I have two snippets of code that tries to convert a float list to a Vector3 or Vector2 list. The idea is to take 2/3 elements at a time from the list and combine them as a vector. The end result is a sequence of vectors.
let rec vec3Seq floatList =
seq {
match floatList with
| x::y::z::tail -> yield Vector3(x,y,z)
yield! vec3Seq tail
| [] -> ()
| _ -> failwith "float array not multiple of 3?"
}
let rec vec2Seq floatList =
seq {
match floatList with
| x::y::tail -> yield Vector2(x,y)
yield! vec2Seq tail
| [] -> ()
| _ -> failwith "float array not multiple of 2?"
}
The code looks very similiar and yet there seems to be no way to extract a common portion. Any ideas?
Here's one approach. I'm not sure how much simpler this really is, but it does abstract some of the repeated logic out.
let rec mkSeq (|P|_|) x =
seq {
match x with
| P(p,tail) ->
yield p
yield! mkSeq (|P|_|) tail
| [] -> ()
| _ -> failwith "List length mismatch" }
let vec3Seq =
mkSeq (function
| x::y::z::tail -> Some(Vector3(x,y,z), tail)
| _ -> None)
As Rex commented, if you want this only for two cases, then you probably won't have any problem if you leave the code as it is. However, if you want to extract a common pattern, then you can write a function that splits a list into sub-list of a specified length (2 or 3 or any other number). Once you do that, you'll only use map to turn each list of the specified length into Vector.
The function for splitting list isn't available in the F# library (as far as I can tell), so you'll have to implement it yourself. It can be done roughly like this:
let divideList n list =
// 'acc' - accumulates the resulting sub-lists (reversed order)
// 'tmp' - stores values of the current sub-list (reversed order)
// 'c' - the length of 'tmp' so far
// 'list' - the remaining elements to process
let rec divideListAux acc tmp c list =
match list with
| x::xs when c = n - 1 ->
// we're adding last element to 'tmp',
// so we reverse it and add it to accumulator
divideListAux ((List.rev (x::tmp))::acc) [] 0 xs
| x::xs ->
// add one more value to 'tmp'
divideListAux acc (x::tmp) (c+1) xs
| [] when c = 0 -> List.rev acc // no more elements and empty 'tmp'
| _ -> failwithf "not multiple of %d" n // non-empty 'tmp'
divideListAux [] [] 0 list
Now, you can use this function to implement your two conversions like this:
seq { for [x; y] in floatList |> divideList 2 -> Vector2(x,y) }
seq { for [x; y; z] in floatList |> divideList 3 -> Vector3(x,y,z) }
This will give a warning, because we're using an incomplete pattern that expects that the returned lists will be of length 2 or 3 respectively, but that's correct expectation, so the code will work fine. I'm also using a brief version of sequence expression the -> does the same thing as do yield, but it can be used only in simple cases like this one.
This is simular to kvb's solution but doesn't use a partial active pattern.
let rec listToSeq convert (list:list<_>) =
seq {
if not(List.isEmpty list) then
let list, vec = convert list
yield vec
yield! listToSeq convert list
}
let vec2Seq = listToSeq (function
| x::y::tail -> tail, Vector2(x,y)
| _ -> failwith "float array not multiple of 2?")
let vec3Seq = listToSeq (function
| x::y::z::tail -> tail, Vector3(x,y,z)
| _ -> failwith "float array not multiple of 3?")
Honestly, what you have is pretty much as good as it can get, although you might be able to make a little more compact using this:
// take 3 [1 .. 5] returns ([1; 2; 3], [4; 5])
let rec take count l =
match count, l with
| 0, xs -> [], xs
| n, x::xs -> let res, xs' = take (count - 1) xs in x::res, xs'
| n, [] -> failwith "Index out of range"
// split 3 [1 .. 6] returns [[1;2;3]; [4;5;6]]
let rec split count l =
seq { match take count l with
| xs, ys -> yield xs; if ys <> [] then yield! split count ys }
let vec3Seq l = split 3 l |> Seq.map (fun [x;y;z] -> Vector3(x, y, z))
let vec2Seq l = split 2 l |> Seq.map (fun [x;y] -> Vector2(x, y))
Now the process of breaking up your lists is moved into its own generic "take" and "split" functions, its much easier to map it to your desired type.
Related
I have a pretty trivial task but I can't figure out how to make the solution prettier.
The goal is taking a List and returning results, based on whether they passed a predicate. The results should be grouped. Here's a simplified example:
Predicate: isEven
Inp : [2; 4; 3; 7; 6; 10; 4; 5]
Out: [[^^^^]......[^^^^^^^^]..]
Here's the code I have so far:
let f p ls =
List.foldBack
(fun el (xs, ys) -> if p el then (el::xs, ys) else ([], xs::ys))
ls ([], [])
|> List.Cons // (1)
|> List.filter (not << List.isEmpty) // (2)
let even x = x % 2 = 0
let ret =
[2; 4; 3; 7; 6; 10; 4; 5]
|> f even
// expected [[2; 4]; [6; 10; 4]]
This code does not seem to be readable that much. Also, I don't like lines (1) and (2). Is there any better solution?
Here is my take. you need a few helper functions first:
// active pattern to choose between even and odd intengers
let (|Even|Odd|) x = if (x % 2) = 0 then Even x else Odd x
// fold function to generate a state tupple of current values and accumulated values
let folder (current, result) x =
match x, current with
| Even x, _ -> x::current, result // even members a added to current list
| Odd x, [] -> current, result // odd members are ignored when current is empty
| Odd x, _ -> [], current::result // odd members starts a new current
// test on data
[2; 4; 3; 7; 6; 10; 4; 5]
|> List.rev // reverse list since numbers are added to start of current
|> List.fold folder ([], []) // perform fold over list
|> function | [],x -> x | y,x -> y::x // check that current is List.empty, otherwise add to result
How about this one?
let folder p l = function
| h::t when p(l) -> (l::h)::t
| []::_ as a -> a
| _ as a -> []::a
let f p ls =
ls
|> List.rev
|> List.fold (fun a l -> folder p l a) [[]]
|> List.filter ((<>) [])
At least the folder is crystal clear and effective, but then you pay the price for this by list reversing.
Here is a recursive solution based on a recursive List.filter
let rec _f p ls =
match ls with
|h::t -> if p(h) then
match f p t with
|rh::rt -> (h::rh)::rt
|[] -> (h::[])::[]
else []::f p t
|[] -> [[]]
let f p ls = _f p ls |> List.filter (fun t -> t <> [])
Having to filter at the end does seem inelegant though.
Here you go. This function should also have fairly good performance.
let groupedFilter (predicate : 'T -> bool) (list : 'T list) =
(([], []), list)
||> List.fold (fun (currentGroup, finishedGroups) el ->
if predicate el then
(el :: currentGroup), finishedGroups
else
match currentGroup with
| [] ->
[], finishedGroups
| _ ->
// This is the first non-matching element
// following a matching element.
// Finish processing the previous group then
// add it to the finished groups list.
[], ((List.rev currentGroup) :: finishedGroups))
// Need to do a little clean-up after the fold.
|> fun (currentGroup, finishedGroups) ->
// If the current group is non-empty, finish it
// and add it to the list of finished groups.
let finishedGroups =
match currentGroup with
| [] -> finishedGroups
| _ ->
(List.rev currentGroup) :: finishedGroups
// Reverse the finished groups list so the grouped
// elements will be in their original order.
List.rev finishedGroups;;
With the list reversing, I would like to go to #seq instead of list.
This version uses mutation (gasp!) internally for efficiency, but may also be a little slower with the overhead of seq. I think it is quite readable though.
let f p (ls) = seq {
let l = System.Collections.Generic.List<'a>()
for el in ls do
if p el then
l.Add el
else
if l.Count > 0 then yield l |> List.ofSeq
l.Clear()
if l.Count > 0 then yield l |> List.ofSeq
}
I can't think of a way to do this elegantly using higher order functions, but here's a solution using a list comprehension. I think it's fairly straightforward to read.
let f p ls =
let rec loop xs =
[ match xs with
| [] -> ()
| x::xs when p x ->
let group, rest = collectGroup [x] xs
yield group
yield! loop rest
| _::xs -> yield! loop xs ]
and collectGroup acc = function
| x::xs when p x -> collectGroup (x::acc) xs
| xs -> List.rev acc, xs
loop ls
I've found this question on hubFS, but that handles a splitting criteria based on individual elements. I'd like to split based on a comparison of adjacent elements, so the type would look like this:
val split = ('T -> 'T -> bool) -> 'T list -> 'T list list
Currently, I am trying to start from Don's imperative solution, but I can't work out how to initialize and use a 'prev' value for comparison. Is fold a better way to go?
//Don's solution for single criteria, copied from hubFS
let SequencesStartingWith n (s:seq<_>) =
seq { use ie = s.GetEnumerator()
let acc = new ResizeArray<_>()
while ie.MoveNext() do
let x = ie.Current
if x = n && acc.Count > 0 then
yield ResizeArray.to_list acc
acc.Clear()
acc.Add x
if acc.Count > 0 then
yield ResizeArray.to_list acc }
This is an interesting problem! I needed to implement exactly this in C# just recently for my article about grouping (because the type signature of the function is pretty similar to groupBy, so it can be used in LINQ query as the group by clause). The C# implementation was quite ugly though.
Anyway, there must be a way to express this function using some simple primitives. It just seems that the F# library doesn't provide any functions that fit for this purpose. I was able to come up with two functions that seem to be generally useful and can be combined together to solve this problem, so here they are:
// Splits a list into two lists using the specified function
// The list is split between two elements for which 'f' returns 'true'
let splitAt f list =
let rec splitAtAux acc list =
match list with
| x::y::ys when f x y -> List.rev (x::acc), y::ys
| x::xs -> splitAtAux (x::acc) xs
| [] -> (List.rev acc), []
splitAtAux [] list
val splitAt : ('a -> 'a -> bool) -> 'a list -> 'a list * 'a list
This is similar to what we want to achieve, but it splits the list only in two pieces (which is a simpler case than splitting the list multiple times). Then we'll need to repeat this operation, which can be done using this function:
// Repeatedly uses 'f' to take several elements of the input list and
// aggregate them into value of type 'b until the remaining list
// (second value returned by 'f') is empty
let foldUntilEmpty f list =
let rec foldUntilEmptyAux acc list =
match f list with
| l, [] -> l::acc |> List.rev
| l, rest -> foldUntilEmptyAux (l::acc) rest
foldUntilEmptyAux [] list
val foldUntilEmpty : ('a list -> 'b * 'a list) -> 'a list -> 'b list
Now we can repeatedly apply splitAt (with some predicate specified as the first argument) on the input list using foldUntilEmpty, which gives us the function we wanted:
let splitAtEvery f list = foldUntilEmpty (splitAt f) list
splitAtEvery (<>) [ 1; 1; 1; 2; 2; 3; 3; 3; 3 ];;
val it : int list list = [[1; 1; 1]; [2; 2]; [3; 3; 3; 3]]
I think that the last step is really nice :-). The first two functions are quite straightforward and may be useful for other things, although they are not as general as functions from the F# core library.
How about:
let splitOn test lst =
List.foldBack (fun el lst ->
match lst with
| [] -> [[el]]
| (x::xs)::ys when not (test el x) -> (el::(x::xs))::ys
| _ -> [el]::lst
) lst []
the foldBack removes the need to reverse the list.
Having thought about this a bit further, I've come up with this solution. I'm not sure that it's very readable (except for me who wrote it).
UPDATE Building on the better matching example in Tomas's answer, here's an improved version which removes the 'code smell' (see edits for previous version), and is slightly more readable (says me).
It still breaks on this (splitOn (<>) []), because of the dreaded value restriction error, but I think that might be inevitable.
(EDIT: Corrected bug spotted by Johan Kullbom, now works correctly for [1;1;2;3]. The problem was eating two elements directly in the first match, this meant I missed a comparison/check.)
//Function for splitting list into list of lists based on comparison of adjacent elements
let splitOn test lst =
let rec loop lst inner outer = //inner=current sublist, outer=list of sublists
match lst with
| x::y::ys when test x y -> loop (y::ys) [] (List.rev (x::inner) :: outer)
| x::xs -> loop xs (x::inner) outer
| _ -> List.rev ((List.rev inner) :: outer)
loop lst [] []
splitOn (fun a b -> b - a > 1) [1]
> val it : [[1]]
splitOn (fun a b -> b - a > 1) [1;3]
> val it : [[1]; [3]]
splitOn (fun a b -> b - a > 1) [1;2;3;4;6;7;8;9;11;12;13;14;15;16;18;19;21]
> val it : [[1; 2; 3; 4]; [6; 7; 8; 9]; [11; 12; 13; 14; 15; 16]; [18; 19]; [21]]
Any thoughts on this, or the partial solution in my question?
"adjacent" immediately makes me think of Seq.pairwise.
let splitAt pred xs =
if Seq.isEmpty xs then
[]
else
xs
|> Seq.pairwise
|> Seq.fold (fun (curr :: rest as lists) (i, j) -> if pred i j then [j] :: lists else (j :: curr) :: rest) [[Seq.head xs]]
|> List.rev
|> List.map List.rev
Example:
[1;1;2;3;3;3;2;1;2;2]
|> splitAt (>)
Gives:
[[1; 1; 2; 3; 3; 3]; [2]; [1; 2; 2]]
I would prefer using List.fold over explicit recursion.
let splitOn pred = function
| [] -> []
| hd :: tl ->
let (outer, inner, _) =
List.fold (fun (outer, inner, prev) curr ->
if pred prev curr
then (List.rev inner) :: outer, [curr], curr
else outer, curr :: inner, curr)
([], [hd], hd)
tl
List.rev ((List.rev inner) :: outer)
I like answers provided by #Joh and #Johan as these solutions seem to be most idiomatic and straightforward. I also like an idea suggested by #Shooton. However, each solution had their own drawbacks.
I was trying to avoid:
Reversing lists
Unsplitting and joining back the temporary results
Complex match instructions
Even Seq.pairwise appeared to be redundant
Checking list for emptiness can be removed in cost of using Unchecked.defaultof<_> below
Here's my version:
let splitWhen f src =
if List.isEmpty src then [] else
src
|> List.foldBack
(fun el (prev, current, rest) ->
if f el prev
then el , [el] , current :: rest
else el , el :: current , rest
)
<| (List.head src, [], []) // Initial value does not matter, dislike using Unchecked.defaultof<_>
|> fun (_, current, rest) -> current :: rest // Merge temporary lists
|> List.filter (not << List.isEmpty) // Drop tail element
(I am aware of this question, but it relates to sequences, which is not my problem here)
Given this input (for example):
let testlist =
[
"*text1";
"*text2";
"text3";
"text4";
"*text5";
"*text6";
"*text7"
]
let pred (s:string) = s.StartsWith("*")
I would like to be able to call MyFunc pred testlist and get this output:
[
["*text1";"*text2"];
["*text5";"*text6";"*text7"]
]
This is my current solution, but I don't really like the nested List.revs (ignore the fact that it takes Seq as input)
let shunt pred sq =
let shunter (prevpick, acc) (pick, a) =
match pick, prevpick with
| (true, true) -> (true, (a :: (List.hd acc)) :: (List.tl acc))
| (false, _) -> (false, acc)
| (true, _) -> (true, [a] :: acc)
sq
|> Seq.map (fun a -> (pred a, a))
|> Seq.fold shunter (false, [])
|> snd
|> List.map List.rev
|> List.rev
there is a List.partition function in the F# core library (in case you wanted to implement this just to have it working and not to learn how to write recursive functions yourself). Using this function, you can write this:
> testlist |> List.partition (fun s -> s.StartsWith("*"))
val it : string list * string list =
(["*text1"; "*text2"; "*text5"; "*text6"; "*text7"], ["text3"; "text4"])
Note that this function returns a tuple instead of returning a list of lists. This is a bit different to what you wanted, but if the predicate returns just true or false, then this makes more sense.
The implementation of partition function that returns tuples is also a bit simpler, so it may be useful for learning purposes:
let partition pred list =
// Helper function, which keeps results collected so
// far in 'accumulator' arguments outTrue and outFalse
let rec partitionAux list outTrue outFalse =
match list with
| [] ->
// We need to reverse the results (as we collected
// them in the opposite order!)
List.rev outTrue, List.rev outFalse
// Append element to one of the lists, depending on 'pred'
| x::xs when pred x -> partitionAux xs (x::outTrue) outFalse
| x::xs -> partitionAux xs outTrue (x::outFalse)
// Run the helper function
partitionAux list [] []
Edit: rev-less version using foldBack added below.
Here's some code that uses lists and tail-recursion:
//divides a list L into chunks for which all elements match pred
let divide pred L =
let rec aux buf acc L =
match L,buf with
//no more input and an empty buffer -> return acc
| [],[] -> List.rev acc
//no more input and a non-empty buffer -> return acc + rest of buffer
| [],buf -> List.rev (List.rev buf :: acc)
//found something that matches pred: put it in the buffer and go to next in list
| h::t,buf when pred h -> aux (h::buf) acc t
//found something that doesn't match pred. Continue but don't add an empty buffer to acc
| h::t,[] -> aux [] acc t
//found input that doesn't match pred. Add buffer to acc and continue with an empty buffer
| h::t,buf -> aux [] (List.rev buf :: acc) t
aux [] [] L
usage:
> divide pred testlist;;
val it : string list list =
[["*text1"; "*text2"]; ["*text5"; "*text6"; "*text7"]]
Using a list as data structure for a buffer means that it always needs to be reversed when outputting the contents. This may not be a problem if individual chunks are modestly sized. If speed/efficiency becomes an issue, you could use a Queue<'a> or a `List<'a>' for the buffers, for which appending is fast. But using these data structures instead of lists also means that you lose the powerful list pattern matching. In my opinion, being able to pattern match lists outweighs the presence of a few List.rev calls.
Here's a streaming version that outputs the result one block at a time. This avoids the List.rev on the accumulator in the previous example:
let dividestream pred L =
let rec aux buf L =
seq { match L, buf with
| [],[] -> ()
| [],buf -> yield List.rev buf
| h::t,buf when pred h -> yield! aux (h::buf) t
| h::t,[] -> yield! aux [] t
| h::t,buf -> yield List.rev buf
yield! aux [] t }
aux [] L
This streaming version avoids the List.rev on the accumulator. Using List.foldBack can be used to avoid reversing the accumulated chunks as well.
update: here's a version using foldBack
//divides a list L into chunks for which all elements match pred
let divide2 pred L =
let f x (acc,buf) =
match pred x,buf with
| true,buf -> (acc,x::buf)
| false,[] -> (acc,[])
| false,buf -> (buf::acc,[])
let rest,remainingBuffer = List.foldBack f L ([],[])
match remainingBuffer with
| [] -> rest
| buf -> buf :: rest
Just reverse the list once up front, and then build the structure in order easily:
let Shunt p l =
let mutable r = List.rev l
let mutable result = []
while not r.IsEmpty do
let mutable thisBatch = []
while not r.IsEmpty && not(p r.Head) do
r <- r.Tail
while not r.IsEmpty && p r.Head do
thisBatch <- r.Head :: thisBatch
r <- r.Tail
if not thisBatch.IsEmpty then
result <- thisBatch :: result
result
The outer while deals with each 'batch', and the first inner while skips over any that don't match the predicate, followed by another while that grabs all those that do and stores them in the current batch. If there was anything in this batch (the final one may be empty), prepend it to the final result.
This is an example where I think locally imperative code is simply superior to a purely functional counterpart. The code above is so easy to write and to reason about.
Another version of shunt:
let shunt pred lst =
let rec tWhile pred lst =
match lst with
| [] -> [], []
| hd :: tl when pred hd -> let taken, rest = tWhile pred tl
(hd :: taken), rest
| lst -> [], lst
let rec collect = function
| [] -> []
| lst -> let taken, rest = tWhile pred lst
taken :: (collect (snd (tWhile (fun x -> not (pred x)) rest)))
collect lst
This one avoids List.rev but it's not tail recursive - so only suitable for small lists.
yet another one...
let partition pred lst =
let rec trec xs cont =
match xs with
| [] -> ([],[]) |> cont
| h::t when pred h -> (fun (y,n) -> h::y,n) >> cont |> trec t
| h::t -> (fun (y,n) -> y,h::n) >> cont |> trec t
trec lst id
then we can define shunt:
let shunt pred lst = lst |> partition pred |> (fun (x,y) -> [x;y])
I need to generate permutations on a given list. I managed to do it like this
let rec Permute (final, arr) =
if List.length arr > 0 then
for x in arr do
let n_final = final # [x]
let rest = arr |> List.filter (fun a -> not (x = a))
Permute (n_final, rest)
else
printfn "%A" final
let DoPermute lst =
Permute ([], lst)
DoPermute lst
There are obvious issues with this code. For example, list elements must be unique. Also, this is more-less a same approach that I would use when generating straight forward implementation in any other language. Is there any better way to implement this in F#.
Thanks!
Here's the solution I gave in my book F# for Scientists (page 166-167):
let rec distribute e = function
| [] -> [[e]]
| x::xs' as xs -> (e::xs)::[for xs in distribute e xs' -> x::xs]
let rec permute = function
| [] -> [[]]
| e::xs -> List.collect (distribute e) (permute xs)
For permutations of small lists, I use the following code:
let distrib e L =
let rec aux pre post =
seq {
match post with
| [] -> yield (L # [e])
| h::t -> yield (List.rev pre # [e] # post)
yield! aux (h::pre) t
}
aux [] L
let rec perms = function
| [] -> Seq.singleton []
| h::t -> Seq.collect (distrib h) (perms t)
It works as follows: the function "distrib" distributes a given element over all positions in a list, example:
distrib 10 [1;2;3] --> [[10;1;2;3];[1;10;2;3];[1;2;10;3];[1;2;3;10]]
The function perms works (recursively) as follows: distribute the head of the list over all permutations of its tail.
The distrib function will get slow for large lists, because it uses the # operator a lot, but for lists of reasonable length (<=10), the code above works fine.
One warning: if your list contains duplicates, the result will contain identical permutations. For example:
perms [1;1;3] = [[1;1;3]; [1;1;3]; [1;3;1]; [1;3;1]; [3;1;1]; [3;1;1]]
The nice thing about this code is that it returns a sequence of permutations, instead of generating them all at once.
Of course, generating permutations with an imperative array-based algorithm will be (much) faster, but this algorithm has served me well in most cases.
Here's another sequence-based version, hopefully more readable than the voted answer.
This version is similar to Jon's version in terms of logic, but uses computation expressions instead of lists. The first function computes all ways to insert an element x in a list l. The second function computes permutations.
You should be able to use this on larger lists (e.g. for brute force searches on all permutations of a set of inputs).
let rec inserts x l =
seq { match l with
| [] -> yield [x]
| y::rest ->
yield x::l
for i in inserts x rest do
yield y::i
}
let rec permutations l =
seq { match l with
| [] -> yield []
| x::rest ->
for p in permutations rest do
yield! inserts x p
}
It depends on what you mean by "better". I'd consider this to be slightly more elegant, but that may be a matter of taste:
(* get the list of possible heads + remaining elements *)
let rec splitList = function
| [x] -> [x,[]]
| x::xs -> (x, xs) :: List.map (fun (y,l) -> y,x::l) (splitList xs)
let rec permutations = function
| [] -> [[]]
| l ->
splitList l
|> List.collect (fun (x,rest) ->
(* permute remaining elements, then prepend head *)
permutations rest |> List.map (fun l -> x::l))
This can handle lists with duplicate elements, though it will result in duplicated permutations.
In the spirit of Cyrl's suggestion, here's a sequence comprehension version
let rec permsOf xs =
match xs with
| [] -> List.toSeq([[]])
| _ -> seq{ for x in xs do
for xs' in permsOf (remove x xs) do
yield (x::xs')}
where remove is a simple function that removes a given element from a list
let rec remove x xs =
match xs with [] -> [] | (x'::xs')-> if x=x' then xs' else x'::(remove x xs')
IMHO the best solution should alleviate the fact that F# is a functional language so imho the solution should be as close to the definition of what we mean as permutation there as possible.
So the permutation is such an instance of list of things where the head of the list is somehow added to the permutation of the rest of the input list.
The erlang solution shows that in a pretty way:
permutations([]) -> [[]];
permutations(L) -> [[H|T] H<- L, T <- permutations( L--[H] ) ].
taken fron the "programming erlang" book
There is a list comprehension operator used, in solution mentioned here by the fellow stackoverflowers there is a helper function which does the similar job
basically I'd vote for the solution without any visible loops etc, just pure function definition
I'm like 11 years late, but still in case anyone needs permutations like I did recently. Here's Array version of permutation func, I believe it's more performant:
[<RequireQualifiedAccess>]
module Array =
let private swap (arr: _[]) i j =
let buf = arr.[i]
arr.[i] <- arr.[j]
arr.[j] <- buf
let permutations arr =
match arr with
| null | [||] -> [||]
| arr ->
let last = arr.Length - 1
let arr = Array.copy arr
let rec perm arr k =
let arr = Array.copy arr
[|
if k = last then
yield arr
else
for i in k .. last do
swap arr k i
yield! perm arr (k + 1)
|]
perm arr 0
Slowly getting the hang of List matching and tail recursion, I needed a function which 'stitches' a list of lists together leaving off intermediate values (easier to show than explain):
merge [[1;2;3];[3;4;5];[5;6;7]] //-> [1;2;3;4;5;6;7]
The code for the List.merge function looks like this:
///Like concat, but removes first value of each inner list except the first one
let merge lst =
let rec loop acc lst =
match lst with
| [] -> acc
| h::t ->
match acc with
| [] -> loop (acc # h) t
| _ -> loop (acc # (List.tl h)) t //first time omit first value
loop [] lst
(OK, it's not quite like concat, because it only handles two levels of list)
Question: How to do this for a Seq of Seqs (without using a mutable flag)?
UPDATE (re comment from Juliet):
My code creates 'paths' composed of 'segments' which are based on an option type:
type SegmentDef = Straight of float | Curve of float * float
let Project sampleinterval segdefs = //('clever' code here)
When I do a List.map (Project 1.) ListOfSegmentDefs, I get back a list where each segment begins on the same point where the previous segment ends. I want to join these lists together to get a Path, keeping only the 'top/tail' of each overlap - but I don't need to do a 'Set', because I know that I don't have any other duplicates.
This is essentially the same as your first solution, but a little more succinct:
let flatten l =
seq {
yield Seq.hd (Seq.hd l) (* first item of first list *)
for a in l do yield! (Seq.skip 1 a) (* other items *)
}
[Edit to add]:
If you need a List version of this code, use append |> Seq.to_list at the end of your method:
let flatten l =
seq {
yield Seq.hd (Seq.hd l) (* first item of first list *)
for a in l do yield! (Seq.skip 1 a) (* other items *)
} |> Seq.to_list
let merge = function
| [] -> []
| xs::xss -> xs # [for _::xs in xss do yield! xs]
or:
let merge = function
| [] -> []
| xs::xss -> xs # List.collect List.tail xss