Im trying to prepare myself for an upcoming re-exam on functional programming. One of the questions I received completelty stumped me.
We are to write a function that takes a sorted list and returns the median value. We can assume the length of the list is not 0 and that it is odd. Ok now this is pretty easy, except we are not allowed to make an explicit second pass over the list to determine its length. No problem I can just use the .length function, except we are also not allowed to use any built-in functions.
This was what I did although I did use the .length function:
let median list =
let rec helper list count =
match list with
|(x::_) when count=0 -> x
|(x::xs) -> helper xs (count-1)
helper list (list.Length/2)
How would I go about getting the length in the required way in order to work out the median?
One way would be to create a function for finding the second half of a list:
let secondHalf l =
let rec aux xs ctr =
match ctr with
| _::_::ctr' ->
match xs with
| _::xs' -> aux xs' ctr'
| [] -> failwith "should never happen!"
| _ -> xs
aux l l
you can then use this to find the median:
let sortedMedian xs = let (m::_) = secondHalf xs in m
Related
Good Morning everyone,
I must do an exercise of Programming, but i'm stuck!
Well, the exercise requires a function that given a list not empty of integers, return the first number with maximum number of occurrences.
For example:
mode [1;2;5;1;2;3;4;5;5;4:5;5] ==> 5
mode [2;1;2;1;1;2] ==> 2
mode [-1;2;1;2;5;-1;5;5;2] ==> 2
mode [7] ==> 7
Important: the exercise must be in functional programming
My idea is:
let rec occurences_counter xs i = match xs with
|[] -> failwith "Error"
|x :: xs when x = i -> 1 + occurences_counter xs i
|x :: xs -> occurences_counter xs i;;
In this function i'm stuck:
let rec mode (l : int list) : int = match l with
|[] -> failwith "Error"
|[x] -> x
|x::y::l when occurences_counter l x >= occurences_counter l y -> x :: mode l
|x::y::l when occurences_counter l y > occurences_counter l x -> y :: mode l;;
Thanks in advance, i'm newbie in programming and in stackoverflow
Sorry for my english
one solution : calculate first a list of couples (number , occurences).
hint : use List.assoc.
Then, loop over that list of couple to find the max occurrence and then return the number.
One suggestion:
your algorithm could be simplified if you sort the list before. This has O(N log(N)) complexity. Then measure the longest sequence of identical numbers.
This is a good strategy because you delegate the hard part of the work to a well known algorithm.
It is probably not the most beautiful code, but here is with what i came up (F#). At first i transform every element to an intermediate format. This format contains the element itself, the position of it occurrence and the amount it occurred.
type T<'a> = {
Element: 'a
Position: int
Occurred: int
}
The idea is that those Records can be added. So you can first transform every element, and then add them together. So a list like
[1;3]
will be first transformed to
[{Element=1;Position=0;Occurred=1}; {Element=3;Position=1;Occurred=1}]
By adding two together you only can add those with the same "Element". The Position with the lower number from both is taken, and Occurred is just added together. So if you for example have
{Element=3;Position=1;Occurred=2} {Element=3;Position=3;Occurred=2}
the result will be
{Element=3;Position=1;Occurred=4}
The idea that i had in mind was a Monoid. But in a real Monoid you had to come up that you also could add different Elements together. By trying some stuff out i feel that the restriction of just adding the same Element where way more easier. I created a small Module with the type. Including some helper functions for creating, adding and comparing.
module Occurred =
type T<'a> = {
Element: 'a
Position: int
Occurred: int
}
let create x pos occ = {Element=x; Position=pos; Occurred=occ}
let sameElements x y = x.Element = y.Element
let add x y =
if not <| sameElements x y then failwith "Cannot add two different Occurred"
create x.Element (min x.Position y.Position) (x.Occurred + y.Occurred)
let compareOccurredPosition x y =
let occ = compare x.Occurred y.Occurred
let pos = compare x.Position y.Position
match occ,pos with
| 0,x -> x * -1
| x,_ -> x
With this setup i now wrote two additional function. One aggregate function that first turns every element into a Occurred.T, group them by x.Element (the result is a list of list). And then it uses List.reduce on the inner list to add the Occurred with the same Element together. The result is a List that Contains only a single Occurred.T for every Element with the first Position and the amount of Occurred items.
let aggregate =
List.mapi (fun i x -> Occurred.create x i 1)
>> List.groupBy (fun occ -> occ.Element)
>> List.map (fun (x,occ) -> List.reduce Occurred.add occ)
You could use that aggregate function to now implement different aggregation logic. In your case you only wanted the one with the highest Occurrences and the lowest position. I wrote another function that did that.
let firstMostOccurred =
List.sortWith (fun x y -> (Occurred.compareOccurredPosition x y) * -1) >> List.head >> (fun x -> x.Element)
One note. Occurred.compareOccurredPosition is written that it sorts everything in ascending order. I think people expecting it in this order to go to the smallest to the biggest element by default. So by default the first element would be the element with the lowest occurrence and the biggest Position. By multiplying the result of it with -1 you turn that function into a descending sorting function. The reason why i did that is that i could use List.head. I also could use List.last to get the last element, but i felt that it would be better not to go through the whole list again just to get the last element. On top of it, you didn't wanted an Occurred.T you wanted the element itself, so i unwrap the Element to get the number.
Here is everything in action
let ll = [
[1;2;5;1;2;3;4;5;5;4;5;5]
[2;1;2;1;1;2]
[-1;2;1;2;5;-1;5;5;2]
[7]
]
ll
|> List.map aggregate
|> List.map firstMostOccurred
|> List.iter (printfn "%d")
This code will now print
5
2
2
7
It has still some rough edges like
Occurred.add throws an exception if you try to add Occurred with different Elements
List.head throws an exception for empty lists
And in both cases no code is written to handle those cases or making sure an exception will not raise.
You need to process you input list while maintaining a state, that stores the number of occurrences of each number. Basically, the state can be a map, where keys are in the domain of list elements, and values are in domain of natural numbers. If you will use Map the algorithm would be of O(NlogN) complexity. You can also use associative list (i.e., a list of type ('key,'value) list) to implement map. This will lead to quadratic complexity. Another approach is to use hash table or an array of the length equal to the size of the input domain. Both will give you a linear complexity.
After you collected the statistics, (i.e., a mapping from element to the number of its occurrences) you need to go through the set of winners, and choose the one, that was first on the list.
In OCaml the solution would look like this:
open Core_kernel.Std
let mode xs : int =
List.fold xs ~init:Int.Map.empty ~f:(fun stat x ->
Map.change stat x (function
| None -> Some 1
| Some n -> Some (n+1))) |>
Map.fold ~init:Int.Map.empty ~f:(fun ~key:x ~data:n modes ->
Map.add_multi modes ~key:n ~data:x) |>
Map.max_elt |> function
| None -> invalid_arg "mode: empty list"
| Some (_,ms) -> List.find_exn xs ~f:(List.mem ms)
The algorithm is the following:
Run through input and compute frequency of each element
Run through statistics and compute spectrum (i.e., a mapping from frequency to elements).
Get the set of elements that has the highest frequency, and find an element in the input list, that is in this set.
For example, if we take sample [1;2;5;1;2;3;4;5;5;4;5;5],
stats = {1 => 2; 2 => 2; 3 => 1; 4 => 2; 5 => 5}
mods = {1 => [3]; 2 => [1;2]; 5 => [5]}
You need to install core library to play with it. Use coretop to play with this function in the toplevel. Or corebuild to compile it, like this:
corebuild test.byte --
if the source code is stored in test.ml
I'm curious as to how the list module/type works in F#, specifically does it optimise this?
let xs = ["1"; "2"; "3"]
let ys = "0"::xs
let zs = ["hello"; "world"]#xs
I've looked over some of the source https://github.com/fsharp/fsharp/blob/68e37d03dfc15f8105aeb0ac70b846f82b364901/src/fsharp/FSharp.Core/prim-types.fs#L3493 seems to be the relevant area.
I would like to know if xs is copied when making ys.
I would have thought it's easy to just point to the existing list if you just cons element.
If you are concatenating I imagine it might be impossible as it would require mutating the last element of the list to point to the next one?
If someone could annotate/paste snippets of code from FSharp.Core that would be ideal.
So the implementation of List is a little odd. It is actually implemented as a discriminated union. From the spec:
type 'T list =
| ([])
| (::) of 'T * 'T list
So you can think of :: as a function that takes two arguments and creates a tuple (which is fast as it is independent of the list size).
# is much more complicated. Here is the implementation:
let (#) l1 l2 =
match l1 with
| [] -> l2
| (h::t) ->
match l2 with
| [] -> l1
| _ ->
let res = [h]
let lastCons = PrivateListHelpers.appendToFreshConsTail res t
PrivateListHelpers.setFreshConsTail lastCons l2;
res
The two weird functions basically mutate the list in place. appendToFreshConsTail copies the list and returns the last element. setFreshConsTail then changes the last element so that its next element is set to l2 rather than [] joining the lists.
I want to make a function that takes an integer list as argument and compares every value and returns the largest value. In C# I would simply iterate through every value in the list, save the largest to a variable and return it, I'm hoping F# works similarly but the syntax is kinda iffy for me, here's what my code looks like. Also max2 is a function that compares 2 values and returns the largest.
let max_list list =
let a = 0 : int
match list with
| head :: tail -> (for i in list do a = max2 i a) a
| [] -> failwith "sry";;
You could use mutable variable and write the code using for loop, just like in C#. However, if you're doing this to learn F# and functional concepts, then it's good idea to use recursion.
In this case, recursive function is a bit longer, but it demonstrates the key concepts including pattern matching - so learning the tricks is something that will be useful when writing more complicated F# code.
The key idea is to write a function that takes the largest value found so far and calls itself recursively until it reaches the end of the list.
let max_list list =
// Inner recursive function that takes the largest value found so far
// and a list to be processed (if it is empty, it returns 'maxSoFar')
let rec loop maxSoFar list =
match list with
// If the head value is greater than what we found so far, use it as new greater
| head::tail when head > maxSoFar -> loop head tail
// If the head is smaller, use the previous maxSoFar value
| _::tail -> loop maxSoFar tail
// At the end, just return the largest value found so far
| [] -> maxSoFar
// Start with head as the greatest and tail as the rest to be processed
// (fails for empty list - but you could match here to give better error)
loop (List.head list) (List.tail list)
As a final note, this will be slow because it uses generic comparison (via an interface). You can make the function faster using let inline max_list list = (...). That way, the code will use native comparison instruction when used with primitive types like int (this is really a special case - the problem only really happens with generic comparison)
Also know that you can write a nice one-liner using reduce:
let max_list list = List.reduce (fun max x -> if x > max then x else max)
If your intention is to be able to find the maximum value of items in a list where the value of the items is found by the function max2 then this approach works:
let findMax list =
list
|> List.map (fun i -> i, max2 i)
|> List.maxBy snd
|> fst
I was reading this post While or Tail Recursion in F#, what to use when? were several people say that the 'functional way' of doing things is by using maps/folds and higher order functions instead of recursing and looping.
I have this function that returns the item at position x in a list:
let rec getPos l c = if c = 0 then List.head l else getPos (List.tail l) (c - 1)
how can it be converted to be more functional?
This is a primitive list function (also known as List.nth).
It is okay to use recursion, especially when creating the basic building blocks. Although it would be nicer with pattern matching instead of if-else, like this:
let rec getPos l c =
match l with
| h::_ when c = 0 -> h
| _::t -> getPos t (c-1)
| [] -> failwith "list too short"
It is possible to express this function with List.fold, however the result is less clear than the recursive version.
I'm not sure what you mean by more functional.
Are you rolling this yourself as a learning exercise?
If not, you could just try this:
> let mylist = [1;2;3;4];;
> let n = 2;;
> mylist.[n];;
Your definition is already pretty functional since it uses a tail-recursive function instead of an imperative loop construct. However, it also looks like something a Scheme programmer might have written because you're using head and tail.
I suspect you're really asking how to write it in a more idiomatic ML style. The answer is to use pattern matching:
let rec getPos list n =
match list with
| hd::tl ->
if n = 0 then hd
else getPos tl (n - 1)
| [] -> failWith "Index out of range."
The recursion on the structure of the list is now revealed in the code. You also get a warning if the pattern matching is non-exhaustive so you're forced to deal with the index too big error.
You're right that functional programming also encourages the use of combinators like map or fold (so called points-free style). But too much of it just leads to unreadable code. I don't think it's warranted in this case.
Of course, Benjol is right, in practice you would just write mylist.[n].
If you'd like to use high-order functions for this, you could do:
let nth n = Seq.take (n+1) >> Seq.fold (fun _ x -> Some x) None
let nth n = Seq.take (n+1) >> Seq.reduce (fun _ x -> x)
But the idea is really to have basic constructions and combine them build whatever you want. Getting the nth element of a sequence is clearly a basic block that you should use. If you want the nth item, as Benjol mentioned, do myList.[n].
For building basic constructions, there's nothing wrong to use recursion or mutable loops (and often, you have to do it this way).
Not as a practical solution, but as an exercise, here is one of the ways to express nth via foldr or, in F# terms, List.foldBack:
let myNth n xs =
let step e f = function |0 -> e |n -> f (n-1)
let error _ = failwith "List is too short"
List.foldBack step xs error n
The :: operator in F# always prepends elements to the list. Is there an operator that appends to the list? I'm guessing that using # operator
[1; 2; 3] # [4]
would be less efficient, than appending one element.
As others said, there is no such operator, because it wouldn't make much sense. I actually think that this is a good thing, because it makes it easier to realize that the operation will not be efficient. In practice, you shouldn't need the operator - there is usually a better way to write the same thing.
Typical scenario: I think that the typical scenario where you could think that you need to append elements to the end is so common that it may be useful to describe it.
Adding elements to the end seems necessary when you're writing a tail-recursive version of a function using the accumulator parameter. For example a (inefficient) implementation of filter function for lists would look like this:
let filter f l =
let rec filterUtil acc l =
match l with
| [] -> acc
| x::xs when f x -> filterUtil (acc # [x]) xs
| x::xs -> filterUtil acc xs
filterUtil [] l
In each step, we need to append one element to the accumulator (which stores elements to be returned as the result). This code can be easily modified to use the :: operator instead of appending elements to the end of the acc list:
let filter f l =
let rec filterUtil acc l =
match l with
| [] -> List.rev acc // (1)
| x::xs when f x -> filterUtil (x::acc) xs // (2)
| x::xs -> filterUtil acc xs
filterUtil [] l
In (2), we're now adding elements to the front of the accumulator and when the function is about to return the result, we reverse the list (1), which is a lot more efficient than appending elements one by one.
Lists in F# are singly-linked and immutable. This means consing onto the front is O(1) (create an element and have it point to an existing list), whereas snocing onto the back is O(N) (as the entire list must be replicated; you can't change the existing final pointer, you must create a whole new list).
If you do need to "append one element to the back", then e.g.
l # [42]
is the way to do it, but this is a code smell.
The cost of appending two standard lists is proportional to the length of the list on the left. In particular, the cost of
xs # [x]
is proportional to the length of xs—it is not a constant cost.
If you want a list-like abstraction with a constant-time append, you can use John Hughes's function representation, which I'll call hlist. I'll try to use OCaml syntax, which I hope is close enough to F#:
type 'a hlist = 'a list -> 'a list (* a John Hughes list *)
let empty : 'a hlist = let id xs = xs in id
let append xs ys = fun tail -> xs (ys tail)
let singleton x = fun tail -> x :: tail
let cons x xs = append (singleton x) xs
let snoc xs x = append xs (singleton x)
let to_list : 'a hlist -> 'a list = fun xs -> xs []
The idea is that you represent a list functionally as a function from "the rest of the elements" to "the final list". This works great if you are going to build up the whole list before you look at any of the elements. Otherwise you'll have to deal with the linear cost of append or use another data structure entirely.
I'm guessing that using # operator [...] would be less efficient, than appending one element.
If it is, it will be a negligible difference. Both appending a single item and concatenating a list to the end are O(n) operations. As a matter of fact I can't think of a single thing that # has to do, which a single-item append function wouldn't.
Maybe you want to use another data structure. We have double-ended queues (or short "Deques") in fsharpx. You can read more about them at http://jackfoxy.com/double-ended-queues-for-fsharp
The efficiency (or lack of) comes from iterating through the list to find the final element. So declaring a new list with [4] is going to be negligible for all but the most trivial scenarios.
Try using a double-ended queue instead of list. I recently added 4 versions of deques (Okasaki's spelling) to FSharpx.Core (Available through NuGet. Source code at FSharpx.Core.Datastructures). See my article about using dequeus Double-ended queues for F#
I've suggested to the F# team the cons operator, ::, and the active pattern discriminator be made available for other data structures with a head/tail signature.3