I've been at this for a few hours now, looking though every website and piece of documentation I could. I can't figure out how to remove one, and only one element (In this case, a string) from an array, keeping any duplicates in tact.
I did find a way, but, it's absolutely atrocious:
let remItem gs item =
if (chkItem gs item) then
let mutable fr = [| |] //temporary array
let mutable don = false //check if we found the element
for i in gs.inventory do
if not (i = item) && don then
fr <- (Array.append fr [|i|])
//add to the temp array until we find our item
elif i = item && don = false then don <- true
//we found it, skip just once so it doesn't get added
elif don then fr <- (Array.append fr [|i|])
//now just add everything else to the temp array
{ gs with inventory = fr }
else gs
I wrote this and I barely know how it works. Please tell me there's a better way to do this. I know the mutable variables aren't needed, but I've written a dozen equally horrendous-looking pure functions and concluded this is the best that I could do. I've tried a lot of the Array.* recursive functions already, I can't seem to make any of those comply with what I want either. I just want to know if it's even possible to do this neatly and purely in F#.
I think the easiest way to do this is to first look for the index (it's an array after all) and then just cut this out - this is (I think) a good compromise between performance and pureness - it's a pure operation but you don't get to much copy-operations:
let remove x (xs : 'a array) =
match Array.tryFindIndex ((=) x) xs with
| Some 0 -> xs.[1..]
| Some i -> Array.append xs.[..i-1] xs.[i+1..]
| None -> xs
please note that you have to take care of it beeing the first index because xs.[..(-1)] will throw an exception (while the other edge-case is ok):
> remove 0 [|1..10|];;
val it : int [] = [|1; 2; 3; 4; 5; 6; 7; 8; 9; 10|]
> remove 1 [|1..10|];;
val it : int [] = [|2; 3; 4; 5; 6; 7; 8; 9; 10|]
> remove 3 [|1..10|];;
val it : int [] = [|1; 2; 4; 5; 6; 7; 8; 9; 10|]
> remove 9 [|1..10|];;
val it : int [] = [|1; 2; 3; 4; 5; 6; 7; 8; 10|]
> remove 10 [|1..10|];;
val it : int [] = [|1; 2; 3; 4; 5; 6; 7; 8; 9|]
> remove 11 [|1..10|];;
val it : int [] = [|1; 2; 3; 4; 5; 6; 7; 8; 9; 10|]
if you need even more performance you could create an empty array and use a more imperative style to copy the parts:
let remove x (xs : 'a array) =
match Array.tryFindIndex ((=) x) xs with
| Some i ->
let res = Array.zeroCreate (xs.Length-1)
if i >= 1 then
System.Array.Copy(xs,0,res,0,i)
if i+1 < xs.Length then
System.Array.Copy(xs,i+1,res,i,xs.Length-i-1)
res
| None -> xs
Remove the first occurrence of a item from a list (taken from http://www.fssnip.net/1T):
let rec remove_first pred lst =
match lst with
| h::t when pred h -> t
| h::t -> h::remove_first pred t
| _ -> []
Usage:
let somelist = [('a',2);('f',7);('a',4);('h',10)]
let removed = somelist |> remove_first (fun (x,y) -> x='a')
// Result is:
// [('f',7);('a',4);('h',10)]
Fold should do the trick:
let remove x a =
Array.fold
(fun (s,found) t ->
if found || t <> x then Array.append s [|t|],found
else s,true) ([||],false) a |> fst
Example usage:
remove 2 [|1; 2; 3; 4; 2; 5|]
val it : int [] = [|1; 3; 4; 2; 5|]
Related
let's take this data:
let a = [10; 11; 12; 13; 14; 0; 15; 16]
I'm trying to do this:
[
let mutable skip = false
for i in 0 .. a.Length - 1 do
if a.[i] = 0 then skip <- true
if not skip then yield a.[i]
]
but I was wondering if List.unfold could be used for this? (and how?)
In practice, I'm getting a sequence of sequences (seq of rows, each holding a seq of columns, from an Excel file), and I want to stop the parsing when I encounter an empty line, but the simplified example illustrates it.
The expression above works, so this is about me learning if unfold could be applied to this.
I would use takeWhile:
let a = [10; 11; 12; 13; 14; 0; 15; 16]
Seq.takeWhile ((<>) 0) a
// |> do your parsing
Yes, you can use List.unfold:
let a = [10; 11; 12; 13; 14; 0; 15; 16]
a
|> List.unfold (function
| [] -> None
| x :: rest -> if x = 0 then None else Some (x, rest)
)
I'm recently new to F# so please bear with me. The problem i have is I'm trying to find only prime numbers.
I've write this code:
let isPrime n =
let rec check i =
i > n/2 || (n % i <> 0 && check (i + 1))
check 2;;
let listNums = List.filter isPrime >> List.length;;
let nums = [ 16; 17; 3; 4; 2; 5; 11; 6; 7; 18; 13; 14; ];;
let countPrimes (x:int) = x |> List.ofSeq |> listNums;;
trying to call
countPrimes nums;;
but this is failed with message:
The type 'int' is not compatible with the type 'seq<'a>'
any help would be appreciated
I found the solution
let isPrime n =
let rec check i =
i > n/2 || (n % i <> 0 && check (i + 1))
check 2;;
let listNums = List.filter isPrime >> List.length;;
let nums = [| 16; 17; 3; 4; 2; 5; 11; 6; 7; 18; 13; 14; |];;
let countPrimes (x:int[]) = x |> List.ofSeq |> listNums;;
countPrimes nums;;
Thanks all!
x |> List.ofSeq
seems to be the problem to me. You are passing an int into a function that requires a list. List.toSeq changes a list into a sequence. You want the function countPrimes to take a List of integers, not simply an integer. Although Carsten is right, listNums already takes a List of integers (edit: and computes the value you want provided isPrime is correct).
You do not need to countPrimes separately. Enough to remove and will work:
let isPrime n =
let rec check i =
i > n/2 || (n % i <> 0 && check (i + 1))
check 2
let nums = [ 16; 17; 3; 4; 2; 5; 11; 6; 7; 18; 13; 14; ]
let listPrime lst =
lst |> List.filter isPrime
nums |> listPrime |> printfn "%A"
Out:
[17; 3; 2; 5; 11; 7; 13]
Link:
https://dotnetfiddle.net/nVXwZ5
I want to create a list of 20 random numbers. I wrote this:
let numberList = [ 1 .. 20 ]
let randoms =
numberList
|> List.map (fun (x) -> System.Random().Next(0,9))
And I got this:
val numberList : int list =
[1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20]
val randoms : int list =
[7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7; 7]
Which makes sense. The problem is that I want to pass in a random number each time the function is evaluated like this:
let numberList = [ 1 .. 20 ]
let randoms =
numberList
|> List.map (fun (Random().Next(0,9)) -> x)
but I am getting a "The pattern discriminator 'Random' is not defined" exception.
Am i approaching this problem the wrong way?
Thanks in advance
In your original version, you create a new Random object at each iteration. As this is seeded with the current time, you always create the same sequence.
You need to create the object outside map like so
let RNG = new System.Random()
numberlist |> List.map (fun x -> RNG.Next(0,9))
Although a more elegant solution would be
let RNG = new System.Random()
let randoms = List.init 20 (fun _ -> RNG.Next(0,9))
Your second version fails because you are trying to treat Random as a pattern which makes no sense in this situation.
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).
how to check if an element is contained within a sequence? I expected some Seq.contains, but i could not find it. Thanks
EDIT:
Or, for an easier task, how to make the diff between two sequences? Like, getting all the elements within a list that doesn not belong to another (or that do)?
Little bit simpler:
let contains x = Seq.exists ((=) x)
Seq.exists
let testseq = seq [ 1; 2; 3; 4 ]
let equalsTwo n = (n = 2)
let containsTwo = Seq.exists equalsTwo testseq
Set is your friend here:
let a = set [0;1;2;3]
let b = set [2;3;4;5]
let c = a - b
let d = b - a
let e = Set.intersect a b
let f = a + b
>
val c : Set<int> = seq [0; 1]
val d : Set<int> = seq [4; 5]
val e : Set<int> = seq [2; 3]
val f : Set<int> = seq [0; 1; 2; 3; ...]
Danny
Seq.exists again, but with slightly different syntax -
let testseq = seq [ 1; 2; 3; 4 ]
let testn = 2
testseq |> Seq.exists (fun x -> x = testn)
See MSDN F#: Seq.exists function: https://msdn.microsoft.com/en-us/library/ee353562.aspx
Lots of other good ones there too!
(Another question, another answer.)
This works, but I don't think that it's the most idomatic way to do it - (you'll need to wait until the US wakes up to find out):
let s1 = seq [ 1; 2; 3; 4 ]
let s2 = seq [ 3; 4; 5; 6 ]
seq {
for a in s1 do
if not (Seq.exists (fun n -> n = a) s2) then
yield a
}