I am new to F# and am having trouble with my code. Its a simple problem to define a function, search, with that take a boolean function and a list and return an index. So for example:
> search (fun x -> x > 10) [ 2; 12; 3; 23; 62; 8; 2 ];;
val it : int = 1
> search (fun s -> s < "horse") [ "pig"; "lion"; "horse"; "cow"; "turkey" ];;
val it : int = 3
What I have as of right now finds the right match but what I cant figure out is how to return a number instead of the rest of the list. I know I'm getting the list instead of a value back because I wrote "if f head then list". What I don't know is what I should put there instead or if what I have is not going to get the result I want.
Below is the code I have written.
let rec search f list =
match list with
| [] -> [-1]
| head::tail ->
if f head then list
else search f tail
Returning a number is easy, you just... return it. Your problem is that you don't have a number to return, because you can't derive it directly from the current state. You have to keep track of the number yourself, using some internal state variable.
When using recursion you change state by calling your function recursively with "modified" arguments. You're already doing that with the list here. To keep internal state in a recursive function you have to introduce another argument, but not expose it outside. You can solve that by using an internal recursive helper function. Here's one that keeps track of the previous item and returns that when it encounters a match:
let search f list =
let rec loop list prev =
match list with
| [] -> None
| head::tail ->
if f head then prev
else loop tail (Some head)
in
loop list None
That's a silly example, but I don't want to just solve your homework for you, because then you wouldn't learn anything. Using this you should be able to figure out how to keep a counter of which position the current item is in, and return that when it matches. Good luck!
You typically define an inner recursive function to help you carry state as you loop, and then call the inner function with an initial state.
let search predicate list =
let rec loop list index =
match list with
| [] -> -1
| head::tail ->
if predicate head then index
else loop tail (index + 1)
loop list 0
I am trying to properly understand how side effects work when traversing a list in F# using monadic style, following Scott's guide here
I have an AsyncSeq of items, and a side-effecting function that can return a Result<'a,'b> (it is saving the items to disk).
I get the general idea - split the head and tail, apply the func to the head. If it returns Ok then recurse through the tail, doing the same thing. If an Error is returned at any point then short circuit and return it.
I also get why Scott's ultimate solution uses foldBack rather than fold - it keeps the output list in the same order as the input as each processed item is prepended to the previous.
I can also follow the logic:
The result from the list's last item (processed first as we are using foldback) will be passed as the accumulator to the next item.
If it is an Error and the next item is Ok, the next item is discarded.
If the next item is an Error, it replaces any previous results and becomes the accumulator.
That means by the time you have recursed over the entire list from right to left and ended up at the start, you either have an Ok of all of the results in the correct order or the most recent Error (which would have been the first to occur if we had gone left to right).
The thing that confuses me is that surely, since we are starting at the end of the list, all side effects of processing every item will take place, even if we only get back the last Error that was created?
This seems to be confirmed here as the print output starts with [5], then [4,5], then [3,4,5] etc.
The thing that confuses me is that this isn't what I see happening when I use AsyncSeq.traverseChoiceAsync from the FSharpx lib (which I wrapped to process Result instead of Choice). I see side effects happening from left to right, stopping on the first error, which is what I want to happen.
It also looks like Scott's non-tail recursive version (which doesn't use foldBack and just recurses over the list) goes from left to right? The same goes for the AsyncSeq version. That would explain why I see it short circuit on the first error but surely if it completes Ok then the output items would be reversed, which is why we normally use foldback?
I feel I am misunderstanding or misreading something obvious! Could someone please explain it to me? :)
Edit:
rmunn has given a really great comprehensive explanation of the AsyncSeq traversal below. The TLDR was that
Scott's initial implementation and the AsyncSeq traverse both do go from left to right as I thought and so only process until they hit an error
they keep their contents in order by prepending the head to the processed tail rather than prepending each processed result to the previous (which is what the built in F# fold does).
foldback would keep things in order but would indeed execute every case (which could take forever with an async seq)
It's pretty simple: traverseChoiceAsync isn't using foldBack. Yes, with foldBack the last item would be processed first, so that by the time you get to the first item and discover that its result is Error you'd have triggered the side effects of every item. Which is, I think, precisely why whoever wrote traverseChoiceAsync in FSharpx chose not to use foldBack, because they wanted to ensure that side effects would be triggered in order, and stop at the first Error (or, in the case of the Choice version of the function, the first Choice2Of2 — but I'll pretend from this point on that that function was written to use the Result type.)
Let's look at the traverseChoieAsync function in the code you linked to, and read through it step-by-step. I'll also rewrite it to use Result instead of Choice, because the two types are basically identical in function but with different names in the DU, and it'll be a little easier to tell what's going on if the DU cases are called Ok and Error instead of Choice1Of2 and Choice2Of2. Here's the original code:
let rec traverseChoiceAsync (f:'a -> Async<Choice<'b, 'e>>) (s:AsyncSeq<'a>) : Async<Choice<AsyncSeq<'b>, 'e>> = async {
let! s = s
match s with
| Nil -> return Choice1Of2 (Nil |> async.Return)
| Cons(a,tl) ->
let! b = f a
match b with
| Choice1Of2 b ->
return! traverseChoiceAsync f tl |> Async.map (Choice.mapl (fun tl -> Cons(b, tl) |> async.Return))
| Choice2Of2 e ->
return Choice2Of2 e }
And here's the original code rewritten to use Result. Note that it's a simple rename, and none of the logic needs to be changed:
let rec traverseResultAsync (f:'a -> Async<Result<'b, 'e>>) (s:AsyncSeq<'a>) : Async<Result<AsyncSeq<'b>, 'e>> = async {
let! s = s
match s with
| Nil -> return Ok (Nil |> async.Return)
| Cons(a,tl) ->
let! b = f a
match b with
| Ok b ->
return! traverseChoiceAsync f tl |> Async.map (Result.map (fun tl -> Cons(b, tl) |> async.Return))
| Error e ->
return Error e }
Now let's step through it. The whole function is wrapped inside an async { } block, so let! inside this function means "unwrap" in an async context (essentially, "await").
let! s = s
This takes the s parameter (of type AsyncSeq<'a>) and unwraps it, binding the result to a local name s that henceforth will shadow the original parameter. When you await the result of an AsyncSeq, what you get is the first element only, while the rest is still wrapped in an async that needs to be further awaited. You can see this by looking at the result of the match expression, or by looking at the definition of the AsyncSeq type:
type AsyncSeq<'T> = Async<AsyncSeqInner<'T>>
and AsyncSeqInner<'T> =
| Nil
| Cons of 'T * AsyncSeq<'T>
So when you do let! x = s when s is of type AsyncSeq<'T>, the value of x will either be Nil (when the sequence has run to its end) or it will be Cons(head, tail) where head is of type 'T and tail is of type AsyncSeq<'T>.
So after this let! s = s line, our local name s now refers to an AsyncSeqInner type, which contains the head item of the sequence (or Nil if the sequence was empty), and the rest of the sequence is still wrapped in an AsyncSeq so it has yet to be evaluated (and, crucially, its side effects have not yet happened).
match s with
| Nil -> return Ok (Nil |> async.Return)
There's a lot happening in this line, so it'll take a bit of unpacking, but the gist is that if the input sequence s had Nil as its head, i.e. had reached its end, then that's not an error, and we return an empty sequence.
Now to unpack. The outer return is in an async keyword, so it takes the Result (whose value is Ok something) and turns it into an Async<Result<something>>. Remembering that the return type of the function is declared as Async<Result<AsyncSeq>>, the inner something is clearly an AsyncSeq type. So what's going on with that Nil |> async.Return? Well, async isn't an F# keyword, it's the name of an instance of AsyncBuilder. Inside a computation expression foo { ... }, return x is translated into foo.Return(x). So calling async.Return x is just the same as writing async { return x }, except that it avoids nesting a computation expression inside another computation expression, which would be a little nasty to try and parse mentally (and I'm not 100% sure the F# compiler allows it syntactically). So Nil |> async.Return is async.Return Nil which means it produces a value of Async<x> where x is the type of the value Nil. And as we just saw, this Nil is a value of type AsyncSeqInner, so Nil |> async.Return produces an Async<AsyncSeqInner>. And another name for Async<AsyncSeqInner> is AsyncSeq. So this whole expression produces an Async<Result<AsyncSeq>> that has the meaning of "We're done here, there are no more items in the sequence, and there was no error".
Phew. Now for the next line:
| Cons(a,tl) ->
Simple: if the next item in the AsyncSeq named s was a Cons, we deconstruct it so that the actual item is now called a, and the tail (another AsyncSeq) is called tl.
let! b = f a
This calls f on the value we just got out of s, and then unwraps the Async part of f's return value, so that b is now a Result<'b, 'e>.
match b with
| Ok b ->
More shadowed names. Inside this branch of the match, b now names a value of type 'b rather than a Result<'b, 'e>.
return! traverseResultAsync f tl |> Async.map (Result.map (fun tl -> Cons(b, tl) |> async.Return))
Hoo boy. That's too much to tackle at once. Let's write this as if the |> operators were lined up on separate lines, and then we'll go through each step one at a time. (Note that I've wrapped an extra pair of parentheses around this, just to clarify that it's the final result of this whole expression that will be passed to the return! keyword).
return! (
traverseResultAsync f tl
|> Async.map (
Result.map (
fun tl -> Cons(b, tl) |> async.Return)))
I'm going to tackle this expression from the inside out. The inner line is:
fun tl -> Cons(b, tl) |> async.Return
The async.Return thing we've already seen. This is a function that takes a tail (we don't currently know, or care, what's inside that tail, except that by the necessity of the type signature of Cons it must be an AsyncSeq) and turns it into an AsyncSeq that is b followed by the tail. I.e., this is like b :: tl in a list: it sticks b onto the front of the AsyncSeq.
One step out from that innermost expression is:
Result.map
Remember that the function map can be thought of in two ways: one is "take a function and run it against whatever is "inside" this wrapper". The other is "take a function that operates on 'T and make it into a function that operates on Wrapper<'T>". (If you don't have both of those clear in your mind yet, https://sidburn.github.io/blog/2016/03/27/understanding-map is a pretty good article to help grok that concept). So what this is doing is taking a function of type AsyncSeq -> AsyncSeq and turning it into a function of type Result<AsyncSeq> -> Result<AsyncSeq>. Alternately, you could think of it as taking a Result<tail> and calling fun tail -> ... against that tail result, then re-wrapping the result of that function in a new Result. Important: Because this is using Result.map (Choice.mapl in the original) we know that if tail is an Error value (or if the Choice was a Choice2Of2 in the original), the function will not be called. So if traverseResultAsync produces a result that starts with an Error value, it's going to produce an <Async<Result<foo>>> where the value of Result<foo> is an Error, and so the value of the tail will be discarded. Keep that in mind for later.
Okay, next step out.
Async.map
Here, we have a Result<AsyncSeq> -> Result<AsyncSeq> function produced by the inner expression, and this converts it to an Async<Result<AsyncSeq>> -> Async<Result<AsyncSeq>> function. We've just talked about this, so we don't need to go over how map works again. Just remember that the effect of this Async<Result<AsyncSeq>> -> Async<Result<AsyncSeq>> function that we've built up will be the following:
Await the outer async.
If the result is Error, return that Error.
If the result is Ok tail, produce an Ok (Cons (b, tail)).
Next line:
traverseResultAsync f tl
I probably should have started with this, because this will actually run first, and then its value will be passed into the Async<Result<AsyncSeq>> -> Async<Result<AsyncSeq>> function that we've just analysed.
So what this whole thing will do is to say "Okay, we took the first part of the AsyncSeq we were handed, and passed it to f, and f produced an Ok result with a value we're calling b. So now we need to process the rest of the sequence similarly, and then, if the rest of the sequence produces an Ok result, we'll stick b on the front of it and return an Ok sequence with contents b :: tail. BUT if the rest of the sequence produces an Error, we'll throw away the value of b and just return that Error unchanged."
return!
This just takes the result we just got (either an Error or an Ok (b :: tail), already wrapped in an Async) and returns it unchanged. But note that the call to traverseResultAsync is NOT tail-recursive, because its value had to be passed into the Async.map (...) expression first.
And now we still have one more bit of traverseResultAsync to look at. Remember when I said "Keep that in mind for later"? Well, that time has arrived.
| Error e ->
return Error e }
Here we're back in the match b with expression. If b was an Error result, then no further recursive calls are made, and the whole traverseResultAsync returns an Async<Result> where the Result value is Error. And if we were currently nested deep inside a recursion (i.e., we're in the return! traverseResultAsync ... expression), then our return value will be Error, which means the result of the "outer" call, as we've kept in mind, will also be Error, discarding any other Ok results that might have happened "before".
Conclusion
And so the effect of all of that is:
Step through the AsyncSeq, calling f on each item in turn.
The first time f returns Error, stop stepping through, throw away any previous Ok results, and return that Error as the result of the whole thing.
If f never returns Error and instead returns Ok b every time, return an Ok result that contains an AsyncSeq of all those b values, in their original order.
Why are they in their original order? Because the logic in the Ok case is:
If sequence was empty, return an empty sequence.
Split into head and tail.
Get value b from f head.
Process the tail.
Stick value b in front of the result of processing the tail.
So if we started with (conceptually) [a1; a2; a3], which actually looks like Cons (a1, Cons (a2, Cons (a3, Nil))) we'll end up with Cons (b1, Cons (b2, Cons (b3, Nil))) which translates to the conceptual sequence [b1; b2; b3].
See #rmunn's great answer above for the explanation. I just wanted to post a little helper for anyone that reads this in the future, it allows you to use the AsyncSeq traverse with Results instead of the old Choice type it was written with:
let traverseResultAsyncM (mapping : 'a -> Async<Result<'b,'c>>) source =
let mapping' =
mapping
>> Async.map (function
| Ok x -> Choice1Of2 x
| Error e -> Choice2Of2 e)
AsyncSeq.traverseChoiceAsync mapping' source
|> Async.map (function
| Choice1Of2 x -> Ok x
| Choice2Of2 e -> Error e)
Also here is a version for non-async mappings:
let traverseResultM (mapping : 'a -> Result<'b,'c>) source =
let mapping' x = async {
return
mapping x
|> function
| Ok x -> Choice1Of2 x
| Error e -> Choice2Of2 e
}
AsyncSeq.traverseChoiceAsync mapping' source
|> Async.map (function
| Choice1Of2 x -> Ok x
| Choice2Of2 e -> Error e)
let memoize (sequence: seq<'a>) =
let cache = Dictionary()
seq {for i in sequence ->
match cache.TryGetValue i with
| true, v -> printf "cached"
| false,_ -> cache.Add(i ,i)
}
I will call my memoize function inside this function :
let isCached (input:seq<'a>) : seq<'a> = memoize input
If the given sequence item is cached it should print cached otherwise it will continue to add sequence value to cache.
Right now I have problems with types.
When I try to call my function like this :
let seq1 = seq { 1 .. 10 }
isCached seq1
It throws an error
"The type int does not match the type unit"
I want my function to work generic even though I return printfn. Is it possible to achieve that? And while adding value to the cache is it appropriate to give the same value to tuple?
eg:
| false,_ -> cache.Add(i ,i)
I think the problem is that your memoize function does not actually return the item from the source sequence as a next element of the returned sequence. Your version only adds items to the cache, but then it returns unit. You can fix that by writing:
let memoize (sequence: seq<'a>) =
let cache = Dictionary()
seq {for i in sequence do
match cache.TryGetValue i with
| true, v -> printf "cached"
| false,_ -> cache.Add(i ,i)
yield i }
I used explicit yield rather than -> because I think that makes the code more readable. With this change, the code runs as expected for me.
Tomas P beat me to the punch, but I'll post this up anyway just in case it helps.
I'm not too sure what you are trying to achieve here, but I'll say a few things that I think might help.
Firstly, the type error. Your isCached function is defined as taking a seq of type 'a, and returning a seq of type 'a. As written in your question, right now it takes a seq of type 'a, and returns a sequence of type unit. If you try modifying the output specification to seq<'b> (or actually just omitting it altogether and letting type inference do it), you should overcome the type error. This probably still won't do what you want, since you aren't actually returning the cache from that function (you can just add cache as the final line to return it). Thus, try something like:
let memoize (sequence: seq<'a>) =
let cache = Dictionary()
for i in sequence do
match cache.TryGetValue i with
| true, v -> printf "cached"
| false,_ -> cache.Add(i ,i)
cache
let isCached (input:seq<'a>) : seq<'b> = memoize input
All this being said, if you are expecting to iterate over the same sequence a lot, it might be best just to use the library function Seq.cache.
Finally, with regards to using the value as the key in the dictionary... There's nothing stopping you from doing that, but it's really fairly pointless. If you already have a value, then you shouldn't need to look it up in the dictionary. If you are just trying to memoize the sequence, then use the index of the given element as the key. Or use the specific input as the key and the output from that input as the value.
I need to pass two list as command line arguments in ocaml.
I used the following code to access it in the program.
let list1=Sys.argv.(1);;
let list2=Sys.argv.(2);;
I need to have the list1 and list2 as list of integers.
I am getting the error
This expression has type string but an expression was expected of type
int list
while processing.
How can I convert that arguments to a list of integers.
The arguments are passed in this format [1;2;3;4] [1;5;6;7]
Sys.argv.(n) will always be a string. You need to parse the string into a list of integers. You could try something like this:
$ ocaml
OCaml version 4.01.0
# #load "str.cma";;
# List.map int_of_string (Str.split (Str.regexp "[^0-9]+") "[1;5;6;7]");;
- : int list = [1; 5; 6; 7]
Of course this doesn't check the input for correct form. It just pulls out sequences of digits by brute force. To do better you need to do some real lexical analysis and simple parsing.
(Maybe this is obvious, but you could also test your function in the toplevel (the OCaml read-eval-print loop). The toplevel will handle the work of making a list from what you type in.)
As Sys.argv is a string array, you need to write your own transcription function.
I guess the simplest way to do this is to use the Genlex module provided by the standard library.
let lexer = Genlex.make_lexer ["["; ";"; "]"; ]
let list_of_string s =
let open Genlex in
let open Stream in
let stream = lexer (of_string s) in
let fail () = failwith "Malformed string" in
let rec aux acc =
match next stream with
| Int i ->
( match next stream with
| Kwd ";" -> aux (i::acc)
| Kwd "]" -> i::acc
| _ -> fail () )
| Kwd "]" -> acc
| _ -> fail ()
in
try
match next stream with
| Kwd "[" -> List.rev (aux [])
| _ -> fail ()
with Stream.Failure -> fail ()
let list1 = list_of_string Sys.argv.(1)
let list2 = list_of_string Sys.argv.(2)
Depending on the OCaml flavor you want to use, some other library may look more interesting. If you like yacc, Menhir may solve your problem in a few lines of code.
I have a recursive function that contains a series of matches that either make the recursive call back to the function, or make a call to failwith.
This is basically a hybrid implementation of the recursive descent parser descibed in Don Syme's Expert F# book (page 180) and the parsing example shown here: http://fsharpforfunandprofit.com/posts/pattern-matching-command-line/
Here is a snippet of my own code.
let rec parseTokenListRec tokenList optionsSoFar =
match tokenList with
| [] -> optionsSoFar
| SOURCE::t ->
match t with
| VALUE x::tt -> parseTokenListRec (returnNonValueTail t) {optionsSoFar with Source = (returnConcatHeadValues t)}
| _ -> failwith "Expected a value after the source argument."
| REGISTRY::t ->
...
A full code listing can be found at http://fssnip.net/nU
The way the code is currently written, when the function has finished working its way through the tokenList, it will return the optionsSoFar record that has been compiled via the object expression {optionsSoFar with Source = (returnConcatHeadValues t)}, or it will throw an exception if an invalid argument is found.
I want to refactor this so that the function does not rely on an exception, but will always return a value of some sort that can be handled by the calling function. The idea I have is to return a discriminated union rather than a record.
This discriminated union would be something like
type Result =
|Success of Options
|Failure of string
The problem I had when I tried to refactor the code was that I couldn't figure out how to get the success value of the DU to initialize via an object expression. Is this possible?
The examples I've looked at on MSDN (http://msdn.microsoft.com/en-us/library/vstudio/dd233237(v=vs.100).aspx), fsharpforfunandprofit (http://fsharpforfunandprofit.com/posts/discriminated-unions/) and elsewhere haven't quite cleared this up for me.
I'm worried that I'm not making any sense here. I'm happy to clarify if needed.
If I understand it correctly, in you current solution, the type of optionsSoFar is Options. The code becomes trickier if you change the type of optionsSoFar to your newly defined Result.
However, I think you do not need to do that - you can keep optionsSoFar : Options and change the function to return Result. This works because you never need to call the function recursively after it fails:
let rec parseTokenListRec tokenList optionsSoFar =
match tokenList with
| [] -> Success optionsSoFar
| SOURCE::t ->
match t with
| VALUE x::tt ->
{optionsSoFar with Source = (returnConcatHeadValues t)}
|> parseTokenListRec (returnNonValueTail t)
| _ -> Failure "Expected a value after the source argument."
| REGISTRY::t -> ...
If you actually wanted to update Source in a Result value, then I'd probably write something like:
module Result =
let map f = function
| Success opt -> f opt
| Failure msg -> Failure msg
Then you could write a transformation as follows:
resultSoFar
|> Result.map (fun opts -> {opts with Source = returnConcatHeadValues t})
|> parseTokenListRec (returnNonValueTail t)