Coming from the F# world, I am used to using |> to pipe data into functions:
[1..10] |> List.filter (fun n -> n % 2 = 0) |> List.map (fun n -> n * n);
I assume that PureScript, being inspired by Haskell, has something similar.
How do I use the pipe operator in PureScript?
Yes, you can use # which is defined in Prelude.
Here is your example, rewritten using #:
http://try.purescript.org/?gist=0448c53ae7dc92278ca7c2bb3743832d&backend=core
module Main where
import Prelude
import Data.List ((..))
import Data.List as List
example = 1..10 # List.filter (\n -> n `mod` 2 == 0)
# map (\n -> n * n)
Here's one way to define the |> operator for use in PureScript; it's defined in exactly the same way as # - i.e. with the same precedence and associativity:-
pipeForwards :: forall a b. a -> (a -> b) -> b
pipeForwards x f = f x
infixl 1 pipeForwards as |>
Related
I'm trying to implement a function that simply counts the number of occurrences of some nat in a bag (just a synonym for a list).
This is what I want to do, but it doesn't work:
Require Import Coq.Lists.List.
Import ListNotations.
Definition bag := list nat.
Fixpoint count (v:nat) (s:bag) : nat :=
match s with
| nil => O
| v :: t => S (count v t)
| _ :: t => count v t
end.
Coq says that the final clause is redundant, i.e., it just treats v as a name for the head instead of the specific v that is passed to the call of count. Is there any way to pattern match on values passed as function arguments? If not, how should I instead write the function?
I got this to work:
Fixpoint count (v:nat) (s:bag) : nat :=
match s with
| nil => O
| h :: t => if (beq_nat v h) then S (count v t) else count v t
end.
But I don't like it. I'd rather pattern match if possible.
Pattern matching is a different construction from equality, meant to discriminate data encoded in form of "inductives", as standard in functional programming.
In particular, pattern matching falls short in many cases, such as when you need potentially infinite patterns.
That being said, a more sensible type for count is the one available in the math-comp library:
count : forall T : Type, pred T -> seq T -> nat
Fixpoint count s := if s is x :: s' then a x + count s' else 0.
You can then build your function as count (pred1 x) where pred1 : forall T : eqType, T -> pred T , that is to say, the unary equality predicate for a fixed element of a type with decidable (computable) equality; pred1 x y <-> x = y.
I found in another exercise that it's OK to open up a match clause on the output of a function. In that case, it was "evenb" from "Basics". In this case, try "eqb".
Well, as v doesn't work in the match, I thought that maybe I could ask whether the head of the list was equal to v. And yes, it worked. This is the code:
Fixpoint count (v : nat) (s : bag) : nat :=
match s with
| nil => 0
| x :: t =>
match x =? v with
| true => S ( count v t )
| false => count v t
end
end.
As stated in the answer to this question, the %hide directive allows one to make an existing name inaccessible:
import Data.String
%hide fib
%default total
fib : Nat -> Nat
fib n = loop n 0 1
where
loop : Nat -> Nat -> Nat -> Nat
loop Z a _ = a
loop (S k) a b = loop k b (a + b)
parseNat : String -> Maybe Nat
parseNat = map cast . parsePositive
response : String -> String
response s = case parseNat s of
Just n => "fib n = " ++ show (fib n)
Nothing => "n ∉ ℕ"
partial main : IO ()
main = repl "n = " ((++ "\n") . response)
This works fine in the code above:
*Main> :exec
n = 10
fib n = 55
However, it does not seem to carry over to the REPL:
*Main> fib 10
Can't disambiguate name: Main.fib, Prelude.Nat.fib
How can I cause the %hide directives from my code to carry over into the REPL?
I think you can't and the only way to invoke your function is to use its fully qualified name, e.g. Main.fib 10 would work.
I was reading the purescript wiki and found following section which explains do in terms of >>=.
What does >>= mean?
Do notation
The do keyword introduces simple syntactic sugar for monadic
expressions.
Here is an example, using the monad for the Maybe type:
maybeSum :: Maybe Number -> Maybe Number -> Maybe Number
maybeSum a b = do
n <- a
m <- b
let result = n + m
return result
maybeSum takes two
values of type Maybe Number and returns their sum if neither number is
Nothing.
When using do notation, there must be a corresponding
instance of the Monad type class for the return type. Statements can
have the following form:
a <- x which desugars to x >>= \a -> ...
x which desugars to x >>= \_ -> ... or just x if this is the last statement.
A let binding let a = x. Note the lack of the in keyword.
The example maybeSum desugars to ::
maybeSum a b =
a >>= \n ->
b >>= \m ->
let result = n + m
in return result
>>= is a function, nothing more. It resides in the Prelude module and has type (>>=) :: forall m a b. (Bind m) => m a -> (a -> m b) -> m b, being an alias for the bind function of the Bind type class. You can find the definitions of the Prelude module in this link, found in the Pursuit package index.
This is closely related to the Monad type class in Haskell, which is a bit easier to find resources. There's a famous question on SO about this concept, which is a good starting point if you're looking to improve your knowledge on the bind function (if you're starting on functional programming now, you can skip it for a while).
Scala offers a List#flatten method for going from List[Option[A]] to List[A].
scala> val list = List(Some(10), None)
list: List[Option[Int]] = List(Some(10), None)
scala> list.flatten
res11: List[Int] = List(10)
I attempted to implement it in Haskell:
flatten :: [Maybe a] -> [a]
flatten xs = map g $ xs >>= f
f :: Maybe a -> [Maybe a]
f x = case x of Just _ -> [x]
Nothing -> []
-- partial function!
g :: Maybe a -> a
g (Just x) = x
However I don't like the fact that g is a partial, i.e. non-total, function.
Is there a total way to write such flatten function?
Your flatten is the same as catMaybes (link) which is defined like this:
catMaybes :: [Maybe a] -> [a]
catMaybes ls = [x | Just x <- ls]
The special syntax Just x <- ls in a list comprehension means to draw an element from ls and discard it if it is not a Just. Otherwise assign x by pattern matching the value against Just x.
A slight modification of the code you have will do the trick:
flatten :: [Maybe a] -> [a]
flatten xs = xs >>= f
f :: Maybe a -> [a]
f x = case x of Just j -> [j]
Nothing -> []
If we extract the value inside of the Just constructor in f, we avoid g altogether.
Incidentally, f already exists as maybeToList and flatten is called catMaybes, both in Data.Maybe.
One could quite easily write a simple recursive function which goes through a list and rejects all the Nothings from the Maybe monad. Here's how I'd do it as a recursive sequence:
flatten :: [Maybe a] -> [a]
flatten [] = []
flatten (Nothing : xs) = flatten xs
flatten (Just x : xs) = x : flatten xs
However, it may be clearer to write it as a fold:
flatten :: [Maybe a] -> [a]
flatten = foldr go []
where go Nothing xs = xs
go (Just x) xs = x : xs
Or, we could use a blindingly elegant solution thanks to #user2407038, which I'd recommend playing around with in GHCi to work out the individual functions' jobs:
flatten :: [Maybe a] -> [a]
flatten = (=<<) (maybe [] (:[])
And it's faster, folded brother:
flatten :: [Maybe a] -> [a]
flatten = foldr (maybe id (:))
Your solution is halfway there. My suggestion if to rewrite your function f to use pattern matching (like my temporary go function), and enclose it in a where statement to keep relevant functions in one place. You've got to remember the differences in function syntax within scala and Haskell.
The big problem you're having is you don't know the differences I've mentioned. Your g function can use pattern matching with multiple patterns:
g :: Maybe a -> [a]
g (Just x) = [x]
g Nothing = []
There you go: your g function is now what you call 'complete', though more accurately, it would be said to have exhaustive patterns.
You can find more about function syntax here.
Is there a concise, idiomatic way how to express function iteration? That is, given a number n and a function f :: a -> a, I'd like to express \x -> f(...(f(x))...) where f is applied n-times.
Of course, I could make my own, recursive function for that, but I'd be interested if there is a way to express it shortly using existing tools or libraries.
So far, I have these ideas:
\n f x -> foldr (const f) x [1..n]
\n -> appEndo . mconcat . replicate n . Endo
but they all use intermediate lists, and aren't very concise.
The shortest one I found so far uses semigroups:
\n f -> appEndo . times1p (n - 1) . Endo,
but it works only for positive numbers (not for 0).
Primarily I'm focused on solutions in Haskell, but I'd be also interested in Scala solutions or even other functional languages.
Because Haskell is influenced by mathematics so much, the definition from the Wikipedia page you've linked to almost directly translates to the language.
Just check this out:
Now in Haskell:
iterateF 0 _ = id
iterateF n f = f . iterateF (n - 1) f
Pretty neat, huh?
So what is this? It's a typical recursion pattern. And how do Haskellers usually treat that? We treat that with folds! So after refactoring we end up with the following translation:
iterateF :: Int -> (a -> a) -> (a -> a)
iterateF n f = foldr (.) id (replicate n f)
or point-free, if you prefer:
iterateF :: Int -> (a -> a) -> (a -> a)
iterateF n = foldr (.) id . replicate n
As you see, there is no notion of the subject function's arguments both in the Wikipedia definition and in the solutions presented here. It is a function on another function, i.e. the subject function is being treated as a value. This is a higher level approach to a problem than implementation involving arguments of the subject function.
Now, concerning your worries about the intermediate lists. From the source code perspective this solution turns out to be very similar to a Scala solution posted by #jmcejuela, but there's a key difference that GHC optimizer throws away the intermediate list entirely, turning the function into a simple recursive loop over the subject function. I don't think it could be optimized any better.
To comfortably inspect the intermediate compiler results for yourself, I recommend to use ghc-core.
In Scala:
Function chain Seq.fill(n)(f)
See scaladoc for Function. Lazy version: Function chain Stream.fill(n)(f)
Although this is not as concise as jmcejuela's answer (which I prefer), there is another way in scala to express such a function without the Function module. It also works when n = 0.
def iterate[T](f: T=>T, n: Int) = (x: T) => (1 to n).foldLeft(x)((res, n) => f(res))
To overcome the creation of a list, one can use explicit recursion, which in reverse requires more static typing.
def iterate[T](f: T=>T, n: Int): T=>T = (x: T) => (if(n == 0) x else iterate(f, n-1)(f(x)))
There is an equivalent solution using pattern matching like the solution in Haskell:
def iterate[T](f: T=>T, n: Int): T=>T = (x: T) => n match {
case 0 => x
case _ => iterate(f, n-1)(f(x))
}
Finally, I prefer the short way of writing it in Caml, where there is no need to define the types of the variables at all.
let iterate f n x = match n with 0->x | n->iterate f (n-1) x;;
let f5 = iterate f 5 in ...
I like pigworker's/tauli's ideas the best, but since they only gave it as a comments, I'm making a CW answer out of it.
\n f x -> iterate f x !! n
or
\n f -> (!! n) . iterate f
perhaps even:
\n -> ((!! n) .) . iterate