Just messing with elixir result got an error that I cannot figure out.Here is my snippet,I implemented a simple Parallel map function for some api calls. func is the function where the actual calls are made and it returns {:ok,result} or {:error,reason} which I handle after the mapping in a different function
Originally
def pmap(collection,func,limit \\ 5000) do
collection
|> Enum.map(&Task.async(func.(&1)))
|> Enum.map(&Task.await(&1,limit))
Got the error so changed it to this for readability
def pmap(collection,func,limit) do
collection
|>Enum.map(fn(x) -> Task.async(func.(x)) end)
|>Enum.map(fn(task) -> Task.await(task,limit) end)
The error I am getting states
[error] Task #PID<0.197.0> started from #PID<0.187.0> terminating
** (BadFunctionError) expected a function, got: {:ok,result}
erlang.apply/2
(elixir) lib/task/supervised.ex:85: Task.Supervised.do_apply/2
(elixir) lib/task/supervised.ex:36: Task.Supervised.reply/5
(stdlib) proc_lib.erl:247: :proc_lib.init_p_do_apply/3
Function: &:erlang.apply/2
From what I gather it is assuming that the actual task collection element is being passed as the function
So I modified the function to
IO.puts "PMAP BEGUN"
tasks = collection
|> Enum.map(fn(x) -> Task.async(func.(x)) end)
answer = Enum.map(tasks,fn(task) -> Task.await(task,limit) end)
IO.puts "PMAP DONE"
answer
The IO.puts were for debugging.So I guess the error happens on the second map as PMAP DONE is never displayed. I still have the same error.
What exactly is wrong here? I have written the same function before almost verbatim and it worked.
The problem you're experiencing happens on line:
|> Enum.map(fn(x) -> Task.async(func.(x)) end)
to make it work you need to wrap function execution with anonymous function, like:
|> Enum.map(fn(x) -> Task.async(fn -> func.(x) end) end)
This due to fact if you run func.(x) this will evaluate, and the result of evaluation will be passed to Task.async/1, while when you wrap it with function, it will be up to Task to execute it.
The hint is indicated in error message:
** (BadFunctionError) expected a function, got: {:ok,result}
Hope this helps!
Related
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)
If I have a method
macro doarray(arr)
if in(:head, fieldnames(typeof(arr))) && arr.head == :vect
println("A Vector")
else
throw(ArgumentError("$(arr) should be a vector"))
end
end
it works if I write this
#doarray([x])
or
#doarray([:x])
but the following code rightly does not work, raising the ArgumentError(i.e. ArgumentError: alist should be a vector).
alist = [:x]
#doarray(alist)
How can I make the above to act similarly as #doarray([x])
Motivation:
I have a recursive macro(say mymacro) which takes a vector, operates on the first value and then calls recursively mymacro with the rest of the vector(say rest_vector). I can create rest_vector, print the value correctly(for debugging) but I don't know how to evaluate rest_vector when I feed it to the mymacro again.
EDIT 1:
I'm trying to implement logic programming in Julia, namely MiniKanren. In the Clojure implementation that I am basing this off, the code is such.
(defmacro fresh
[var-vec & clauses]
(if (empty? var-vec)
`(lconj+ ~#clauses)
`(call-fresh (fn [~(first var-vec)]
(fresh [~#(rest var-vec)]
~#clauses)))))
My failing Julia code based on that is below. I apologize if it does not make sense as I am trying to understand macros by implementing it.
macro fresh(varvec, clauses...)
if isempty(varvec.args)
:(lconjplus($(esc(clauses))))
else
varvecrest = varvec.args[2:end]
return quote
fn = $(esc(varvec.args[1])) -> #fresh($(varvecvest), $(esc(clauses)))
callfresh(fn)
end
end
end
The error I get when I run the code #fresh([x, y], ===(x, 42))(you can disregard ===(x, 42) for this discussion)
ERROR: LoadError: LoadError: UndefVarError: varvecvest not defined
The problem line is fn = $(esc(varvec.args[1])) -> #fresh($(varvecvest), $(esc(clauses)))
If I understand your problem correctly it is better to call a function (not a macro) inside a macro that will operate on AST passed to the macro. Here is a simple example how you could do it:
function recarray(arr)
println("head: ", popfirst!(arr.args))
isempty(arr.args) || recarray(arr)
end
macro doarray(arr)
if in(:head, fieldnames(typeof(arr))) && arr.head == :vect
println("A Vector")
recarray(arr)
else
throw(ArgumentError("$(arr) should be a vector"))
end
end
Of course in this example we do not do anything useful. If you specified what exactly you want to achieve then I might suggest something more specific.
Suppose I have a function such as:
query_server : Server.t -> string Or_error.t Deferred.t
Then I produce a list of deferred queries:
let queries : string Or_error.t Deferred.t list = List.map servers ~f:query_server
How to get the result of the first query that doesn't fail (or some error otherwise). Basically, I'd like a function such as:
any_non_error : 'a Or_error.t Deferred.t list -> 'a Or_error.t
Also, I'm not sure how to somehow aggregate the errors. Maybe my function needs an extra parameter such as Error.t -> Error.t -> Error.t or is there a standard way to combine errors?
A simple approach would be to use Deferred.List that contains list operations lifted into the Async monad, basically a container interface in the Kleisli category. We will try each server in order until the first one is ready, e.g.,
let first_non_error =
Deferred.List.find ~f:(fun s -> query_server s >>| Result.is_ok)
Of course, it is not any_non_error, as the processing is sequential. Also, we are losing the error information (though the latter is very easy to fix).
So to make it parallel, we will employ the following strategy. We will have two deferred computations, the first will run all queries in parallel and wait until all are ready, the second will become determined as soon as an Ok result is received. If the first one happens before the last one, then it means that all servers failed. So let's try:
let query_servers servers =
let a_success,got_success = Pipe.create () in
let all_errors = Deferred.List.map ~how:`Parallel servers ~f:(fun s ->
query_server s >>| function
| Error err as e -> e
| Ok x as ok -> Pipe.write_without_pushback x; ok) in
Deferred.any [
Deferred.any all_errors;
Pipe.read a_success >>= function
| `Ok x -> Ok x
| `Eof -> assert false
]
This snippet of F# code
let rec reformat = new EventHandler(fun _ _ ->
b.TextChanged.RemoveHandler reformat
b |> ScrollParser.rewrite_contents_of_rtb
b.TextChanged.AddHandler reformat
)
b.TextChanged.AddHandler reformat
results in the following warning:
traynote.fs(62,41): warning FS0040: This and other recursive references to the object(s) being defined will be checked for initialization-soundness at runtime through the use of a delayed reference. This is because you are defining one or more recursive objects, rather than recursive functions. This warning may be suppressed by using '#nowarn "40"' or '--nowarn:40'.
Is there a way in which the code can be rewritten to avoid this warning? Or is there no kosher way of having recursive objects in F#?
Your code is a perfectly fine way to construct a recursive object. The compiler emits a warning, because it cannot guarantee that the reference won't be accessed before it is initialized (which would cause a runtime error). However, if you know that EventHandler does not call the provided lambda function during the construction (it does not), then you can safely ignore the warning.
To give an example where the warning actually shows a problem, you can try the following code:
type Evil(f) =
let n = f()
member x.N = n + 1
let rec e = Evil(fun () ->
printfn "%d" (e:Evil).N; 1)
The Evil class takes a function in a constructor and calls it during the construction. As a result, the recursive reference in the lambda function tries to access e before it is set to a value (and you'll get a runtime error). However, especially when working with event handlers, this is not an issue (and you get the warnning when you're using recursive objects correctly).
If you want to get rid of the warning, you can rewrite the code using explicit ref values and using null, but then you'll be in the same danger of a runtime error, just without the warning and with uglier code:
let foo (evt:IEvent<_, _>) =
let eh = ref null
eh := new EventHandler(fun _ _ ->
evt.RemoveHandler(!eh) )
evt.AddHandler(!eh)
My code for display all days in this year.
I don't understand why if NewSec =< EndSec -> init:stop() end did not execute the first time in run_calendar?
I expect init:stop() could be executed first time but it is not.
What is wrong?
Code:
-module(cal).
-export([main/0]).
main() ->
StartSec = calendar:datetime_to_gregorian_seconds({{2009,1,1},{0,0,0}}),
EndSec = calendar:datetime_to_gregorian_seconds({{2009,12,31},{0,0,0}}),
run_calendar(StartSec,EndSec).
run_calendar(CurSec, EndSec) ->
{Date,_Time} = calendar:gregorian_seconds_to_datetime(CurSec),
io:format("~p~n", [Date]),
NewSec = CurSec + 60*60*24,
if NewSec =< EndSec -> init:stop() end,
run_calendar(NewSec, EndSec).
Result:
wk# erlc cal.erl
wk# erl -noshell -s cal main
{2009,1,1}
{2009,1,2}
{2009,1,3}
{2009,1,4}
{2009,1,5}
...
{2009,12,22}
{2009,12,23}
{2009,12,24}
{2009,12,25}
{2009,12,26}
{2009,12,27}
{2009,12,28}
{2009,12,29}
{2009,12,30}
{2009,12,31}
wk#
I believe that init:stop() is an asynchronous process that will attempt to shut down the runtime smoothly. According to the docs, "All applications are taken down smoothly, all code is unloaded, and all ports are closed before the system terminates."
It probably takes a while to actually stop, because you have an actively running process. If you change "init:stop()" to "exit(stop)", it will terminate immediately:
3> cal:main().
{2009,1,1}
** exception exit: stop
in function cal:run_calendar/2
Init:stop is asynchronous and it will take time to quit. An alternate way would be to wrap up the test in the call itself and use pattern matching to terminate the loop:
-module(cal).
-export([main/0]).
main() ->
StartSec = calendar:datetime_to_gregorian_seconds({{2009,1,1},{0,0,0}}),
EndSec = calendar:datetime_to_gregorian_seconds({{2009,12,31},{0,0,0}}),
run_calendar(false, StartSec, EndSec).
run_calendar(true, _StartSec, _EndSec) ->
finished;
run_calendar(false, CurSec, EndSec) ->
{Date,_Time} = calendar:gregorian_seconds_to_datetime(CurSec),
io:format("~p~n", [Date]),
NewSec = CurSec + 60*60*24,
run_calendar(NewSec =< EndSec, NewSec, EndSec).
(or something similar, hopefully you get the idea)
You have a mistake in your if statement
You said
if NewSec =< EndSec -> init:stop() end,
This is incorrect. You have to write something like:
if
A =< B -> do something ...;
true -> do something else
end
The if syntax is
if
Condition1 -> Actions1;
Condition2 -> Actions2;
...
end
One of these conditions must always be true.
Why is this?
Erlang is a functional language, not a statement language. In an functional
language every expression must have a value. if is an expression, so it must have a value.
The value of (if 2 > 1 -> 3 end) is 3 but what is the value of
(if 1 > 2 -> 3 end) - answer it has no value - but it must have a value
everything must have a value.
In a statement language everything is evaluated for its side effect -so this would
be a valid construction.
In Erlang you will generate an exception.
So your code generates an exception - which you don't trap so you don't see it and
init:stop() never gets called ...