I'm worried when don't know when you can use "Seq" , "seq" . Can you tell me which defferences are ?
This's my code . Why dont't use "seq" ?
let s = ResizeArray<float>()
s.Add(1.1)
s.Add(2.2)
s.Add(3.3)
s.Add(4.4)
s |> Seq.iter (fun x -> printfn("%f") x )
Seq is a module that contains functions that work with seq values:
Seq.map string [ 1; 2 ]
Seq.sum [ 1; 2 ]
seq is a type name:
let f1 (xs : seq<int>) = ()
let f2 (xs : int seq) = ()
seq is also a function that converts something like a list into the type seq:
seq [ 1; 2 ]
seq { ... } is a computation expression:
seq { yield 1; yield 2 }
You use the uppercase Seq in all cases except in type annotation.
For example:
let (x:seq<int>) =
[1..10]
|> Seq.map (fun t -> t + 1)
Edit: Please refer to recommended answer, as my answer is incomplete.
Related
Supposed there is a list:
let lst = [1;2;3]
And a curried function:
let addAll a b c =
a + b + c
How can I input the parameters for the curried function coveniently using the elements in list lst?
One way of doing this is:
addAll (lst |> List.item 0) (lst |> List.item 1) (lst |> List.item 2)
But this doesn't scale very well! Also, it's boring.
It is hard to say from the limited example what your actual use case is. Lists are designed to contain a varying number of items and functions take constant number of items, so the two do not match well. It might make more sense to use a tuple rather than a list:
let tup = (1,2,3)
let addAll (a, b, c) =
a + b + c
addAll tup
Tuples contain fixed number of items, but they can be easily constructed and deconstructed and allow you to pass all parameters to your function at once.
You can also do what you asked about using reflection, but this may break in future versions of F# and it is almost never a good design for a simple case like this. It is also slow and as you can see from the number of downcasts and boxing, it is also not very safe:
let lst = [1;2;3]
let addAll a b c =
a + b + c
let addAllVal = addAll
let f = addAllVal.GetType().GetMethod("Invoke", [| typeof<int>; typeof<int>; typeof<int> |])
let res = f.Invoke(addAllVal, Array.map box (Array.ofList lst)) :?> int
Another option is to use pattern matching:
let lst = [1;2;3]
match lst with [ a ; b; c] -> addAll a b c |_-> 0
returns 6.
If lst does not have exactly 3 elements then it returns 0 but you can change it to handle other cases:
let callAddAll lst =
match lst with
| [ ] -> 0
| [ a ] -> addAll a 0 0
| [ a ; b ] -> addAll a b 0
| [ a ; b ; c ] -> addAll a b c
| a :: b :: c :: rest -> addAll a b c // ignore rest
[ ] |> callAddAll |> printfn "lst = %d" // = 0
[1 ] |> callAddAll |> printfn "lst = %d" // = 1
[1;2 ] |> callAddAll |> printfn "lst = %d" // = 3
[1;2;3 ] |> callAddAll |> printfn "lst = %d" // = 6
[1;2;3;4] |> callAddAll |> printfn "lst = %d" // = 6
I want a tool for testing Rx components that would work like this:
Given an order of the events specified as a 'v seq and a key selector function (keySelector :: 'v -> 'k) I want to create a Map<'k, IObservable<'k>> where the guarantee is that the groupped observables yield the values in the global order defined by the above enumerable.
For example:
makeObservables isEven [1;2;3;4;5;6]
...should produce
{ true : -2-4-6|,
false: 1-3-5| }
This is my attempt looks like this:
open System
open System.Reactive.Linq
open FSharp.Control.Reactive
let subscribeAfter (o1: IObservable<'a>) (o2 : IObservable<'b>) : IObservable<'b> =
fun (observer : IObserver<'b>) ->
let tempObserver = { new IObserver<'a> with
member this.OnNext x = ()
member this.OnError e = observer.OnError e
member this.OnCompleted () = o2 |> Observable.subscribeObserver observer |> ignore
}
o1.Subscribe tempObserver
|> Observable.Create
let makeObservables (keySelector : 'a -> 'k) (xs : 'a seq) : Map<'k, IObservable<'a>> =
let makeDependencies : ('k * IObservable<'a>) seq -> ('k * IObservable<'a>) seq =
let makeDep ((_, o1), (k2, o2)) = (k2, subscribeAfter o1 o2)
Seq.pairwise
>> Seq.map makeDep
let makeObservable x = (keySelector x, Observable.single x)
let firstItem =
Seq.head xs
|> makeObservable
|> Seq.singleton
let dependentObservables =
xs
|> Seq.map makeObservable
|> makeDependencies
dependentObservables
|> Seq.append firstItem
|> Seq.groupBy fst
|> Seq.map (fun (k, obs) -> (k, obs |> Seq.map snd |> Observable.concatSeq))
|> Map.ofSeq
[<EntryPoint>]
let main argv =
let isEven x = (x % 2 = 0)
let splits : Map<bool, IObservable<int>> =
[1;2;3;4;5]
|> makeObservables isEven
use subscription =
splits
|> Map.toSeq
|> Seq.map snd
|> Observable.mergeSeq
|> Observable.subscribe (printfn "%A")
Console.ReadKey() |> ignore
0 // return an integer exit code
...but the results are not as expected and the observed values are not in the global order.
Apparently the items in each group are yield correctly but when the groups are merged its more like a concat then a merge
The expected output is: 1 2 3 4 5
...but the actual output is 1 3 5 2 4
What am I doing wrong?
Thanks!
You describe wanting this:
{ true : -2-4-6|,
false: 1-3-5| }
But you're really creating this:
{ true : 246|,
false: 135| }
Since there's no time gaps between the items in the observables, the merge basically has a constant race condition. Rx guarantees that element 1 of a given sequence will fire before element 2, but Merge offers no guarantees around cases like this.
You need to introduce time gaps into your observables if you want Merge to be able to re-sequence in the original order.
I don't understand behavior of distinctBy in this snippet:
let s = [123; 231; 321]
let s1 = s |> Seq.map (string >> Seq.sort)
let s2 = s |> Seq.distinctBy (string >> Seq.sort)
which produces:
s1 = seq [seq ['1'; '2'; '3']; seq ['1'; '2'; '3']; seq ['1'; '2'; '3']]
as expected, but:
s2 = seq [123; 231; 321]
where I expected only one element, because the 3 keys are identical. Which part I got wrong?
F# does not compare Sequences for equality see this example
(123 |> string |> Seq.sort) = (123 |> string |> Seq.sort)
val it : bool = false
I imagine this is to allow support for infinite sequences.
You can fix this by mapping to lists
let s = [123; 231; 321] |> Seq.distinctBy (string >> Seq.sort >> Seq.toList);;
val s : seq<int>
> s;;
val it : seq<int> = seq [123]
Seq.sort probably doesn't implement comparison logic. So the underlying implementation sees three distinct objects.
Similar if you did the following:
object.ReferenceEquals("1", 1.ToString());
I want to create a function with the signature seq<#seq<'a>> ->seq<seq<'a>> that acts like a Zip method taking a sequence of an arbitrary number of input sequences (instead of 2 or 3 as in Zip2 and Zip3) and returning a sequence of sequences instead of tuples as a result.
That is, given the following input:
[[1;2;3];
[4;5;6];
[7;8;9]]
it will return the result:
[[1;4;7];
[2;5;8];
[3;6;9]]
except with sequences instead of lists.
I am very new to F#, but I have created a function that does what I want, but I know it can be improved. It's not tail recursive and it seems like it could be simpler, but I don't know how yet. I also haven't found a good way to get the signature the way I want (accepting, e.g., an int list list as input) without a second function.
I know this could be implemented using enumerators directly, but I'm interested in doing it in a functional manner.
Here's my code:
let private Tail seq = Seq.skip 1 seq
let private HasLengthNoMoreThan n = Seq.skip n >> Seq.isEmpty
let rec ZipN_core = function
| seqs when seqs |> Seq.isEmpty -> Seq.empty
| seqs when seqs |> Seq.exists Seq.isEmpty -> Seq.empty
| seqs ->
let head = seqs |> Seq.map Seq.head
let tail = seqs |> Seq.map Tail |> ZipN_core
Seq.append (Seq.singleton head) tail
// Required to change the signature of the parameter from seq<seq<'a> to seq<#seq<'a>>
let ZipN seqs = seqs |> Seq.map (fun x -> x |> Seq.map (fun y -> y)) |> ZipN_core
let zipn items = items |> Matrix.Generic.ofSeq |> Matrix.Generic.transpose
Or, if you really want to write it yourself:
let zipn items =
let rec loop items =
seq {
match items with
| [] -> ()
| _ ->
match zipOne ([], []) items with
| Some(xs, rest) ->
yield xs
yield! loop rest
| None -> ()
}
and zipOne (acc, rest) = function
| [] -> Some(List.rev acc, List.rev rest)
| []::_ -> None
| (x::xs)::ys -> zipOne (x::acc, xs::rest) ys
loop items
Since this seems to be the canonical answer for writing a zipn in f#, I wanted to add a "pure" seq solution that preserves laziness and doesn't force us to load our full source sequences in memory at once like the Matrix.transpose function. There are scenarios where this is very important because it's a) faster and b) works with sequences that contain 100s of MBs of data!
This is probably the most un-idiomatic f# code I've written in a while but it gets the job done (and hey, why would there be sequence expressions in f# if you couldn't use them for writing procedural code in a functional language).
let seqdata = seq {
yield Seq.ofList [ 1; 2; 3 ]
yield Seq.ofList [ 4; 5; 6 ]
yield Seq.ofList [ 7; 8; 9 ]
}
let zipnSeq (src:seq<seq<'a>>) = seq {
let enumerators = src |> Seq.map (fun x -> x.GetEnumerator()) |> Seq.toArray
if (enumerators.Length > 0) then
try
while(enumerators |> Array.forall(fun x -> x.MoveNext())) do
yield enumerators |> Array.map( fun x -> x.Current)
finally
enumerators |> Array.iter (fun x -> x.Dispose())
}
zipnSeq seqdata |> Seq.toArray
val it : int [] [] = [|[|1; 4; 7|]; [|2; 5; 8|]; [|3; 6; 9|]|]
By the way, the traditional matrix transpose is much more terse than #Daniel's answer. Though, it requires a list or LazyList that both will eventually have the full sequence in memory.
let rec transpose =
function
| (_ :: _) :: _ as M -> List.map List.head M :: transpose (List.map List.tail M)
| _ -> []
To handle having sub-lists of different lengths, I've used option types to spot if we've run out of elements.
let split = function
| [] -> None, []
| h::t -> Some(h), t
let rec zipN listOfLists =
seq { let splitted = listOfLists |> List.map split
let anyMore = splitted |> Seq.exists (fun (f, _) -> f.IsSome)
if anyMore then
yield splitted |> List.map fst
let rest = splitted |> List.map snd
yield! rest |> zipN }
This would map
let ll = [ [ 1; 2; 3 ];
[ 4; 5; 6 ];
[ 7; 8; 9 ] ]
to
seq
[seq [Some 1; Some 4; Some 7]; seq [Some 2; Some 5; Some 8];
seq [Some 3; Some 6; Some 9]]
and
let ll = [ [ 1; 2; 3 ];
[ 4; 5; 6 ];
[ 7; 8 ] ]
to
seq
[seq [Some 1; Some 4; Some 7]; seq [Some 2; Some 5; Some 8];
seq [Some 3; Some 6; null]]
This takes a different approach to yours, but avoids using some of the operations that you had before (e.g. Seq.skip, Seq.append), which you should be careful with.
I realize that this answer is not very efficient, but I do like its succinctness:
[[1;2;3]; [4;5;6]; [7;8;9]]
|> Seq.collect Seq.indexed
|> Seq.groupBy fst
|> Seq.map (snd >> Seq.map snd);;
Another option:
let zipN ls =
let rec loop (a,b) =
match b with
|l when List.head l = [] -> a
|l ->
let x1,x2 =
(([],[]),l)
||> List.fold (fun acc elem ->
match acc,elem with
|(ah,at),eh::et -> ah#[eh],at#[et]
|_ -> acc)
loop (a#[x1],x2)
loop ([],ls)
I should split seq<a> into seq<seq<a>> by an attribute of the elements. If this attribute equals by a given value it must be 'splitted' at that point. How can I do that in FSharp?
It should be nice to pass a 'function' to it that returns a bool if must be splitted at that item or no.
Sample:
Input sequence: seq: {1,2,3,4,1,5,6,7,1,9}
It should be splitted at every items when it equals 1, so the result should be:
seq
{
seq{1,2,3,4}
seq{1,5,6,7}
seq{1,9}
}
All you're really doing is grouping--creating a new group each time a value is encountered.
let splitBy f input =
let i = ref 0
input
|> Seq.map (fun x ->
if f x then incr i
!i, x)
|> Seq.groupBy fst
|> Seq.map (fun (_, b) -> Seq.map snd b)
Example
let items = seq [1;2;3;4;1;5;6;7;1;9]
items |> splitBy ((=) 1)
Again, shorter, with Stephen's nice improvements:
let splitBy f input =
let i = ref 0
input
|> Seq.groupBy (fun x ->
if f x then incr i
!i)
|> Seq.map snd
Unfortunately, writing functions that work with sequences (the seq<'T> type) is a bit difficult. They do not nicely work with functional concepts like pattern matching on lists. Instead, you have to use the GetEnumerator method and the resulting IEnumerator<'T> type. This often makes the code quite imperative. In this case, I'd write the following:
let splitUsing special (input:seq<_>) = seq {
use en = input.GetEnumerator()
let finished = ref false
let start = ref true
let rec taking () = seq {
if not (en.MoveNext()) then finished := true
elif en.Current = special then start := true
else
yield en.Current
yield! taking() }
yield taking()
while not (!finished) do
yield Seq.concat [ Seq.singleton special; taking()] }
I wouldn't recommend using the functional style (e.g. using Seq.skip and Seq.head), because this is quite inefficient - it creates a chain of sequences that take value from other sequence and just return it (so there is usually O(N^2) complexity).
Alternatively, you could write this using a computation builder for working with IEnumerator<'T>, but that's not standard. You can find it here, if you want to play with it.
The following is an impure implementation but yields immutable sequences lazily:
let unflatten f s = seq {
let buffer = ResizeArray()
let flush() = seq {
if buffer.Count > 0 then
yield Seq.readonly (buffer.ToArray())
buffer.Clear() }
for item in s do
if f item then yield! flush()
buffer.Add(item)
yield! flush() }
f is the function used to test whether an element should be a split point:
[1;2;3;4;1;5;6;7;1;9] |> unflatten (fun item -> item = 1)
Probably no the most efficient solution, but this works:
let takeAndSkipWhile f s = Seq.takeWhile f s, Seq.skipWhile f s
let takeAndSkipUntil f = takeAndSkipWhile (f >> not)
let rec splitOn f s =
if Seq.isEmpty s then
Seq.empty
else
let pre, post =
if f (Seq.head s) then
takeAndSkipUntil f (Seq.skip 1 s)
|> fun (a, b) ->
Seq.append [Seq.head s] a, b
else
takeAndSkipUntil f s
if Seq.isEmpty pre then
Seq.singleton post
else
Seq.append [pre] (splitOn f post)
splitOn ((=) 1) [1;2;3;4;1;5;6;7;1;9] // int list is compatible with seq<int>
The type of splitOn is ('a -> bool) -> seq<'a> -> seq>. I haven't tested it on many inputs, but it seems to work.
In case you are looking for something which actually works like split as an string split (i.e the item is not included on which the predicate returns true) the below is what I came up with.. tried to be as functional as possible :)
let fromEnum (input : 'a IEnumerator) =
seq {
while input.MoveNext() do
yield input.Current
}
let getMore (input : 'a IEnumerator) =
if input.MoveNext() = false then None
else Some ((input |> fromEnum) |> Seq.append [input.Current])
let splitBy (f : 'a -> bool) (input : 'a seq) =
use s = input.GetEnumerator()
let rec loop (acc : 'a seq seq) =
match s |> getMore with
| None -> acc
| Some x ->[x |> Seq.takeWhile (f >> not) |> Seq.toList |> List.toSeq]
|> Seq.append acc
|> loop
loop Seq.empty |> Seq.filter (Seq.isEmpty >> not)
seq [1;2;3;4;1;5;6;7;1;9;5;5;1]
|> splitBy ( (=) 1) |> printfn "%A"