I have a sorted sequence and want to go through it and return the unique entries in the sequence. I can do it using the following function, but it uses reference variables and I don't think it's the correct way of solving the problem.
let takeFirstCell sectors =
let currentRNCId = ref -1
let currentCellId = ref -1
seq {
for sector in sectors do
if sector.RNCId <> !currentRNCId || sector.CellId <> !currentCellId then
currentRNCId := sector.RNCId
currentCellId := sector.CellId
yield sector
}
How can I do this in a functional way?
[1;1;1;2;2;2;3;3;3]
|> Seq.distinctBy id
|> printfn "%A"
Seq.distinct (1::[1..5]) returns seq [1; 2; 3; 4; 5]. Is that what you meant?
distinct and distinctBy both use Dictionary and therefore require hashing and a bit of memory for storing unique items. If your sequence is already sorted, you can use the following approach (similar to yours). It's nearly twice as fast and has constant memory use, making it usable for sequences of any size.
let distinctWithoutHash (items:seq<_>) =
seq {
use e = items.GetEnumerator()
if e.MoveNext() then
let prev = ref e.Current
yield !prev
while e.MoveNext() do
if e.Current <> !prev then
yield e.Current
prev := e.Current
}
let items = Seq.init 1000000 (fun i -> i / 2)
let test f = items |> f |> (Seq.length >> printfn "%d")
test Seq.distinct //Real: 00:00:01.038, CPU: 00:00:01.435, GC gen0: 47, gen1: 1, gen2: 1
test distinctWithoutHash //Real: 00:00:00.622, CPU: 00:00:00.624, GC gen0: 44, gen1: 0, gen2: 0
I couldn't figure out a way to use mutables instead of refs (short of hand-coding an enumerator), which I'm sure would speed it up considerably (I tried it--it makes no difference).
Just initialize a unique collection (like a set) with the sequence like this:
set [1; 2; 3; 3; 4; 5; 5];;
=> val it : Set<int> = set [1; 2; 3; 4; 5]
In my case I could not use Seq.distinct because I needed to preserve order of list elements.
I used solution from http://ocaml.org/learn/tutorials/99problems.html.
I think it is quite short
let rec compress = function
| a :: (b :: _ as t) -> if a = b then compress t else a :: compress t
| smaller -> smaller
The solution below, preserves order of elements and returns only the first occurance of an element in a generic list. Of course this generates a new List with the redundant items removed.
// **** Returns a list having subsequent redundant elements removed
let removeDuplicates(lst : 'a list) =
let f item acc =
match acc with
| [] -> [item]
| _ ->
match List.exists(fun x -> x = item) acc with
| false -> item :: acc
| true -> acc
lst
|> List.rev
|> fun x -> List.foldBack f x []
|> List.rev
// **** END OF FUNCTION removeDuplicates
val removeDuplicates : 'a list -> 'a list when 'a : equality
val testList : int list = [1; 4; 3; 1; 2; 2; 1; 1; 3; 4; 3]
val tryAbove : int list = [1; 4; 3; 2]
Related
Can anyone help me with this problem?
"Realize a function which duplicate each item in a list. You can use List.map"
IN F# sharp language.
And also
"Use the List.init function to generate a list of n random natural numbers between 0 and m."
let makeCopy elem Count =
match Count with
| 0 -> []
| 1 -> elem
let rec dupeElem row count =
match row with
| [] -> []
| hd::tl -> (makeCopy hd count) # dupeElem tl count
//let xs = [1; 2; 3]
//xs |> List.collect (fun x -> List.replicate 3 x)
//val it : int list = [1; 1; 2; 2; 3; 3]
Duplicating the items is pretty straight forward, you just need a recursive function that walks the list.
Generating the random numbers is where you can use List.init to create a new list. You can use the .NET Random class to generate the random numbers you're after.
This gives up the following functions:
let rec duplicateItems list =
match list with
| [] -> []
| head :: tail -> head :: head :: duplicateItems tail
let makeRandomList count upperBound =
let random = Random()
List.init count (fun i -> random.Next(0, upperBound))
You can now generate a random list and pipe it into the duplicate function:
let numbers = makeRandomList 10 20 |> duplicateItems
NOTE: duplicateItems is not tail recursive, so for really large lists this might be an issue. You can get around this by treating the data to duplicate as a sequence:
let duplicateSequence sequence =
seq {
for a in sequence do
yield a
yield a
}
Now we just need to pipe the result into Seq.toList:
let numbers = makeRandomList 10 20 |> duplicateSequence |> Seq.toList
We could also have written makeRandom to return a sequence rather than a list. This would have made the whole computation lazy up until the point we call Seq.toList.
which duplicate each item in a list. You can use List.map
I think your own solution with List.collect is fine. But here's one with List.map:
> let dupe x = List.map (fun s -> [s;s]) x |> List.concat
val dupe : x:'a list -> 'a list
> dupe [1;2;3];;
val it : int list = [1; 1; 2; 2; 3; 3]
Use the List.init function to generate a list of n random natural numbers between 0 and m
I'll show you the general idea, then you can work out the rest, I'm sure. This basically works:
> let rand n max = let r = Random() in List.init n (fun _ -> r.Next(0, max));;
val rand : n:int -> max:int -> int list
> rand 10 12;;
val it : int list = [11; 11; 10; 11; 6; 1; 3; 6; 8; 2]
I'm trying to learn F# by rewriting some C# algorithms I have into idiomatic F#.
One of the first functions I'm trying to rewrite is a batchesOf where:
[1..17] |> batchesOf 5
Which would split the sequence into batches with a max of five in each, i.e:
[[1; 2; 3; 4; 5]; [6; 7; 8; 9; 10]; [11; 12; 13; 14; 15]; [16; 17]]
My first attempt at doing this is kind of ugly where I've resorted to using a mutable ref object after running into errors trying to use mutable type inside the closure. Using ref is particularly unpleasant since to dereference it you have to use the ! operator which when inside a condition expression can be counter intuitive to some devs who will read it as logical not. Another problem I ran into is where Seq.skip and Seq.take are not like their Linq aliases in that they will throw an error if size exceeds the size of the sequence.
let batchesOf size (sequence: _ seq) : _ list seq =
seq {
let s = ref sequence
while not (!s |> Seq.isEmpty) do
yield !s |> Seq.truncate size |> List.ofSeq
s := System.Linq.Enumerable.Skip(!s, size)
}
Anyway what would be the most elegant/idiomatic way to rewrite this in F#? Keeping the original behaviour but preferably without the ref mutable variable.
Implementing this function using the seq<_> type idiomatically is difficult - the type is inherently mutable, so there is no simple nice functional way. Your version is quite inefficient, because it uses Skip repeatedly on the sequence. A better imperative option would be to use GetEnumerator and just iterate over elements using IEnumerator. You can find various imperative options in this snippet: http://fssnip.net/1o
If you're learning F#, then it is better to try writing the function using F# list type. This way, you can use idiomatic functional style. Then you can write batchesOf using pattern matching with recursion and accumulator argument like this:
let batchesOf size input =
// Inner function that does the actual work.
// 'input' is the remaining part of the list, 'num' is the number of elements
// in a current batch, which is stored in 'batch'. Finally, 'acc' is a list of
// batches (in a reverse order)
let rec loop input num batch acc =
match input with
| [] ->
// We've reached the end - add current batch to the list of all
// batches if it is not empty and return batch (in the right order)
if batch <> [] then (List.rev batch)::acc else acc
|> List.rev
| x::xs when num = size - 1 ->
// We've reached the end of the batch - add the last element
// and add batch to the list of batches.
loop xs 0 [] ((List.rev (x::batch))::acc)
| x::xs ->
// Take one element from the input and add it to the current batch
loop xs (num + 1) (x::batch) acc
loop input 0 [] []
As a footnote, the imperative version can be made a bit nicer using computation expression for working with IEnumerator, but that's not standard and it is quite advanced trick (for example, see http://fssnip.net/37).
A friend asked me this a while back. Here's a recycled answer. This works and is pure:
let batchesOf n =
Seq.mapi (fun i v -> i / n, v) >>
Seq.groupBy fst >>
Seq.map snd >>
Seq.map (Seq.map snd)
Or an impure version:
let batchesOf n =
let i = ref -1
Seq.groupBy (fun _ -> i := !i + 1; !i / n) >> Seq.map snd
These produce a seq<seq<'a>>. If you really must have an 'a list list as in your sample then just add ... |> Seq.map (List.ofSeq) |> List.ofSeq as in:
> [1..17] |> batchesOf 5 |> Seq.map (List.ofSeq) |> List.ofSeq;;
val it : int list list = [[1; 2; 3; 4; 5]; [6; 7; 8; 9; 10]; [11; 12; 13; 14; 15]; [16; 17]]
Hope that helps!
This can be done without recursion if you want
[0..20]
|> Seq.mapi (fun i elem -> (i/size),elem)
|> Seq.groupBy (fun (a,_) -> a)
|> Seq.map (fun (_,se) -> se |> Seq.map (snd));;
val it : seq<seq<int>> =
seq
[seq [0; 1; 2; 3; ...]; seq [5; 6; 7; 8; ...]; seq [10; 11; 12; 13; ...];
seq [15; 16; 17; 18; ...]; ...]
Depending on how you think this may be easier to understand. Tomas' solution is probably more idiomatic F# though
Hurray, we can use List.chunkBySize, Seq.chunkBySize and Array.chunkBySize in F# 4, as mentioned by Brad Collins and Scott Wlaschin.
This isn't perhaps idiomatic but it works:
let batchesOf n l =
let _, _, temp', res' = List.fold (fun (i, n, temp, res) hd ->
if i < n then
(i + 1, n, hd :: temp, res)
else
(1, i, [hd], (List.rev temp) :: res))
(0, n, [], []) l
(List.rev temp') :: res' |> List.rev
Here's a simple implementation for sequences:
let chunks size (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop i acc =
seq {
if i = size then
yield (List.rev acc)
yield! loop 0 []
elif e.MoveNext() then
yield! loop (i+1) (e.Current::acc)
else
yield (List.rev acc)
}
if size = 0 then invalidArg "size" "must be greater than zero"
if Seq.isEmpty items then Seq.empty else loop 0 []
let s = Seq.init 10 id
chunks 3 s
//output: seq [[0; 1; 2]; [3; 4; 5]; [6; 7; 8]; [9]]
My method involves converting the list to an array and recursively chunking the array:
let batchesOf (sz:int) lt =
let arr = List.toArray lt
let rec bite curr =
if (curr + sz - 1 ) >= arr.Length then
[Array.toList arr.[ curr .. (arr.Length - 1)]]
else
let curr1 = curr + sz
(Array.toList (arr.[curr .. (curr + sz - 1)])) :: (bite curr1)
bite 0
batchesOf 5 [1 .. 17]
[[1; 2; 3; 4; 5]; [6; 7; 8; 9; 10]; [11; 12; 13; 14; 15]; [16; 17]]
I found this to be a quite terse solution:
let partition n (stream:seq<_>) = seq {
let enum = stream.GetEnumerator()
let rec collect n partition =
if n = 1 || not (enum.MoveNext()) then
partition
else
collect (n-1) (partition # [enum.Current])
while enum.MoveNext() do
yield collect n [enum.Current]
}
It works on a sequence and produces a sequence. The output sequence consists of lists of n elements from the input sequence.
You can solve your task with analog of Clojure partition library function below:
let partition n step coll =
let rec split ss =
seq {
yield(ss |> Seq.truncate n)
if Seq.length(ss |> Seq.truncate (step+1)) > step then
yield! split <| (ss |> Seq.skip step)
}
split coll
Being used as partition 5 5 it will provide you with sought batchesOf 5 functionality:
[1..17] |> partition 5 5;;
val it : seq<seq<int>> =
seq
[seq [1; 2; 3; 4; ...]; seq [6; 7; 8; 9; ...]; seq [11; 12; 13; 14; ...];
seq [16; 17]]
As a premium by playing with n and step you can use it for slicing overlapping batches aka sliding windows, and even apply to infinite sequences, like below:
Seq.initInfinite(fun x -> x) |> partition 4 1;;
val it : seq<seq<int>> =
seq
[seq [0; 1; 2; 3]; seq [1; 2; 3; 4]; seq [2; 3; 4; 5]; seq [3; 4; 5; 6];
...]
Consider it as a prototype only as it does many redundant evaluations on the source sequence and not likely fit for production purposes.
This version passes all my tests I could think of including ones for lazy evaluation and single sequence evaluation:
let batchIn batchLength sequence =
let padding = seq { for i in 1 .. batchLength -> None }
let wrapped = sequence |> Seq.map Some
Seq.concat [wrapped; padding]
|> Seq.windowed batchLength
|> Seq.mapi (fun i el -> (i, el))
|> Seq.filter (fun t -> fst t % batchLength = 0)
|> Seq.map snd
|> Seq.map (Seq.choose id)
|> Seq.filter (fun el -> not (Seq.isEmpty el))
I am still quite new to F# so if I'm missing anything - please do correct me, it will be greatly appreciated.
I need to create an infinite sequence containing a subsequence of elements which repeats infinitely.
[1; 2; 3; 4; 1; 2; 3; 4; 1; 2; 3; 4; ...]
So I wrote this:
let l = [1; 2; 3; 4]
let s = seq { while true do yield! l }
Is there a standard way (function) to do this?
I think that your approach is good in this scenario. There is no built-in function to implement repetition, but if you need to repeat sequences often, you can define one yourself and make it available in the Seq module:
module Seq =
let repeat items =
seq { while true do yield! items }
Then you can nicely write Seq.repeat [ 1 .. 4 ], as if repeat was a standard F# library function, because F# IntelliSense shows both functions from your Seq module and from the Seq module as if they were defined in a single module.
Aside from your implementation, you can also use recursive sequence expression, which is another quite common pattern when generating sequences. Using while is in some ways imperative (although you don't need any state for simple repetitions) compared to functional recursion:
let rec repeat items =
seq { yield! items
yield! repeat items }
This approach is better when you want to keep some state while generating. For example, generating all numbers 1 .. using while would not be so nice, because you'd need mutable state. Using recursion, you can write the same thing as:
let rec numbersFrom n =
seq { yield n
yield! numbersFrom (n + 1) }
I don't think there's an idiom for this, and what you have is fine, but here are some alternatives.
If you change your subsequence to an array, you can do
let a = [|1; 2; 3; 4|]
let s = Seq.initInfinite (fun i -> a.[i % a.Length])
Using what you have, you could also do
let l = [1; 2; 3; 4]
let s = Seq.initInfinite (fun _ -> l) |> Seq.concat
but it's no shorter.
Similar to Daniel's answer, but encapsulating it into a function, and pretending that function is in the Seq module:
module Seq =
let infiniteOf repeatedList =
Seq.initInfinite (fun _ -> repeatedList)
|> Seq.concat
// Tests
let intList = [1; 2; 3; 4]
let charList = ['a'; 'b'; 'c'; 'd']
let objList = [(new System.Object()); (new System.Object()); (new System.Object()); (new System.Object())]
do
Seq.infiniteOf intList |> Seq.take 20 |> Seq.iter (fun item -> printfn "%A" item)
Seq.infiniteOf charList |> Seq.take 20 |> Seq.iter (fun item -> printfn "%A" item)
Seq.infiniteOf objList |> Seq.take 20 |> Seq.iter (fun item -> printfn "%A" item)
This will do it as a (more-or-less) one-liner, without having to create any helper objects.
let s = seq { while true do
for i in 1 .. 4 -> i }
In F#, imagine we have an array of bytes representing pixel data with three bytes per pixel in RGB order:
[| 255; 0; 0; //Solid red
0; 255; 0; //Solid green
0; 0; 255; //Solid blue
1; 72; 9;
34; 15; 155
... |]
I'm having a hard time knowing how to functionally operate on this data as-is, since a single item is really a consecutive block of three elements in the array.
So, I need to first group the triples in the array into something like this:
[|
[| 255; 0; 0 |];
[| 0; 255; 0 |];
[| 0; 0; 255 |];
[| 1; 72; 9 |];
[| 34; 15; 155 |]
... |]
Now, gathering up the triples into sub-arrays is easy enough to do with a for loop, but I'm curious--is there a functional way to gather up groups of array elements in F#? My ultimate goal is not simply to convert the data as illustrated above, but to solve the problem in a more declarative and functional manner. But I have yet to find an example of how to do this without an imperative loop.
kvb's answer may not give you what you want. Seq.windowed returns a sliding window of values, e.g., [1; 2; 3; 4] becomes [[1; 2; 3]; [2; 3; 4]]. It seems like you want it split into contiguous chunks. The following function takes a list and returns a list of triples ('T list -> ('T * 'T * 'T) list).
let toTriples list =
let rec aux f = function
| a :: b :: c :: rest -> aux (fun acc -> f ((a, b, c) :: acc)) rest
| _ -> f []
aux id list
Here's the inverse:
let ofTriples triples =
let rec aux f = function
| (a, b, c) :: rest -> aux (fun acc -> f (a :: b :: c :: acc)) rest
| [] -> f []
aux id triples
EDIT
If you're dealing with huge amounts of data, here's a sequence-based approach with constant memory use (all the options and tuples it creates have a negative impact on GC--see below for a better version):
let (|Next|_|) (e:IEnumerator<_>) =
if e.MoveNext() then Some e.Current
else None
let (|Triple|_|) = function
| Next a & Next b & Next c -> Some (a, b, c) //change to [|a;b;c|] if you like
| _ -> None
let toSeqTriples (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop() =
seq {
match e with
| Triple (a, b, c) ->
yield a, b, c
yield! loop()
| _ -> ()
}
loop()
EDIT 2
ebb's question about memory use prompted me to test and I found toSeqTriples to be slow and cause surprisingly frequent GCs. The following version fixes those issues and is almost 4x faster than the list-based version.
let toSeqTriplesFast (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop() =
seq {
if e.MoveNext() then
let a = e.Current
if e.MoveNext() then
let b = e.Current
if e.MoveNext() then
let c = e.Current
yield (a, b, c)
yield! loop()
}
loop()
This has relatively constant memory usage vs a list or array-based approach because a) if you have a seq to start with the entire sequence doesn't have to be slurped into a list/array; and b) it also returns a sequence, making it lazy, and avoiding allocating yet another list/array.
I need to first group the triples in the array into something like this:
If you know they will always be triples then representing then as a tuple int * int * int is more "typeful" than using an array because it conveys the fact that there are only ever exactly three elements.
Other people have described various ways to massage the data but I would actually recommend not bothering (unless there is more to this than you have described). I would opt for a function to destructure your array as-is instead:
let get i = a.[3*i], a.[3*i+1], a.[3*i+2]
If you really want to change the representation then you can now do:
let b = Array.init (a.Length/3) get
The answer really depends upon what you want to do next though...
(Hat tip: Scott Wlaschin) As of F# 4.0, you can use Array.chunkBySize(). It does exactly what you want:
let bs = [| 255; 0; 0; //Solid red
0; 255; 0; //Solid green
0; 0; 255; //Solid blue
1; 72; 9;
34; 15; 155 |]
let grouped = bs |> Array.chunkBySize 3
// [| [|255; 0; 0|]
// [| 0; 255; 0|]
// [| 0; 0; 255|]
// [| 1; 72; 9|]
// [| 34; 15; 155|] |]
The List and Seq modules also have chunkBySize() in F# 4.0. As of this writing, the docs at MSDN don't show chunkBySize() anywhere, but it's there if you're using F# 4.0.
UPDATE: As pointed out by Daniel, this answer is wrong because it creates a sliding window.
You can use the Seq.windowed function from the library. E.g.
let rgbPix = rawValues |> Seq.windowed 3
This returns a sequence rather than an array, so if you need random access, you could follow that with a call to Seq.toArray.
Another approach, that takes and yields arrays directly:
let splitArrays n arr =
match Array.length arr with
| 0 ->
invalidArg "arr" "array is empty"
| x when x % n <> 0 ->
invalidArg "arr" "array length is not evenly divisible by n"
| arrLen ->
let ret = arrLen / n |> Array.zeroCreate
let rec loop idx =
ret.[idx] <- Array.sub arr (idx * n) n
match idx + 1 with
| idx' when idx' <> ret.Length -> loop idx'
| _ -> ret
loop 0
Or, yet another:
let splitArray n arr =
match Array.length arr with
| 0 ->
invalidArg "arr" "array is empty"
| x when x % n <> 0 ->
invalidArg "arr" "array length is not evenly divisible by n"
| arrLen ->
let rec loop idx = seq {
yield Array.sub arr idx n
let idx' = idx + n
if idx' <> arrLen then
yield! loop idx' }
loop 0 |> Seq.toArray
Say I have a sequence of 100 elements. Every 10th element I want a new list of the previous 10 elements. In this case I will end up with a list of 10 sublists.
Seq.take(10) looks promising, how can I repeatedly call it to return a list of lists?
now there's Seq.chunkBySize available:
[1;2;3;4;5] |> Seq.chunkBySize 2 = seq [[|1; 2|]; [|3; 4|]; [|5|]]
This is not bad:
let splitEach n s =
seq {
let r = ResizeArray<_>()
for x in s do
r.Add(x)
if r.Count = n then
yield r.ToArray()
r.Clear()
if r.Count <> 0 then
yield r.ToArray()
}
let s = splitEach 5 [1..17]
for a in s do
printfn "%A" a
(*
[|1; 2; 3; 4; 5|]
[|6; 7; 8; 9; 10|]
[|11; 12; 13; 14; 15|]
[|16; 17|]
*)
I have an evolution of three solutions. None of them preserves the ordering of the input elements, which is hopefully OK.
My first solution is quite ugly (making use of ref cells):
//[[4; 3; 2; 1; 0]; [9; 8; 7; 6; 5]; [14; 13; 12; 11; 10]; [17; 16; 15]]
let solution1 =
let split s n =
let i = ref 0
let lst = ref []
seq {
for item in s do
if !i = n then
yield !lst
lst := [item]
i := 1
else
lst := item::(!lst)
i := !i+1
yield !lst
} |> Seq.toList
split {0..17} 5
My second solution factors out the use of ref cells in the first solution, but consequently forces the use of direct IEnumerator access (push in one side, pop out the other)!
//[[17; 16; 15]; [14; 13; 12; 11; 10]; [9; 8; 7; 6; 5]; [4; 3; 2; 1; 0]]
let solution2 =
let split (s:seq<_>) n =
let e = s.GetEnumerator()
let rec each lstlst lst i =
if e.MoveNext() |> not then
lst::lstlst
elif i = n then
each (lst::lstlst) [e.Current] 1
else
each lstlst ((e.Current)::lst) (i+1)
each [] [] 0
split {0..17} 5
My third solution is based on the second solution except it "cheats" by taking a list as input instead of a seq, which enables the most elegant solution using pattern matching as Tomas points out is lacking with seq (which is why we were forced to use direct IEnumerator access).
//[[17; 16; 15]; [14; 13; 12; 11; 10]; [9; 8; 7; 6; 5]; [4; 3; 2; 1; 0]]
let solution3 =
let split inputList n =
let rec each inputList lstlst lst i =
match inputList with
| [] -> (lst::lstlst)
| cur::inputList ->
if i = n then
each inputList (lst::lstlst) [cur] 1
else
each inputList lstlst (cur::lst) (i+1)
each inputList [] [] 0
split [0..17] 5
If preserving the ordering of the elements is important, you can use List.rev for this purpose. For example, in solution2, change the last line of the split function to:
each [] [] 0 |> List.rev |> List.map List.rev
Out of the top of my head:
let rec split size list =
if List.length list < size then
[list]
else
(list |> Seq.take size |> Seq.toList) :: (list |> Seq.skip size |> Seq.toList |> split size)
Perhaps this simple pure implementation might be useful:
let splitAt n xs = (Seq.truncate n xs, if Seq.length xs < n then Seq.empty else Seq.skip n xs)
let rec chunk n xs =
if Seq.isEmpty xs then Seq.empty
else
let (ys,zs) = splitAt n xs
Seq.append (Seq.singleton ys) (chunk n zs)
For example:
> chunk 10 [1..100];;
val it : seq<seq<int>> =
seq
[seq [1; 2; 3; 4; ...]; seq [11; 12; 13; 14; ...];
seq [21; 22; 23; 24; ...]; seq [31; 32; 33; 34; ...]; ...]
> chunk 5 [1..12];;
val it : seq<seq<int>> =
seq [seq [1; 2; 3; 4; ...]; seq [6; 7; 8; 9; ...]; seq [11; 12]]
If in doubt, use fold.
let split n = let one, append, empty = Seq.singleton, Seq.append, Seq.empty
Seq.fold (fun (m, cur, acc) x ->
if m = n then (1, one x, append acc (one cur))
else (m+1, append cur (one x), acc))
(0, empty, empty)
>> fun (_, cur, acc) -> append acc (one cur)
This has the advantage of being fully functional, yet touch each element of the input sequence only once(*) (as opposed to the Seq.take + Seq.skip solutions proposed above).
(*) Assuming O(1) Seq.append. I should certainly hope so.
I found this to be easily the fastest:
let windowChunk n xs =
let range = [0 .. Seq.length xs]
Seq.windowed n xs |> Seq.zip range
|> Seq.filter (fun d -> (fst d) % n = 0)
|> Seq.map(fun x -> (snd x))
i.e. window the list, zip with a list of integers, remove all the overlapping elements, and then drop the integer portion of the tuple.
I think that the solution from Brian is probably the most reasonable simple option. A probelm with sequences is that they cannot be easily processed with the usual pattern matching (like functional lists). One option to avoid that would be to use LazyList from F# PowerPack.
Another option is to define a computation builder for working with IEnumerator type. I wrote something like that recently - you can get it here. Then you can write something like:
let splitEach chunkSize (s:seq<_>) =
Enumerator.toSeq (fun () ->
let en = s.GetEnumerator()
let rec loop n acc = iter {
let! item = en
match item with
| Some(item) when n = 1 ->
yield item::acc |> List.rev
yield! loop chunkSize []
| Some(item) ->
yield! loop (n - 1) (item::acc)
| None -> yield acc |> List.rev }
loop chunkSize [] )
This enables using some functional patterns for list processing - most notably, you can write this as a usual recursive function (similar to the one you would write for lists/lazy lists), but it is imperative under the cover (the let! constructo of iter takes the next element and modifies the enumerator).