I've been trying to implement a function that takes a list of integers and then return a list of lists of integers which are non-decreasing.
i.e
let ls = [ 1;2;3;5;6;3;2;5;6;2]
I should get [[1;2;3;5;6];[3];[2;5;6];[2]]
How should i approach this ? i'm a total noob at functional programming.
I can think of the steps needed:
1. Start a new sublist, compare each element with the one next to it. if it is greater then add to list. if not, start a new list and so on.
From what I've learned so far from the book Functional Programming with f# ( which i just started a few days ago), I could possibly use pattern matching and a recursive function maybe to go through the list comparing two elements
something like this :
let rec nonDecreasing list =
match list with
| (x,y) :: xs when x <= y ->
how would I go about to create the sublists using pattern matching ?
or have i approached the question wrongly?
Since there's already a solution using fold, here's another answer using foldBack, so you don't have to reverse it. Now you can backout a pure recursive solution.
let splitByInc x lls = // x is an item from the list, lls is a list of lists
match lls with
| y::xs -> // split the list of lists into head and tail
match y with
| h::_ when x <= h -> (x::y)::xs // take the head, and compare it with x, then cons it together with the rest
| _ -> [x]::lls // in the other case cons the single item with the rest of the list of lists
| _ -> [[x]] // nothing else to do, return the whole thing
let ls = [ 1;2;3;5;6;3;2;5;6;3]
List.foldBack splitByInc ls [] //foldBack needs a folder function, a list and a starting state
Edit:
Here's a really simplified example, you could write a recursive sum and compare it with the fold version:
let sumList x y =
x + y
List.foldBack sumList ls 0 //36
To better understand what splitByInc does, try it out with these examples:
splitByInc 4 [[5;6;7]] // matches (x::y)::xs
splitByInc 4 [] // matches [[x]]
splitByInc 4 [[1;2;3]] // matches [x]::lls
That's basically the same answer as the one given by #s952163 but maybe more readable by removing the nested match and also more general by adding a comparison function to do the "packing".
let packWhile predicate list =
let folder item = function
| [] -> [[ item ]]
| (subHead :: _ as subList) :: accTail
when predicate item subHead -> (item :: subList) :: accTail
| accList -> [ item ] :: accList
List.foldBack folder list []
// usage (you can replace (<=) by (fun x y -> x <= y) if it's clearer for you)
packWhile (<=) [1;2;3;5;6;3;2;5;6;3]
// you can also define a function to bake-in the comparison
let packIncreasing list = packWhile (<=) list
packIncreasing [1;2;3;5;6;3;2;5;6;3]
I'd use a fold, where your 'State is a tuple containing the previous value, the list of lists, and the current non-decreasing list you're working on.
let ls = [ 1;2;3;5;6;3;2;5;6;3]
let _, listOfLists, currList =
((Int32.MinValue, [], []), ls) ||>
List.fold(fun (prev, listOfLists, currList) t ->
if t < prev then //decreasing, so store your currList and start a new one
t, currList::listOfLists, [t]
else //just add t to your currList
t, listOfLists, t::currList)
let listOfLists = currList::listOfLists //cleanup: append final sublist
let final = List.rev(List.map List.rev listOfLists) //cleanup: reverse everything
printfn "%A" final
Note you'll have to clean up, adding the final list to the list-of-lists, and then reversing the full list-of-lists and each sublist once you've done the fold.
Related
I have recently started learning f# and I have a problem with a task like the one in the subject line. I managed to solve this task but not using a recursive function. I have tried to convert my function to a recursive function but it does not work because in the function I create arrays which elements I then change. Please advise me how to convert my function to a recursive function or how else to perform this task.
let list = [8;4;3;3;5;9;-7]
let comp (a,b) = if a>b then a elif b = a then a else b
let maks (b: _ list) =
let x = b.Length
if x % 2 = 0 then
let tab = Array.create ((x/2)) 0
for i = 0 to (x/2)-1 do
tab.[i] <- (comp(b.Item(2*i),b.Item(2*i+1)))
let newlist = tab |> Array.toList
newlist
else
let tab = Array.create (((x-1)/2)+1) 0
tab.[(((x-1)/2))] <- b.Item(x-1)
for i = 0 to ((x-1)/2)-1 do
tab.[i] <- (comp(b.Item(2*i),b.Item(2*i+1)))
let newlist = tab |> Array.toList
newlist
It is worth noting that, if you were doing this not for learning purposes, there is a nice way of doing this using the chunkBySize function:
list
|> List.chunkBySize 2
|> List.map (fun l -> comp(l.[0], l.[l.Length-1]))
This splits the list into chunks of size at most 2. For each chunk, you can then compare the first element with the last element and that is the result you wanted.
If this is a homework question, I don't want to give away the answer, so consider this pseudocode solution instead:
If the list contains at least two elements:
Answer a new list consisting of:
The greater of the first two elements, followed by
Recursively applying the function to the rest of the list
Else the list contains less than two elements:
Answer the list unchanged
Hint: F#'s pattern matching ability makes this easy to implement.
Thanks to your guidance I managed to create the following function:
let rec maks2 (b: _ list,newlist: _ list,i:int) =
let x = b.Length
if x >= 2 then
if x % 2 = 0 then
if i < ((x/2)-1)+1 then
let d = (porownaj(b.Item(2*i),b.Item(2*i+1)))
let list2 = d::newlist
maks2(b,list2,i+1)
else
newlist
else
if i < ((x/2)-1)+1 then
let d = (porownaj(b.Item(2*i),b.Item(2*i+1)))
let list2 = d::newlist
maks2(b,list2,i+1)
else
let list3 = b.Item(x-1)::newlist
list3
else
b
The function works correctly, it takes as arguments list, empty list and index.
The only problem is that the returned list is reversed, i.e. values that should be at the end are at the beginning. How to add items to the end of the list?
You can use pattern matching to match and check/extract lists in one step.A typical recursive function, would look like:
let rec adjGreater xs =
match xs with
| [] -> []
| [x] -> [x]
| x::y::rest -> (if x >= y then x else y) :: adjGreater rest
It checks wether the list is empty, has one element, or has two elements and the remaining list in rest.
Then it builds a new list by either using x or y as the first element, and then compute the result of the remaing rest recursivly.
This is not tail-recursive. A tail-call optimized version would be, that instead of using the result of the recursive call. You would create a new list, and pass the computed valuke so far, to the recursive function. Usually this way, you want to create a inner recursive loop function.
As you only can add values to the top of a list, you then need to reverse the result of the recursive function like this:
let adjGreater xs =
let rec loop xs result =
match xs with
| [] -> result
| [x] -> x :: result
| x::y::rest -> loop rest ((if x >= y then x else y) :: result)
List.rev (loop xs [])
I'm trying to write a tail-recursion function that will look at a list of distinct words, a list of all words, and return a list with the count of occurrences of each word. I'm actually reading the words out of files in a directory, but I can't seem to get the tail-recursion to compile. This is what I have so far:
let countOccurence (word:string) list =
List.filter (fun x -> x.Equals(word)) list
//(all words being a list of all words across several files)
let distinctWords = allWords |> Seq.distinct
let rec wordCloud distinct (all:string list) acc =
match distinct with
| head :: tail -> wordCloud distinct tail Array.append(acc, (countOccurence head all)) //<- What am I doing with my life?
| [] -> 0
I realize this is probably a fairly straightforward question, but I've been banging my head for an hour on this final piece of the puzzle. Any thoughts?
There are several issues with the statement as given:
Use of Array.append to manipulate lists
Typos
Incorrect use of whitespace to group things together
Try expressing the logic as a series of steps instead of putting everything into a single, unreadable line of code. Here's what I did to understand the problems with the above expression:
let rec wordCloud distinct (all:string list) acc =
match distinct with
| head :: tail ->
let count = countOccurence head all
let acc' = acc |> List.append count
wordCloud distinct tail acc'
| [] -> 0
This compiles, but I don't know if it does what you want it to do...
Notice the replacement of Array.append with List.append.
This is still tail recursive, since the call to wordCloud sits in the tail position.
After several hours more work, I came up with this:
let countOccurance (word:string) list =
let count = List.filter (fun x -> word.Equals(x)) list
(word, count.Length)
let distinctWords = allWords |> Seq.distinct |> Seq.toList
let print (tup:string*int) =
match tup with
| (a,b) -> printfn "%A: %A" a b
let rec wordCloud distinct (all:string list) (acc:(string*int) list) =
match distinct with
| [] -> acc
| head :: tail ->
let accumSoFar = acc # [(countOccurance head all)]
wordCloud tail all accumSoFar
let acc = []
let cloud = (wordCloud distinctWords allWords acc)
let rec printTup (tupList:(string*int) list) =
match tupList with
| [] -> 0
| head :: tail ->
printfn "%A" head
printTup tail
printTup cloud
This problem actually has a pretty straightforward solution, if you take a step back and simply type in what you want to do.
/// When you want to make a tag cloud...
let makeTagCloud (words: string list) =
// ...take a list of all words...
words
// ...then walk along the list...
|> List.fold (fun cloud word ->
// ...and check if you've seen that word...
match cloud |> Map.tryFind word with
// ...if you have, bump the count...
| Some count -> cloud |> Map.add word (count+1)
// ...if not, add it to the map...
| None -> cloud |> Map.add word 1) Map.empty
// ...and change the map back into a list when you are done.
|> Map.toList
Reads like poetry ;)
I'm trying to solve tasks from 99 Haskell problems in F#.
The task #7 looks pretty simple, and the solution can be found in lots of places. Except the fact that the first several solutions that I've tried and found by googling (e.g. https://github.com/paks/99-FSharp-Problems/blob/master/P01to10/Solutions.fs) are wrong.
My example is pretty simple.
I'm trying to build extremely deep nested structure and fold it
type NestedList<'a> =
| Elem of 'a
| NestedList of NestedList<'a> list
let flatten list =
//
(* StackOverflowException
| Elem(a) as i -> [a]
| NestedList(nest) -> nest |> Seq.map myFlatten |> List.concat
*)
// Both are failed with stackoverflowexception too https://github.com/paks/99-FSharp-Problems/blob/master/P01to10/Solutions.fs
let insideGen count =
let rec insideGen' count agg =
match count with
| 0 -> agg
| _ ->
insideGen' (count-1) (NestedList([Elem(count); agg]))
insideGen' count (Elem(-1))
let z = insideGen 50000
let res = flatten z
I've tried to rewrite solution in CPS style, but eiter I'm doing something wrong or look into incorrect direction - everything that I've tried isn't working.
Any advices?
p.s. Haskell solution, at least on nested structure with 50000 nested levels is working slowly, but without stack overflow.
Here's a CPS version that doesn't overflow using your test.
let flatten lst =
let rec loop k = function
| [] -> k []
| (Elem x)::tl -> loop (fun ys -> k (x::ys)) tl
| (NestedList xs)::tl -> loop (fun ys -> loop (fun zs -> k (zs # ys)) xs) tl
loop id [lst]
EDIT
A much more readable way to write this would be:
let flatten lst =
let results = ResizeArray()
let rec loop = function
| [] -> ()
| h::tl ->
match h with
| Elem x -> results.Add(x)
| NestedList xs -> loop xs
loop tl
loop [lst]
List.ofSeq results
Disclaimer - I'm not a deep F# programmer and this will not be idiomatic.
If your stack is overflowing, it means that you don't have a tail recursive solution. It also means that you are choosing to use stack memory for state. Traditionally, you want to exchange heap memory for stack memory since heap memory is in comparatively large supply. So the trick is to model a stack.
I'm going to define a virtual machine that is a stack. Each stack element will be a state nugget for traversing a list which will include the list and a program counter, which is the current element to examine and will be a tuple of a NestedList<'a> list * int. The list is the current list being traversed. The int is the current position in the list.
type NestedList<'a> =
| Elem of 'a
| Nested of NestedList<'a> list
let flatten l =
let rec listMachine instructions result =
match instructions with
| [] -> result
| (currList, currPC) :: tail ->
if currPC >= List.length currList then listMachine tail result
else
match List.nth currList currPC with
| Elem(a) -> listMachine ((currList, currPC + 1 ) :: tail) (result # [ a ])
| Nested(l) -> listMachine ((l, 0) :: (currList, currPC + 1) :: instructions.Tail) result
match l with
| Elem(a) -> [ a ]
| Nested(ll) -> listMachine [ (ll, 0) ] []
What have I done? I've written a tail-recursive function that operates of "Little Lisper" style code - if my instruction list is empty, return my accumulated result. If not, operate on the top of the stack. I bind a convenience variable to the top and if the PC is at the end, I recurse on the tail of the stack (pop) with the current result. Otherwise, I look at the current element in the list. If it's an Elem, I recurse, advancing the PC and appending the Elem onto the list. If it's not an elem, I recurse, by pushing a new stack with the NestedList followed by the current stack elem with the PC advanced by 1 and everything else.
Define the function max2 that takes two integers as arguments and returns the largest of them.
I did this: let max2 x y = if x < y then y else x this I belive is correct
Then define the function max_list that returns the largest of the elements in a nonempty list of integers by calling max2. For the empty list, it should abort with an error message ( raising an exception)
I did this: let list = [3;4] let max_list = if list.IsEmpty then 0 else max2 list.Item(0) list.Item(1) but this wont work if the list is more then two elements. I dont want to use any object-orientated stuff. What is the correct answer?
The correct answer is that you should read about recursion with lists.
F# list is built up gradually using empty list [] and cons (::) constructor. For example,
[3; 4] is a syntactic sugar for 3::4::[]. We often use pattern matching on lists in writing recursive functions.
Here is a recursive function following your requirements closely:
let rec max_list xs =
match xs with
// The function aborts with an error message on empty lists
| [] -> invalidArg "xs" "Empty list"
// Return immediately on a singleton list
| [x] -> x
// xs has at least two elements, call max_list
// on the bigger element of the first two ones and the rest of the list
| x1::x2::xs' -> max_list((max2 x1 x2)::xs')
On a side note, there is a built-in generic max function which also works on integers.
A simple recursive solution:
let max2 x y = if x < y then y else x
let max_list list =
let rec loop hi list =
match list with
| h::t -> let hi = max2 h hi
loop hi t
| [] -> hi
match list with
| h::t -> loop h t
| [] -> invalidArg "list" "Empty list"
Test in FSI:
> max_list [3;4;5;1;2;9;0];;
val it : int = 9
For each element in the list, compare it to the previous highest ('hi'). Pass the new highest and the rest of the list into the loop function, until the input list is empty. Then just return 'hi'.
I've trying to learn F#. I'm a complete beginner, so this might be a walkover for you guys :)
I have the following function:
let removeEven l =
let n = List.length l;
let list_ = [];
let seq_ = seq { for x in 1..n do if x % 2 <> 0 then yield List.nth l (x-1)}
for x in seq_ do
let list_ = list_ # [x];
list_;
It takes a list, and return a new list containing all the numbers, which is placed at an odd index in the original list, so removeEven [x1;x2;x3] = [x1;x3]
However, I get my already favourite error-message: Incomplete construct at or before this point in expression...
If I add a print to the end of the line, instead of list_:
...
print_any list_;
the problem is fixed. But I do not want to print the list, I want to return it!
What causes this? Why can't I return my list?
To answer your question first, the compiler complains because there is a problem inside the for loop. In F#, let serves to declare values (that are immutable and cannot be changed later in the program). It isn't a statement as in C# - let can be only used as part of another expression. For example:
let n = 10
n + n
Actually means that you want the n symbol to refer to the value 10 in the expression n + n. The problem with your code is that you're using let without any expression (probably because you want to use mutable variables):
for x in seq_ do
let list_ = list_ # [x] // This isn't assignment!
list_
The problematic line is an incomplete expression - using let in this way isn't allowed, because it doesn't contain any expression (the list_ value will not be accessed from any code). You can use mutable variable to correct your code:
let mutable list_ = [] // declared as 'mutable'
let seq_ = seq { for x in 1..n do if x % 2 <> 0 then yield List.nth l (x-1)}
for x in seq_ do
list_ <- list_ # [x] // assignment using '<-'
Now, this should work, but it isn't really functional, because you're using imperative mutation. Moreover, appending elements using # is really inefficient thing to do in functional languages. So, if you want to make your code functional, you'll probably need to use different approach. Both of the other answers show a great approach, although I prefer the example by Joel, because indexing into a list (in the solution by Chaos) also isn't very functional (there is no pointer arithmetic, so it will be also slower).
Probably the most classical functional solution would be to use the List.fold function, which aggregates all elements of the list into a single result, walking from the left to the right:
[1;2;3;4;5]
|> List.fold (fun (flag, res) el ->
if flag then (not flag, el::res) else (not flag, res)) (true, [])
|> snd |> List.rev
Here, the state used during the aggregation is a Boolean flag specifying whether to include the next element (during each step, we flip the flag by returning not flag). The second element is the list aggregated so far (we add element by el::res only when the flag is set. After fold returns, we use snd to get the second element of the tuple (the aggregated list) and reverse it using List.rev, because it was collected in the reversed order (this is more efficient than appending to the end using res#[el]).
Edit: If I understand your requirements correctly, here's a version of your function done functional rather than imperative style, that removes elements with odd indexes.
let removeEven list =
list
|> Seq.mapi (fun i x -> (i, x))
|> Seq.filter (fun (i, x) -> i % 2 = 0)
|> Seq.map snd
|> List.ofSeq
> removeEven ['a'; 'b'; 'c'; 'd'];;
val it : char list = ['a'; 'c']
I think this is what you are looking for.
let removeEven list =
let maxIndex = (List.length list) - 1;
seq { for i in 0..2..maxIndex -> list.[i] }
|> Seq.toList
Tests
val removeEven : 'a list -> 'a list
> removeEven [1;2;3;4;5;6];;
val it : int list = [1; 3; 5]
> removeEven [1;2;3;4;5];;
val it : int list = [1; 3; 5]
> removeEven [1;2;3;4];;
val it : int list = [1; 3]
> removeEven [1;2;3];;
val it : int list = [1; 3]
> removeEven [1;2];;
val it : int list = [1]
> removeEven [1];;
val it : int list = [1]
You can try a pattern-matching approach. I haven't used F# in a while and I can't test things right now, but it would be something like this:
let rec curse sofar ls =
match ls with
| even :: odd :: tl -> curse (even :: sofar) tl
| even :: [] -> curse (even :: sofar) []
| [] -> List.rev sofar
curse [] [ 1; 2; 3; 4; 5 ]
This recursively picks off the even elements. I think. I would probably use Joel Mueller's approach though. I don't remember if there is an index-based filter function, but that would probably be the ideal to use, or to make if it doesn't exist in the libraries.
But in general lists aren't really meant as index-type things. That's what arrays are for. If you consider what kind of algorithm would require a list having its even elements removed, maybe it's possible that in the steps prior to this requirement, the elements can be paired up in tuples, like this:
[ (1,2); (3,4) ]
That would make it trivial to get the even-"indexed" elements out:
thelist |> List.map fst // take first element from each tuple
There's a variety of options if the input list isn't guaranteed to have an even number of elements.
Yet another alternative, which (by my reckoning) is slightly slower than Joel's, but it's shorter :)
let removeEven list =
list
|> Seq.mapi (fun i x -> (i, x))
|> Seq.choose (fun (i,x) -> if i % 2 = 0 then Some(x) else None)
|> List.ofSeq