Is there standard function to enumerate an F# sequence that works like Python's enumerate()?
It's very easy to write from scratch:
let enumerate (sq : seq<'T>) = seq {
let rec loop (e : IEnumerator<'T>) index = seq {
if e.MoveNext() then
yield (index, e.Current)
yield! loop e (index+1)
}
use enum = sq.GetEnumerator()
yield! loop enum 0
}
but I don't want to reinvent the wheel.
PS: also, I tried
let seasons = ["Spring"; "Summer"; "Fall"; "Winter"]
for x in Seq.zip [0..100000] seasons do
printfn "%A" x
but this [0..10000] part looks ugly.
I think maybe you want Seq.mapi or Seq.iteri.
http://msdn.microsoft.com/en-us/library/ee340431.aspx
http://msdn.microsoft.com/en-us/library/ee370541
This is what you want:
module Seq =
let inline enumerate source = Seq.mapi (fun i x -> i,x) source
> ["a"; "b"] |> Seq.enumerate;;
val it : seq<int * string> = seq [(0, "a"); (1, "b")]
Or Hot 'n Spicy with FSharpx:
let enumerate source = Seq.mapi (curry id) source
Well, actually, in FSharpx it's already available as Seq.index.
Related
I have two lists listA and listB where I want to return true if listB contains any element also in listA.
let listA = ["A";"B";"C"]
let listB = ["D";"E";"A"]
Should return true in this case. I feel like this should be easy to solve and I'm missing something fundamental somewhere.
For example, why can't I do like this?
let testIntersect = for elem in listA do List.exists (fun x -> x = elem) listB
You can't write something like your example code because a plain for doesn't return a result, it just evaluates an expression for its side-effects. You could write the code in a for comprehension:
let testIntersect listA listB =
[for elem in listA do yield List.exists (fun x -> x = elem) listB]
Of course, this then returns a bool list rather than a single bool.
val testIntersect :
listA:seq<'a> -> listB:'a list -> bool list when 'a : equality
let listA = ["A";"B";"C"]
let listB = ["D";"E";"A"]
testIntersect listA listB
val it : bool list = [true; false; false]
So, we can use the List.exists function to ensure that a true occurs at least once:
let testIntersect listA listB =
[for elem in listA do yield List.exists (fun x -> x = elem) listB]
|> List.exists id
val testIntersect :
listA:seq<'a> -> listB:'a list -> bool list when 'a : equality
val listA : string list = ["A"; "B"; "C"]
val listB : string list = ["D"; "E"; "A"]
val it : bool = false
It's pretty inefficient to solve this problem using List though, it's better to use Set. With Set, you can calculate intersection in O(log N * log M) time rather than O(N*M).
let testSetIntersect listA listB =
Set.intersect (Set.ofList listA) (Set.ofList listB)
|> Set.isEmpty
|> not
One function that you could use is List.except, which is not yet documented (!) but can be seen in this pull request that was merged a couple of years ago. You'd probably use it like this:
let testIntersect a b =
let b' = b |> List.except a
// If b' is shorter than b, then b contained at least one element of a
List.length b' < List.length b
However, this runs through list B about three times, once to do the except algorithm and once each to do both the length calls. So another approach might be to do what you did, but turn list A into a set so that the exists call won't be O(N):
let testIntersect a b =
let setA = a |> Set.ofList
match b |> List.tryFind (fun x -> setA |> Set.contains x) with
| Some _ -> true
| None -> false
The reason I used tryFind is because List.find would throw an exception if the predicate didn't match any items of the list.
Edit: An even better approach is to use List.exists, which I temporarily forgot about (thanks to Honza Brestan for reminding me about it):
let testIntersect a b =
let setA = a |> Set.ofList
b |> List.exists (fun x -> setA |> Set.contains x)
Which, of course, is pretty much what you were originally wanting to do in your testIntersect code sample. The only difference is that you were using the for ... in syntax in your code sample, which wouldn't work. In F#, the for loop is exclusively for expressions that return unit (and thus, probably have side effects). If you want to return a value, the for loop won't do that. So using the functions that do return value, like List.exists, is the approach you want to take.
let testIntersect listA listB =
(Set.ofList listA) - (Set.ofList listB) |> Set.isEmpty |> not
I'm new to functional programming and am working on a project in F#.
I've run a problem: I have a list of type string list list and I need to build separate lists based on the middle element of each string list. For example:
[["1";"b";"2"];["2";"a";"0"];["3";"b";"4"];["3";"a";"5"]]
Would be broken into 2 lists similar to the following:
let a = [["2";"0"];["3";"5"]]
let b = [["1";"2"];["3";"4"]]
I tried to use let a = [for [x;y;z] in myList do yield [x;z]] but am having trouble adding in the condition of y = "b", for instance.
Any help would be greatly appreciated
let myList = [["1";"b";"2"];["2";"a";"0"];["3";"b";"4"];["3";"a";"5"]]
let a = [for [x;y;z] in myList do if y="a" then yield [x;z]]
let b = [for [x;y;z] in myList do if y="b" then yield [x;z]]
You're trying to split a list by its middle element. What is the expected behaviour when your list does not have 3 elements?
In the answer provided by Functional_S, you'll see the wiggly lines under the [x;y;z] in
let a = [for [x;y;z] in myList do if y="a" then yield [x;z]]
The compiler says "Incomplete pattern matches". Rather than now adding extra checks to handle empty lists, lists of length 2, etc, consider changing the design of your data types. If you have data that always contains of 3 elements, then use a data structure that has exactly 3 elements. Tuples are an obvious choice here, or use a record.
let myList = [ ("1","b","2"); ("2","a","0"); ("3","b","4"); ("3","a","5") ]
let splitByMiddle =
myList
|> List.groupBy (fun (_, middle, _) -> middle)
|> List.map (fun (middle, elems) -> middle, elems |> List.map (fun (l, _, r) -> l, r))
If you execute that in interactive, you'll get:
val splitByMiddle : (string * (string * string) list) list =
[("b", [("1", "2"); ("3", "4")]); ("a", [("2", "0"); ("3", "5")])]
An alternative would be:
let splitByMiddle =
myList
|> List.map (fun (l, middle, r) -> middle, (l, r))
|> List.groupBy fst
|> List.map (fun (middle, elems) -> middle, elems |> List.map snd)
I find that F# is really at its peak performance when you model your domain as closely as possible with your datatypes. In languages like Matlab, vectors and matrics are your number one work horse, you'd put everything into lists. But in F#, defining data types comes so cheaply (in terms of typing effort) - and once you've done so, the compiler is your best friend to remind you of possible corner cases your code is not covering.
In that light: I see all your middle elements are string, whereas the left/right elements are integers. Maybe your domain is better modelled by this record?
type R =
{
Left: int
Right: int
Middle: string
}
let create (l, m, r) = { Left = l; Right = r; Middle = m}
let myList = [ create(1,"b",2); create(2,"a",0); create(3,"b",4); create(3,"a",5) ]
let splitByMiddle =
myList
|> List.groupBy (fun r -> r.Middle)
This will give you:
val splitByMiddle : (string * R list) list =
[("b", [{Left = 1;
Middle = "b";
Right = 2;}; {Left = 3;
Middle = "b";
Right = 4;}]); ("a", [{Left = 2;
Middle = "a";
Right = 0;}; {Left = 3;
Middle = "a";
Right = 5;}])]
What is the best way to map the following:
[|"A"; "B"; "C"; "D"|]
to
[|("","A","B"); ("A","B","C"); ("B","C","D"); ("C","D","")|]
?
I'd prefer approach based on idiomatic combinators: attach an empty string to both ends of array, use Seq.windowed 3 to produce sequence of sliding arrays of three items, and finally map each sequence element into tuple and convert sequence back to array:
let conv source =
Array.concat [[|String.Empty|]; source; [|String.Empty|]] |> Seq.windowed 3
|> Seq.map (fun x -> x.[0],x.[1],x.[2]) |> Seq.toArray
Testing conv [|"A";"B";"C";"D"|] under FSI yields:
val it : (string * string * string) [] =
[|("", "A", "B"); ("A", "B", "C"); ("B", "C", "D"); ("C", "D", "")|]
I would do this:
let neighbors (xs: _ []) =
let get i = if i>=0 && i<xs.Length then xs.[i] else ""
Array.mapi (fun i x -> get(i-1), x, get(i+1)) xs
This is the first time I've looked at f# so pardon my attempt if it isn't correct. I also don't have f# so I can't test this. But based on Array.map...
let data = [|"A"; "B"; "C"; "D"|]
let convert = data |> Array.map (fun x -> (char(int x - 1), x, char(int x + 1)))
printfn "Converting to tupels by using map = %A" convert
Hopefully it will point you in the right direction.
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"
a little rusty from my Scheme days, I'd like to take 2 lists: one of numbers and one of strings, and fold them together into a single string where each pair is written like "{(ushort)5, "bla bla bla"},\n". I have most of it, i'm just not sure how to write the Fold properly:
let splitter = [|","|]
let indexes =
indexStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let values =
valueStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let pairs = List.zip indexes values
printfn "%A" pairs
let result = pairs |> Seq.fold
(fun acc a -> String.Format("{0}, \{(ushort){1}, \"{2}\"\}\n",
acc, (List.nth a 0), (List.nth a 1)))
Your missing two things. The initial state of the fold which is an empty string and you can't use list comprehension on tuples in F#.
let splitter = [|","|]
let indexes =
indexStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let values =
valueStr.Split(splitter, System.StringSplitOptions.None) |> Seq.toList
let pairs = List.zip indexes values
printfn "%A" pairs
let result =
pairs
|> Seq.fold (fun acc (index, value) ->
String.Format("{0}{{(ushort){1}, \"{2}\"}},\n", acc, index, value)) ""
fold2 version
let result =
List.fold2
(fun acc index value ->
String.Format("{0}{{(ushort){1}, \"{2}\"}},\n", acc, index, value))
""
indexes
values
If you are concerned with speed you may want to use string builder since it doesn't create a new string every time you append.
let result =
List.fold2
(fun (sb:StringBuilder) index value ->
sb.AppendFormat("{{(ushort){0}, \"{1}\"}},\n", index, value))
(StringBuilder())
indexes
values
|> string
Fold probably isn't the best method for this task. Its a lot easier to map and concat like this:
let l1 = "a,b,c,d,e".Split([|','|])
let l2 = "1,2,3,4,5".Split([|','|])
let pairs =
Seq.zip l1 l2
|> Seq.map (fun (x, y) -> sprintf "(ushort)%s, \"%s\"" x y)
|> String.concat "\n"
I think you want List.fold2. For some reason the List module has a fold2 member but Seq doesn't. Then you can dispense with the zip entirely.
The types of your named variables and the type of the result you hope for are all implicit, so it's difficult to help, but if you are trying to accumulate a list of strings you might consider something along the lines of
let result = pairs |> Seq.fold
(fun prev (l, r) ->
String.Format("{0}, \{(ushort){1}, \"{2}\"\}\n", prev, l, r)
"" pairs
My F#/Caml is very rusty so I may have the order of arguments wrong. Also note your string formation is quadratic; in my own code I would go with something more along these lines:
let strings =
List.fold2 (fun ss l r ->
String.format ("\{(ushort){0}, \"{1}\"\}\n", l, r) :: ss)
[] indexes values
let result = String.concat ", " strings
This won't cost you quadratic time and it's a little easier to follow. I've checked MSDN and believe I have the correct order of arguments on fold2.
Keep in mind I know Caml not F# and so I may have details or order of arguments wrong.
Perhaps this:
let strBuilder = new StringBuilder()
for (i,v) in Seq.zip indexes values do
strBuilder.Append(String.Format("{{(ushort){0}, \"{1}\"}},\n", i,v))
|> ignore
with F# sometimes is better go imperative...
map2 or fold2 is the right way to go. Here's my take, using the (||>) operator:
let l1 = [| "a"; "b"; "c"; "d"; "e" |]
let l2 = [| "1"; "2"; "3"; "4"; "5" |]
let pairs = (l1, l2) ||> Seq.map2 (sprintf ("(ushort)%s, \"%s\""))
|> String.concat "\n"