In F# I want to transform a list of chars into a string. Consider the following code:
let lChars = ['a';'b';'c']
If I simply do lChars.ToString, I get "['a';'b';'c']". I'm trying to get "abc". I realize I could probably do a List.reduce to get the effect I'm looking for but it seems like there should be some primitive built into the library to do this.
To give a little context to this, I'm doing some manipulation on individual characters in a string and when I'm done, I want to display the resulting string.
I've tried googling this and no joy that way. Do I need to just bite the bullet and build a List.reduce expression to do this transformation or is there some more elegant way to do this?
Have you tried
System.String.Concat(Array.ofList(lChars))
How many ways can you build a string in F#?
Here's another handful:
let chars = ['H';'e';'l';'l';'o';',';' ';'w';'o';'r';'l';'d';'!']
//Using an array builder
let hw1 = new string [|for c in chars -> c|]
//StringBuilder-Lisp-like approach
open System.Text
let hw2 =
string (List.fold (fun (sb:StringBuilder) (c:char) -> sb.Append(c))
(new StringBuilder())
chars)
//Continuation passing style
let hw3 =
let rec aux L k =
match L with
| [] -> k ""
| h::t -> aux t (fun rest -> k (string h + rest) )
aux chars id
Edit: timings may be interesting? I turned hw1..3 into functions and fed them a list of 500000 random characters:
hw1: 51ms
hw2: 16ms
hw3: er... long enough to grow a beard? I think it just ate all of my memory.
Didn't see this one here, so:
let stringFromCharList (cl : char list) =
String.concat "" <| List.map string cl
"" is just an empty string.
FSI output:
> stringFromCharList ['a'..'d'];;
val it : string = "abcd"
EDIT:
Didn't like this syntax coming back to this so here's a more canonically functional one:
['a'..'z'] |> List.map string |> List.reduce (+)
['a';'b';'c'] |> List.fold_left (fun acc c -> acc ^ (string c)) ""
Edited:
Here is yet another funny way to do your task:
type t =
| N
| S of string
static member Zero
with get() = N
static member (+) (a: t, b: t) =
match a,b with
| S a, S b -> S (a+b)
| N, _ -> b
| _, N -> a
let string_of_t = function
|N -> ""
|S s -> s
let t_of_char c = S (string c)
['a'; 'b'; 'c'] |> List.map t_of_char |> List.sum |> string_of_t
Sadly, just extending System.String with 'Zero' member does not allow to use List.sum with strings.
Edited (answer to Juilet):
Yes, you are right, left fold is slow. But i know more slow right fold :) :
#r "FSharp.PowerPack"
List.fold_right (String.make 1 >> (^)) ['a';'b';'c'] ""
and of course there is fast and simple:
new System.String(List.to_array ['1';'2';'3'])
And i used 'sprintf' seems to me easier:
let t = "Not what you might expect"
let r = [ for i in "aeiou" -> i]
let q = [for a in t do if not (List.exists (fun x -> x=a) r) then yield a]
let rec m = function [] -> "" | h::t -> (sprintf "%c" h) + (m t)
printfn "%A" (m q)
The following solution works for me:
let charList = ["H";"E";"L";"L";"O"]
let rec buildString list =
match list with
| [] -> ""
| head::tail -> head + (buildString tail)
let resultBuildString = buildString charList
[|'w'; 'i'; 'l'; 'l'|]
|> Array.map string
|> Array.reduce (+)
or as someone else posted:
System.String.Concat([|'w'; 'i'; 'l'; 'l'|])
Related
I need a method using List.fold that will count the number of vowels in a given string.
So far I have this method.
let vowels = ['a';'e';'i';'o';'u']
let isVowel =
fun c -> vowels |> List.contains c
let count =
String.filter isVowel
>> String.length
printfn "%A" (count "aaaa")
It works fine, but I cannot figure out how to make a List.fold one using the same isVowel method. Here is my try.
The code below does not work, its purpose is to reflect my idea. Since the fold method applies the isVowel() which returns a true/false on each char in the string, if the condition is true it will add 1 to the accumulator which is the 0 parameter. When I try to use if else insinde the anon function I get an error.
let isVowel x =
match x with
| 'a' -> true
| 'e' -> true
| 'i' -> true
| 'o' -> true
| 'u' -> true
| _ -> false
let countNumOfVowels =
List.fold (fun (isVowel) (x) -> x + 1) 0 ["aaaa"]
You are trying to fold a list but your source is actually a string.
A string can be interpreted as a sequence of chars if you use Seq.fold:
"abracadabra" |> Seq.fold (fun i c -> if isVowel c then i + 1 else i) 0
// val it : int = 5
This is what I was looking for. Thanks Gus!
let countNumOfVowels str =
List.fold (fun (x: int) (c: char) -> if (isVowel c) then x + 1 else x) 0 (Seq.toList str)
countNumOfVowels "Hello"
> countNumOfVowels "Hello";;
val it : int = 2
What I am trying to do is to remove duplicates of a specific given char in a string but letting the first char to remain. I.e:
let myStr = "hi. my .name."
//a function that gets a string and the element to be removed in the string
someFunc myStr "."
where someFunc returns the string showen as below:
"hi. my name"
It is easy to remove duplicates from a string, but is there a way to remove the duplicates but letting the first duplicated element remain in the string?
Here's one approach:
let keepFirst c s =
Seq.mapFold (fun k c' -> (c', k||c<>c'), k&&c<>c') true s
|> fst
|> Seq.filter snd
|> Seq.map fst
|> Array.ofSeq
|> System.String
let example = keepFirst '.' "hi. my .name."
let someFunc (str : string) c =
let parts = str.Split([| c |])
if Array.length parts > 1 then
seq {
yield Array.head parts
yield string c
yield! Array.tail parts
}
|> String.concat ""
else
str
Note that the character is given as char instead of a string.
let someFunc chr (str:string) =
let rec loop (a: char list) b = function
| [] -> a |> List.rev |> System.String.Concat
| h::t when h = chr -> if b then loop a b t
else loop (h::a) true t
| h::t -> loop (h::a) b t
loop [] false (str.ToCharArray() |> Array.toList)
Note that the character is given as char instead of a string.
Edit: Another way would be using regular expressions
open System.Text.RegularExpressions
let someOtherFunc c s =
let pat = Regex.Escape(c)
Regex.Replace(s, sprintf "(?<=%s.*)%s" pat pat, "")
Note that, in this case the character is given as string.
Edit 2:
let oneMoreFunc (c:char) (s:string) =
let pred = (<>) c
[ s |> Seq.takeWhile pred
seq [c]
s |> Seq.skipWhile pred |> Seq.filter pred ]
|> Seq.concat
|> System.String.Concat
When devising a function, think about gains from making its arguments generic. To pass state through the iteration, barring mutable variables, Seq.scan could be a weapon of choice. It folds into a tuple of new state and an option, then Seq.choose strips out the state and the unwanted elements.
In terms of functional building blocks, make it accept a predicate function 'a -> bool and let it return a function seq<'a> -> seq<'a>.
let filterDuplicates predicate =
Seq.scan (fun (flag, _) x ->
let p = predicate x in flag || p,
if flag && p then None else Some x ) (false, None)
>> Seq.choose snd
This can then easily reused to do other things as well, like 0 together with odd numbers.
filterDuplicates (fun i -> i % 2 = 0) [0..10]
// val it : seq<int> = seq [0; 1; 3; 5; ...]
Supplied with a call to the equality operator and fed into the constructor of System.String, you'll get near the signature you want, char -> seq<char> -> System.String.
let filterDuplicatesOfChar what s =
System.String(Array.ofSeq <| filterDuplicates ((=) what) s)
filterDuplicatesOfChar '.' "hi. my .name."
// val it : string = "hi. my name"
The title basically says it all. I am aware that through the use of regexp via OCaml's Str module, this task is quite simple - however, suppose you are only allowed to use the standard library and Scanf module. I am interested in taking a string looks like this:
"12 34 555 6 23 34 5663 234 ..."
And returning an array that looks like this
[|12; 34; 555; 6; 23; 34; 5663; 234; |]
Can someone help me out? I found the Scanf documentation (available http://caml.inria.fr/pub/docs/manual-ocaml/libref/Scanf.html) to be pretty unhelpful in understanding how to use the module
Not sure how idiomatic it is, but it works:
let parse_integers s =
let stream = (Scanning.from_string s) in
let rec do_parse acc =
try
do_parse (Scanf.bscanf stream " %d " (fun x -> x :: acc))
with
Scan_failure _ -> acc
| End_of_file -> acc
in Array.of_list (List.rev (do_parse []));;
A little test:
# parse_integers " 20 3 22";;
- : int array = [|20; 3; 22|]
(Update)
As explained in comments, the above code is not tail-recursive, whereas the following one is:
...
let rec do_parse acc =
match (Scanf.bscanf stream " %d " (fun x -> x :: acc))
with
| xs -> do_parse xs
| exception Scan_failure _ -> acc
| exception End_of_file -> acc
in ...
let rec f acc s =
if s="" then
Array.of_list (List.rev acc)
else
Scanf.sscanf s "%d %[^\n]" (fun n s-> f (n::acc) s)
;;
f [] "12 34 555 6 23 34 5663 234";;
# - : int array = [|12; 34; 555; 6; 23; 34; 5663; 234|]
The format string for all the Scanf functions specifies a fixed number of values; thus, you can't expect to use a function from Scanf to read a variable number of values from a string.
If you break your string into pieces, you can use Scanf.sscanf s "%d" (fun x -> x) to translate each piece into an int. However, the function int_of_string is much simpler for that purpose.
I'd say you should start by breaking the string into pieces.
I found a great haskell solution (source) for generating a Hofstadter sequence:
hofstadter = unfoldr (\(r:s:ss) -> Just (r, r+s:delete (r+s) ss)) [1..]
Now, I am trying to write such a solution in F#, too. Unfortunately (I am not really familar to F#) I had no success so far.
My problem is, that when I use a sequence in F#, it seems not to be possible to remove an element (like it is done in the haskell solution).
Other data structures like arrays, list or set which allow to remove elements are not generating an infinite sequence, but operate on certain elements, only.
So my question: Is it possible in F# to generate an infinite sequence, where elements are deleted?
Some stuff I tried so far:
Infinite sequence of numbers:
let infinite =
Seq.unfold( fun state -> Some( state, state + 1) ) 1
Hofstadter sequence - not working, because there is no del keyword and there are more syntax errors
let hofstadter =
Seq.unfold( fun (r :: s :: ss) -> Some( r, r+s, del (r+s) ss)) infinite
I thought about using Seq.filter, but found no solution, either.
I think you need more than a delete function on sequence. Your example requires pattern matching on inifinite collections, which sequence doesn't support.
The F# counterpart of Haskell list is LazyList from F# PowerPack. LazyList is also potentially infinite and it supports pattern matching, which helps you to implement delete easily.
Here is a faithful translation:
open Microsoft.FSharp.Collections.LazyList
let delete x xs =
let rec loop x xs = seq {
match xs with
| Nil -> yield! xs
| Cons(x', xs') when x = x' -> yield! xs'
| Cons(x', xs') ->
yield x'
yield! loop x xs'
}
ofSeq (loop x xs)
let hofstadter =
1I |> unfold (fun state -> Some(state, state + 1I))
|> unfold (function | (Cons(r, Cons(s, ss))) ->
Some(r, cons (r+s) (delete (r+s) ss))
| _ -> None)
|> toSeq
There are a few interesting things here:
Use sequence expression to implement delete to ensure that the function is tail-recursive. A non-tail-recursive version should be easy.
Use BigInteger; if you don't need too many elements, using int and Seq.initInfinite is more efficient.
Add a case returning None to ensure exhaustive pattern matching.
At last I convert LazyList to sequence. It gives better interoperability with .NET collections.
Implementing delete on sequence is uglier. If you are curious, take a look at Remove a single non-unique value from a sequence in F# for reference.
pad's solution is nice but, likely due to the way LazyList is implemented, stack overflows somewhere between 3-4K numbers. For curiosity's sake I wrote a version built around a generator function (unit -> 'a) which is called repeatedly to get the next element (to work around the unwieldiness of IEnumerable). I was able to get the first 10K numbers (haven't tried beyond that).
let hofstadter() =
let delete x f =
let found = ref false
let rec loop() =
let y = f()
if not !found && x = y
then found := true; loop()
else y
loop
let cons x f =
let first = ref true
fun () ->
if !first
then first := false; x
else f()
let next =
let i = ref 0
fun () -> incr i; !i
Seq.unfold (fun next ->
let r = next()
let s = next()
Some(r, (cons (r+s) (delete (r+s) next)))) next
In fact, you can use filter and a design that follows the haskell solution (but, as #pad says, you don't have pattern matching on sequences; so I used lisp-style destruction):
let infinite = Seq.initInfinite (fun i -> i+1)
let generator = fun ss -> let (r, st) = (Seq.head ss, Seq.skip 1 ss)
let (s, stt) = (Seq.head st, Seq.skip 1 st)
let srps = seq [ r + s ]
let filtered = Seq.filter (fun t -> (r + s) <> t) stt
Some (r, Seq.append srps filtered)
let hofstadter = Seq.unfold generator infinite
let t10 = Seq.take 10 hofstadter |> Seq.toList
// val t10 : int list = [1; 3; 7; 12; 18; 26; 35; 45; 56; 69]
I make no claims about efficiency though!
I have a list as let a = ["q0,x";"q1,y"]; which is of type string list.
I want to make it as [("q0","x");("q1","y")]; which is a list of (string * string) tuples.
How do I do that??
You can use module Str and the function split :
let split =
List.map (fun str ->
match Str.split (Str.regexp ",") str with
| a :: b :: _ -> a, b
| _ -> assert false (* should not happen *))