I am currently doing reasonably well in functional programming using F#. I tend, however, to do a lot of programming using recursion, when it seems that there are better idioms in the F#/functional programming community. So in the spirit of learning, is there a better/more idiomatic way of writing the function below without recursion?
let rec convert line =
if line.[0..1] = " " then
match convert line.[2..] with
| (i, subline) -> (i+1, subline)
else
(0, line)
with results such as:
> convert "asdf";;
val it : int * string = (0, "asdf")
> convert " asdf";;
val it : int * string = (1, "asdf")
> convert " asdf";;
val it : int * string = (3, "asdf")
Recursion is the basic mechanism for writing loops in functional languages, so if you need to iterate over characters (as you do in your sample), then recursion is what you need.
If you want to improve your code, then you should probably avoid using line.[2..] because that is going to be inefficient (strings are not designed for this kind of processing). It is better to convert the string to a list and then process it:
let convert (line:string) =
let rec loop acc line =
match line with
| ' '::' '::rest -> loop (acc + 1) rest
| _ -> (acc, line)
loop 0 (List.ofSeq line)
You can use various functions from the standard library to implement this in a more shorter way, but they are usually recursive too (you just do not see the recursion!), so I think using functions like Seq.unfold and Seq.fold is still recursive (and it looks way more complex than your code).
A more concise approach using standard libraries is to use the TrimLeft method (see comments), or using standard F# library functions, do something like this:
let convert (line:string) =
// Count the number of spaces at the beginning
let spaces = line |> Seq.takeWhile (fun c -> c = ' ') |> Seq.length
// Divide by two - we want to count & skip two-spaces only
let count = spaces / 2
// Get substring starting after all removed two-spaces
count, line.[(count * 2) ..]
EDIT Regarding the performance of string vs. list processing, the problem is that slicing allocates a new string (because that is how strings are represented on the .NET platform), while slicing a list just changes a reference. Here is a simple test:
let rec countList n s =
match s with
| x::xs -> countList (n + 1) xs
| _ -> n
let rec countString n (s:string) =
if s.Length = 0 then n
else countString (n + 1) (s.[1 ..])
let l = [ for i in 1 .. 10000 -> 'x' ]
let s = new System.String('x', 10000)
#time
for i in 0 .. 100 do countList 0 l |> ignore // 0.002 sec (on my machine)
for i in 0 .. 100 do countString 0 s |> ignore // 5.720 sec (on my machine)
Because you traverse the string in a non-uniform way, a recursive solution is much more suitable in this example. I would rewrite your tail-recursive solution for readability as follows:
let convert (line: string) =
let rec loop i line =
match line.[0..1] with
| " " -> loop (i+1) line.[2..]
| _ -> i, line
loop 0 line
Since you asked, here is a (bizarre) non-recursive solution :).
let convert (line: string) =
(0, line) |> Seq.unfold (fun (i, line) ->
let subline = line.[2..]
match line.[0..1] with
| " " -> Some((i+1, subline), (i+1, subline))
| _ -> None)
|> Seq.fold (fun _ x -> x) (0, line)
Using tail recursion, it can be written as
let rec convert_ acc line =
if line.[0..1] <> " " then
(acc, line)
else
convert_ (acc + 1) line.[2..]
let convert = convert_ 0
still looking for a non-recursive answer, though.
Here's a faster way to write your function -- it checks the characters explicitly instead of using string slicing (which, as Tomas said, is slow); it's also tail-recursive. Finally, it uses a StringBuilder to create the "filtered" string, which will provide better performance once your input string reaches a decent length (though it'd be a bit slower for very small strings due to the overhead of creating the StringBuilder).
let convert' str =
let strLen = String.length str
let sb = System.Text.StringBuilder strLen
let rec convertRec (count, idx) =
match strLen - idx with
| 0 ->
count, sb.ToString ()
| 1 ->
// Append the last character in the string to the StringBuilder.
sb.Append str.[idx] |> ignore
convertRec (count, idx + 1)
| _ ->
if str.[idx] = ' ' && str.[idx + 1] = ' ' then
convertRec (count + 1, idx + 2)
else
sb.Append str.[idx] |> ignore
convertRec (count, idx + 1)
// Call the internal, recursive implementation.
convertRec (0, 0)
Related
I am absolute OCaml beginner. I want to create a function that repeats characters 20 times.
This is the function, but it does not work because of an error.
let string20 s =
let n = 20 in
s ^ string20 s (n - 1);;
string20 "u";;
I want to run like this
# string20 "u"
- : string = "uuuuuuuuuuuuuuuuuuuu"
Your function string20 takes one parameter but you are calling it recursively with 2 parameters.
The basic ideas are in there, but not quite in the right form. One way to proceed is to separate out the 2-parameter function as a separate "helper" function. As #PierreG points out, you'll need to delcare the helper function as a recursive function.
let rec string n s =
if n = 0 then "" else s ^ string (n - 1) s
let string20 = string 20
It is a common pattern to separate a function into a "fixed" part and inductive part. In this case, a nested helper function is needed to do the real recursive work in a new scope while we want to fix an input string s as a constant so we can use to append to s2. s2 is an accumulator that build up the train of strings over time while c is an inductor counting down to 1 toward the base case.
let repeat s n =
let rec helper s1 n1 =
if n1 = 0 then s1 else helper (s1 ^ s) (n1 - 1)
in helper "" n
A non-tail call versions is more straightforward since you won't need a helper function at all:
let rec repeat s n =
if n = 0 then "" else s ^ repeat s (n - 1)
On the side note, one very fun thing about a functional language with first-class functions like Ocaml is currying (or partial application). In this case you can create a function named repeat that takes two arguments n of type int and s of type string as above and partially apply it to either n or s like this:
# (* top-level *)
# let repeat_foo = repeat "foo";;
# repeat_foo 5;;
- : bytes = "foofoofoofoofoo" (* top-level output *)
if the n argument was labeled as below:
let rec repeat ?(n = 0) s =
if n = 0 then "" else s ^ repeat s (n - 1)
The order of application can be exploited, making the function more flexible:
# (* top-level *)
# let repeat_10 = repeat ~n:10;;
# repeat_10 "foo";;
- : bytes = "foofoofoofoofoofoofoofoofoofoo" (* top-level output *)
See my post Currying Exercise in JavaScript (though it is in JavaScript but pretty simple to follow) and this lambda calculus primer.
Recursive functions in Ocaml are defined with let rec
As pointed out in the comments you've defined your function to take one parameter but you're trying to recursively call with two.
You probably want something like this:
let rec stringn s n =
match n with
1 -> s
| _ -> s ^ stringn s (n - 1)
;;
I'm currently generating a sequence in a similar way to:
migrators
|> Seq.map (fun m -> m())
The migrator function is ultimately returning a discriminated union like:
type MigratorResult =
| Success of string * TimeSpan
| Error of string * Exception
I want to stop the map once I encounter my first Error but I need to include the Error in the final sequence.
I have something like the following to display a final message to the user
match results |> List.rev with
| [] -> "No results equals no migrators"
| head :: _ ->
match head with
| Success (dt, t) -> "All migrators succeeded"
| Error (dt, ex) -> "Migration halted owing to error"
So I need:
A way to stop the mapping when one of the map steps produces an Error
A way to have that error be the final element added to the sequence
I appreciate there may be a different sequence method other than map that will do this, I'm new to F# and searching online hasn't yielded anything as yet!
I guess there are multiple approaches here, but one way would be to use unfold:
migrators
|> Seq.unfold (fun ms ->
match ms with
| m :: tl ->
match m () with
| Success res -> Some (Success res, tl)
| Error res -> Some (Error res, [])
| [] -> None)
|> List.ofSeq
Note the List.ofSeq at the end, that's just there for realizing the sequence. A different way to go would be to use sequence comprehensions, some might say it results in a clearer code.
The ugly things Tomaš alludes to are 1) mutable state, and 2) manipulation of the underlying enumerator. A higher-order function which returns up to and including when the predicate holds would then look like this:
module Seq =
let takeUntil pred (xs : _ seq) = seq{
use en = xs.GetEnumerator()
let flag = ref true
while !flag && en.MoveNext() do
flag := not <| pred en.Current
yield en.Current }
seq{1..10} |> Seq.takeUntil (fun x -> x % 5 = 0)
|> Seq.toList
// val it : int list = [1; 2; 3; 4; 5]
For your specific application, you'd map the cases of the DU to a boolean.
(migrators : seq<MigratorResult>)
|> Seq.takeUntil (function Success _ -> false | Error _ -> true)
I think the answer from #scrwtp is probably the nicest way to do this if your input is reasonably small (and you can turn it into an F# list to use pattern matching). I'll add one more version, which works when your input is just a sequence and you do not want to turn it into a list.
Essentially, you want to do something that's almost like Seq.takeWhile, but it gives you one additional item at the end (the one, for which the predicate fails).
To use a simpler example, the following returns all numbers from a sequence until one that is divisible by 5:
let nums = [ 2 .. 10 ]
nums
|> Seq.map (fun m -> m % 5)
|> Seq.takeWhile (fun n -> n <> 0)
So, you basically just need to look one element ahead - to do this, you could use Seq.pairwise which gives you the current and the next element in the sequence"
nums
|> Seq.map (fun m -> m % 5)
|> Seq.pairwise // Get sequence of pairs with the next value
|> Seq.takeWhile (fun (p, n) -> p <> 0) // Look at the next value for test
|> Seq.mapi (fun i (p, n) -> // For the first item, we return both
if i = 0 then [p;n] else [n]) // for all other, we return the second
|> Seq.concat
The only ugly thing here is that you then need to flatten the sequence again using mapi and concat.
This is not very nice, so a good thing to do would be to define your own higher-order function like Seq.takeUntilAfter that encapsulates the behavior you need (and hides all the ugly things). Then your code could just use the function and look nice & readable (and you can experiment with other ways of implementing this).
I just begin OCaml (and functional programming) today and I'm trying to code a function that count the number of occurrences of "value" into an array (tab).
I tried :
let rec count_occ tab value =
let rec count_rec idx time = function
| tab.length - 1 -> time
| _ when tab.(indice) == value-> count_rec (idx + 1) (time + 1)
| _ -> count_rec (indice + 1) time
in
count_rec 0 0
;;
Unfortunately, it doesn't compile because of a syntax error, and I don't find the solution.
let rec count_occ tab value =
This rec above is not necessary.
let rec count_rec idx time = function
| tab.length - 1 -> time
You cannot match against an expression. You want to use guards like you did on the next line, or if statements to test something like this. tab.length also does not exist as tab is an array, not a record with a length field. You want Array.length tab.
Really though, you don't want the function at all. function is the same as fun x -> match x with, and would imply that count_rec has type, int -> int -> int -> int.
| _ when tab.(indice) == value-> count_rec (idx + 1) (time + 1)
indices is not declared; lets assume you meant idx. Also, == is physical equality, you really want =.
| _ -> count_rec (indice + 1) time
in
count_rec 0 0
You're off to a good start, the basics of your recursion are correct although one edge case is incorrect, but a minor issue you should be able to resolve once you have the syntactic issues fixed.
finnaly I post my final code :
let count_occ tab value =
let rec count_rec idx time =
if (Array.length tab) = idx then
time
else if (tab.(idx)) = value then
count_rec (idx + 1) (time + 1)
else
count_rec (idx + 1) time
in
count_rec 0 0
;;
let _ as s = "abc" in s ^ "def"
So how should understand this?
I guess it is some kind of let pattern = expression thing?
First, what's the meaning/purpose/logic of let pattern = expression?
Also, in pattern matching, I know there is pattern as identifier usage, in let _ as s = "abc" in s ^ "def", _ is pattern, but behind as, it is an expression s = "abc" in s ^ "def", not an identifier, right?
edit:
finally, how about this: (fun (1 | 2) as i -> i + 1) 2, is this correct?
I know it is wrong, but why? fun pattern -> expression is allowed, right?
I really got lost here.
The grouping is let (_ as s) = "abc" -- which is just a convoluted way of saying let s = "abc", because as with a wildcard pattern _ in front is pretty much useless.
The expression let pattern = expr1 in expr2 is pretty central to OCaml. If the pattern is just a name, it lets you name an expression. This is like a local variable in other language. If the pattern is more complicated, it lets you destructure expr1, i.e., it lets you give names to its components.
In your expression, behind as is just an identifier: s. I suspect your confusion all comes down to this one thing. The expression can be parenthesized as:
let (_ as s) = "abc" in s ^ "def"
as Andreas Rossberg shows.
Your final example is correct if you add some parentheses. The compiler/toplevel rightly complains that your function is partial; i.e., it doesn't know what to do with most ints,
only with 1 and 2.
Edit: here's a session that shows how to add the parentheses to your final example:
$ ocaml
OCaml version 4.00.0
# (fun (1 | 2) as i -> i + 1) 2;;
Error: Syntax error
# (fun ((1 | 2) as i) -> i + 1) 2;;
Warning 8: this pattern-matching is not exhaustive.
Here is an example of a value that is not matched:
0
- : int = 3
#
Edit 2: here's a session that shows how to remove the warning by specifying an exhaustive set of patterns.
$ ocaml
OCaml version 4.00.0
# (function ((1|2) as i) -> i + 1 | _ -> -1) 2;;
- : int = 3
# (function ((1|2) as i) -> i + 1 | _ -> -1) 3;;
- : int = -1
#
Take this example code (ignore it being horribly inefficient for the moment)
let listToString (lst:list<'a>) = ;;' prettify fix
let rec inner (lst:list<'a>) buffer = ;;' prettify fix
match List.length lst with
| 0 -> buffer
| _ -> inner (List.tl lst) (buffer + ((List.hd lst).ToString()))
inner lst ""
This is a common pattern I keep coming across in F#, I need to have an inner function who recurses itself over some value - and I only need this function once, is there in any way possible to call a lambda from within it self (some magic keyword or something) ? I would like the code to look something like this:
let listToString2 (lst:list<'a>) = ;;' prettify fix
( fun
(lst:list<'a>) buffer -> match List.length lst with ;;' prettify fix
| 0 -> buffer
| _ -> ##RECURSE## (List.tl lst) (buffer + ((List.hd lst).ToString()))
) lst ""
But as you might expect there is no way to refer to the anonymous function within itself, which is needed where I put ##RECURSE##
Yes, it's possible using so called y-combinators (or fixed-point combinators). Ex:
let rec fix f x = f (fix f) x
let fact f = function
| 0 -> 1
| x -> x * f (x-1)
let _ = (fix fact) 5 (* evaluates to "120" *)
I don't know articles for F# but this haskell entry might also be helpful.
But: I wouldn't use them if there is any alternative - They're quite hard to understand.
Your code (omit the type annotations here) is a standard construct and much more expressive.
let listToString lst =
let rec loop acc = function
| [] -> acc
| x::xs -> loop (acc ^ (string x)) xs
loop "" lst
Note that although you say you use the function only once, technically you refer to it by name twice, which is why it makes sense to give it a name.