reached the
end for sure, end *)
if List.length tl < n then List.rev (avgn :: acc)
(* recursive call on the rest of the list (without the head) *)
else aux (avgn :: acc) tl
in aux [] l
Example: split [1;3;2;4;7;9];;
Output: ([1;3;7;9], [2;4])
I'm new to F# and I can't figure it out.
Can't use the partition built in function.
This is what I have so far:
let rec split xs =
match xs with
| [] -> [], []
| xs -> xs, []
| xh::xt -> let odds, evens = split xt
if (xh % 2) = 0 then xh::odds, xh::evens
else xh::odds, evens
Fixed code:
let rec split xs =
match xs with
| [] -> [], []
| xh::xt -> let odds, evens = split xt
if (xh % 2) = 0 then odds, xh::evens
else xh::odds, evens
*Thanks to #TheInnerLight for pointing out my errors: unreachable case and unnecessarily modifying odds
You can use the built-in List.partition function
let splitOddEven xs =
xs |> List.partition (fun x -> x % 2 <> 0)
splitOddEven [1;3;2;4;7;9];;
val it : int list * int list = ([1; 3; 7; 9], [2; 4])
If you want a recursive implementation, I'd probably go for a tail recursive implementation like this:
let splitOddEven xs =
let rec splitOddEvenRec oddAcc evenAcc xs =
match xs with
| [] -> oddAcc, evenAcc
| xh::xt ->
if (xh % 2) = 0 then splitOddEvenRec oddAcc (xh :: evenAcc) xt
else splitOddEvenRec (xh :: oddAcc) evenAcc xt
splitOddEvenRec [] [] xs
splitOddEven [1;3;2;4;7;9]
Note that this will give you the two resulting lists in reverse order so you might wish to reverse them yourself.
I have this function:
let encode list =
let rec aux count acc = function
| [] -> [] (* Caso a lista esteja vazia*)
| [x] -> (count+1, x) :: acc
| a :: (b :: _ as t) ->
if a = b then aux (count + 1) acc t
else aux 0 ((count+1,a) :: acc) t in
List.rev (aux 0 [] list)
;;
and with this input:
let test = encode ["a";"a";"a";"a";"b";"f";"f";"c";"c";"a";"a";"d";"e";"e";"e";"e"];;
And i have this Output:
val test : (int * string) list =
[(4, "a"); (1, "b"); (2, "f"); (2, "c"); (2, "a"); (1, "d"); (4, "e")]
but the "a" is a repeated and "f" need to be at final!
I need a output like:
val test : (int * string) list =
[(6, "a"); (1, "b"); (2, "c"); (1, "d"); (4, "e"); (2, "f")]
Can anybody help, please?! Thanks!
You are counting repeated adjacent values, so-called run-length encoding. It appears you want to count occurrences across the whole input. You can either sort the input beforehand, or you can use a more complicated data structure (such as a Map) to keep track of your counts.
Something like this:
let encode xs =
let f acc x =
let n = try M.find x acc with Not_found -> 0 in
M.add x (n+1) acc in
let ans = (List.fold_left f M.empty) xs in
M.bindings ans ;;
# encode ["a";"a";"a";"a";"b";"f";"f";"c";"c";"a";"a";"d";"e";"e";"e";"e"];;
- : (M.key * int) list =
[("a", 6); ("b", 1); ("c", 2); ("d", 1); ("e", 4); ("f", 2)]