I find this documentation in the basic Haskell libraries:
zip :: [a] -> [b] -> [(a, b)]
zip takes two lists and returns a list of corresponding pairs. If one input list is short, excess elements of the longer list are discarded.
zip3 :: [a] -> [b] -> [c] -> [(a, b, c)]
zip3 takes three lists and returns a list of triples, analogous to zip.
zip4 :: [a] -> [b] -> [c] -> [d] -> [(a, b, c, d)]
The zip4 function takes four lists and returns a list of quadruples, analogous to zip.
[...snip...]
unzip :: [(a, b)] -> ([a], [b])
unzip transforms a list of pairs into a list of first components and a list of second components.
unzip3 :: [(a, b, c)] -> ([a], [b], [c])
The unzip3 function takes a list of triples and returns three lists, analogous to unzip.
unzip4 :: [(a, b, c, d)] -> ([a], [b], [c], [d])
The unzip4 function takes a list of quadruples and returns four lists, analogous to unzip.
... and so on, up to zip7 and unzip7.
Is this a fundamental limitation of Haskell's type system? Or is there a way to implement zip and unzip once, to work on different configurations of input?
This is one very useful aspect of applicatives. Check out ZipList which is just a newtype wrapper around a simple list. The reason for the wrapper is that ZipList has an applicative instance for, you guessed it, zipping lists together. Then, if you want zip7 as bs cs ds es fs gs hs, you can just do something like
(,,,,,,) <$> as <*> bs <*> cs <*> ds <*> es <*> fs <*> gs <*> hs
As you can tell, this mechanism is meant to be also for extending zipWith, which is a general case of zip. To be honest, I think we should rip out all of the zipN functions and teach people the above instead. zip itself is fine, but beyond that...
Template Haskell solution
As the comments and other answers indicate, this is not a particularly satisfying answer. The one thing I was expecting someone else to implement was a TemplateHaskell version of zip and unzip. As no one has done so yet, here it is.
All it does is mechanically produce AST for zip or unzip functions. The idea behind zip is to use ZipList and behind unzip is to use foldr:
zip as ... zs === \as ... zs -> getZipList $ (, ... ,) <$> ZipList as <*> ... <*> ZipList zs
unzip === foldr (\ (a, ... ,z) ~(as, ... ,zs) -> (a:as, ... ,z:zs) ) ([], ... ,[])
The implementation looks like this.
{-# LANGUAGE TemplateHaskell #-}
module Zip (zip, unzip) where
import Prelude hiding (zip, unzip)
import Language.Haskell.TH
import Control.Monad
import Control.Applicative (ZipList(..))
-- | Given number, produces the `zip` function of corresponding arity
zip :: Int -> Q Exp
zip n = do
lists <- replicateM n (newName "xs")
lamE (varP <$> lists)
[| getZipList $
$(foldl (\a b -> [| $a <*> ZipList $(varE b) |])
[| pure $(conE (tupleDataName n)) |]
lists) |]
-- | Given number, produces the `unzip` function of corresponding arity
unzip :: Int -> Q Exp
unzip n = do
heads <- replicateM n (newName "x")
tails <- replicateM n (newName "xs")
[| foldr (\ $(tupP (varP <$> heads)) ~ $(tupP (varP <$> tails)) ->
$(tupE (zipWith (\x xs -> [| $x : $xs |])
(varE <$> heads)
(varE <$> tails))))
$(tupE (replicate n [| [] |])) |]
You can try this at GHCi:
ghci> :set -XTemplateHaskell
ghci> $(zip 3) [1..10] "abcd" [4,6..]
[(1,'a',4),(2,'b',6),(3,'c',8),(4,'d',10)]
ghci> $(unzip 3) [(1,'a',4),(2,'b',6),(3,'c',8),(4,'d',10)]
([1,2,3,4],"abcd",[4,6,8,10])
This is a zipN function that depends on the machinery of the generics-sop package:
{-# language TypeFamilies #-}
{-# language DataKinds #-}
{-# language TypeApplications #-}
import Control.Applicative
import Generics.SOP
-- "a" is some single-constructor product type, like some form of n-ary tuple
-- "xs" is a type-level list of the types of the elements of "a"
zipN :: (Generic a, Code a ~ '[ xs ]) => NP [] xs -> [a]
zipN np = to . SOP . Z <$> getZipList (hsequence (hliftA ZipList np))
main :: IO ()
main = do
let zipped = zipN #(_,_,_) ([1,2,3,4,5,6] :* ['a','b','c'] :* [True,False] :* Nil)
print $ zipped
The result:
[(1,'a',True),(2,'b',False)]
This solution has two disadvantages:
You have to wrap the argument lists in the special NP type from generics-sop that is constructed with :* and Nil.
You need to specify somehow that the result value is a list of tuples, and not a list of some other Generic-compatible type. Here, it is done with the #(_,_,_) type application.
2-ary, 3-ary.. n-ary tuples are all distinct data types, so you can't handle them uniformly directly, but you can introduce a type class that provides an interface that allows to define generic zip and unzip. Here is how it looks for generic unzip:
class Tuple t where
type Map (f :: * -> *) t
nilMap :: Proxy t -> (forall a. f a) -> Map f t
consMap :: (forall a. a -> f a -> f a) -> t -> Map f t -> Map f t
Map maps all types in a tuple type with f. nilMap constructs a Mapped tuple that contains empty values (I have no idea why Haskell requires that Proxy t there). consMap receives a function, a tuple and a Mapped tuple and zip the tuples with the function pointwise. Here is how instances look for 2- and 3-tuples:
instance Tuple (a, b) where
type Map f (a, b) = (f a, f b)
nilMap _ a = (a, a)
consMap f (x, y) (a, b) = (f x a, f y b)
instance Tuple (a, b, c) where
type Map f (a, b, c) = (f a, f b, f c)
nilMap _ a = (a, a, a)
consMap f (x, y, z) (a, b, c) = (f x a, f y b, f z c)
The gunzip itself:
gunzip :: forall t. Tuple t => [t] -> Map [] t
gunzip [] = nilMap (Proxy :: Proxy t) []
gunzip (p:ps) = consMap (:) p (gunzip ps)
This looks a lot like transpose:
transpose :: [[a]] -> [[a]]
transpose [] = repeat [] -- `gunzip` handles this case better
transpose (xs:xss) = zipWith (:) xs (transpose xss)
which it basically is, except with tuples. gunzip can be equivalently defined in terms of foldr as follows:
gunzip :: forall t. Tuple t => [t] -> Map [] t
gunzip = foldr (consMap (:)) $ nilMap (Proxy :: Proxy t) []
To define generic zip we need a type class of splittable data types (is there something like this on Hackage?).
class Splittable f g where
split :: f a -> g a (f a)
E.g. for lists we have
newtype MaybeBoth a b = MaybeBoth { getMaybeBoth :: Maybe (a, b) }
instance Splittable [] MaybeBoth where
split [] = MaybeBoth Nothing
split (x:xs) = MaybeBoth (Just (x, xs))
And here is what we add to the Tuple type class:
splitMap :: (Biapplicative g, Splittable f g) => Proxy (f t) -> Map f t -> g t (Map f t)
The Biapplicative g constraint ensures that it's possible to combine g a b and g c d into g (a, c) (b, d). For 2- and 3- tuples it looks like this:
splitMap _ (a, b) = biliftA2 (,) (,) (split a) (split b)
splitMap _ (a, b, c) = biliftA3 (,,) (,,) (split a) (split b) (split c)
After providing a Biapplicative instance for MaybeBoth
instance Biapplicative MaybeBoth where
bipure x y = MaybeBoth $ Just (x, y)
MaybeBoth f <<*>> MaybeBoth a = MaybeBoth $ uncurry (***) <$> f <*> a
we can finally define gzip:
gzip :: forall t. Tuple t => Map [] t -> [t]
gzip a = maybe [] (\(p, a') -> p : gzip a') . getMaybeBoth $ splitMap (Proxy :: Proxy [t]) a
It repeteadly cuts first elements of lists in a tuple, forms a tuple from them and prepends it to the result.
It should be possible to generalize gunzip by adding a dual to Splittable (Uniteable or something like that), but I'll stop here.
EDIT: I couldn't stop.
You are right that these functions (zip2, zip3 etc.) are all instances of the same pattern and in an ideal world, they should be implementable generically. By the way, as an exercise to the reader, figure out what zip1 and zip0 should be ;).
However, it is hard to implement zipN generically, because the common pattern between all the different cases is rather non-trivial. This does not mean it's impossible to implement it generically, but you'll need some of the more advanced type system features of Haskell GHC to do it.
To be more concrete, zip2, zip3 etc. all have a different number of arguments, making this an instance of "arity-generic programming" (the arity of a function is its number of arguments). As you might expect in the world of functional programming, there is an interesting research paper that covers precisely this topic ("arity-generic programming"), and conveniently, one of their main examples is... zipWithN. It doesn't directly answer your question because it uses Agda rather than Haskell, but you might still find it interesting. In any case, similar ideas can be implemented in terms of one or more of Haskell's GHC's more advanced type-system features (TypeFamilies and DataKinds come to mind). PDF version here.
By the way, this is just about an arity-generic zipWithN. For an arity-generic zipN, you probably need some support from the compiler, particularly an arity-generic interface to the tuple constructor, which I suspect might not be in GHC. This is what I believe augustss's comment to the question and chepner's comment to Alec's answer refer to.
Related
Shouldn’t this definition be allowed in a lazy language like Haskell in which functions are curried?
apply f [] = f
apply f (x:xs) = apply (f x) xs
It’s basically a function that applies the given function to the given list of arguments and is very easily done in Lisp for example.
Are there any workarounds?
It is hard to give a static type to the apply function, since its type depends on the type of the (possibly heterogeneous) list argument. There are at least two ways one way to write this function in Haskell that I can think of:
Using reflection
We can defer type checking of the application until runtime:
import Data.Dynamic
import Data.Typeable
apply :: Dynamic -> [Dynamic] -> Dynamic
apply f [] = f
apply f (x:xs) = apply (f `dynApp` x) xs
Note that now the Haskell program may fail with a type error at runtime.
Via type class recursion
Using the semi-standard Text.Printf trick (invented by augustss, IIRC), a solution can be coded up in this style (exercise). It may not be very useful though, and still requires some trick to hide the types in the list.
Edit: I couldn't come up with a way to write this, without using dynamic types or hlists/existentials. Would love to see an example
I like Sjoerd Visscher's reply, but the extensions -- especially IncoherentInstances, used in this case to make partial application possible -- might be a bit daunting. Here's a solution that doesn't require any extensions.
First, we define a datatype of functions that know what to do with any number of arguments. You should read a here as being the "argument type", and b as being the "return type".
data ListF a b = Cons b (ListF a (a -> b))
Then we can write some (Haskell) functions that munge these (variadic) functions. I use the F suffix for any functions that happen to be in the Prelude.
headF :: ListF a b -> b
headF (Cons b _) = b
mapF :: (b -> c) -> ListF a b -> ListF a c
mapF f (Cons v fs) = Cons (f v) (mapF (f.) fs)
partialApply :: ListF a b -> [a] -> ListF a b
partialApply fs [] = fs
partialApply (Cons f fs) (x:xs) = partialApply (mapF ($x) fs) xs
apply :: ListF a b -> [a] -> b
apply f xs = headF (partialApply f xs)
For example, the sum function could be thought of as a variadic function:
sumF :: Num a => ListF a a
sumF = Cons 0 (mapF (+) sumF)
sumExample = apply sumF [3, 4, 5]
However, we also want to be able to deal with normal functions, which don't necessarily know what to do with any number of arguments. So, what to do? Well, like Lisp, we can throw an exception at runtime. Below, we'll use f as a simple example of a non-variadic function.
f True True True = 32
f True True False = 67
f _ _ _ = 9
tooMany = error "too many arguments"
tooFew = error "too few arguments"
lift0 v = Cons v tooMany
lift1 f = Cons tooFew (lift0 f)
lift2 f = Cons tooFew (lift1 f)
lift3 f = Cons tooFew (lift2 f)
fF1 = lift3 f
fExample1 = apply fF1 [True, True, True]
fExample2 = apply fF1 [True, False]
fExample3 = apply (partialApply fF1 [True, False]) [False]
Of course, if you don't like the boilerplate of defining lift0, lift1, lift2, lift3, etc. separately, then you need to enable some extensions. But you can get quite far without them!
Here is how you can generalize to a single lift function. First, we define some standard type-level numbers:
{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances, FlexibleContexts, TypeFamilies, UndecidableInstances #-}
data Z = Z
newtype S n = S n
Then introduce the typeclass for lifting. You should read the type I n a b as "n copies of a as arguments, then a return type of b".
class Lift n a b where
type I n a b :: *
lift :: n -> I n a b -> ListF a b
instance Lift Z a b where
type I Z a b = b
lift _ b = Cons b tooMany
instance (Lift n a (a -> b), I n a (a -> b) ~ (a -> I n a b)) => Lift (S n) a b where
type I (S n) a b = a -> I n a b
lift (S n) f = Cons tooFew (lift n f)
And here's the examples using f from before, rewritten using the generalized lift:
fF2 = lift (S (S (S Z))) f
fExample4 = apply fF2 [True, True, True]
fExample5 = apply fF2 [True, False]
fExample6 = apply (partialApply fF2 [True, False]) [False]
No, it cannot. f and f x are different types. Due to the statically typed nature of haskell, it can't take any function. It has to take a specific type of function.
Suppose f is passed in with type a -> b -> c. Then f x has type b -> c. But a -> b -> c must have the same type as a -> b. Hence a function of type a -> (b -> c) must be a function of type a -> b. So b must be the same as b -> c, which is an infinite type b -> b -> b -> ... -> c. It cannot exist. (continue to substitute b -> c for b)
Here's one way to do it in GHC. You'll need some type annotations here and there to convince GHC that it's all going to work out.
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE UndecidableInstances #-}
{-# LANGUAGE IncoherentInstances #-}
class Apply f a r | f -> a r where
apply :: f -> [a] -> r
instance Apply f a r => Apply (a -> f) a r where
apply f (a:as) = apply (f a) as
instance Apply r a r where
apply r _ = r
test = apply ((+) :: Int -> Int -> Int) [1::Int,2]
apply' :: (a -> a -> a) -> [a] -> a
apply' = apply
test' = apply' (+) [1,2]
This code is a good illustration of the differences between static and dynamic type-checking. With static type-checking, the compiler can't be sure that apply f really is being passed arguments that f expects, so it rejects the program. In lisp, the checking is done at runtime and the program might fail then.
I am not sure how much this would be helpful as I am writing this in F# but I think this can be easily done in Haskell too:
type 'a RecFunction = RecFunction of ('a -> 'a RecFunction)
let rec apply (f: 'a RecFunction) (lst: 'a list) =
match (lst,f) with
| ([],_) -> f
| ((x::xs), RecFunction z) -> apply (z x) xs
In this case the "f" in question is defined using a discriminated union which allows recursive data type definition. This can be used to solved the mentioned problem I guess.
With the help and input of some others I defined a way to achieve this (well, sort of, with a custom list type) which is a bit different from the previous answers. This is an old question, but it seems to still be visited so I will add the approach for completeness.
We use one extension (GADTs), with a list type a bit similar to Daniel Wagner's, but with a tagging function type rather than a Peano number. Let's go through the code in pieces. First we set the extension and define the list type. The datatype is polymorphic so in this formulation arguments don't have to have the same type.
{-# LANGUAGE GADTs #-}
-- n represents function type, o represents output type
data LApp n o where
-- no arguments applied (function and output type are the same)
End :: LApp o o
-- intentional similarity to ($)
(:$) :: a -> LApp m o -> LApp (a -> m) o
infixr 5 :$ -- same as :
Let's define a function that can take a list like this and apply it to a function. There is some type trickery here: the function has type n, a call to listApply will only compile if this type matches the n tag on our list type. By leaving our output type o unspecified, we leave some freedom in this (when creating the list we don't have to immediately entirely fix the kind of function it can be applied to).
-- the apply function
listApply :: n -> LApp n o -> o
listApply fun End = fun
listApply fun (p :$ l) = listApply (fun p) l
That's it! We can now apply functions to arguments stored in our list type. Expected more? :)
-- showing off the power of AppL
main = do print . listApply reverse $ "yrruC .B lleksaH" :$ End
print . listApply (*) $ 1/2 :$ pi :$ End
print . listApply ($) $ head :$ [1..] :$ End
print $ listApply True End
Unfortunately we are kind of locked in to our list type, we can't just convert normal lists to use them with listApply. I suspect this is a fundamental issue with the type checker (types end up depending on the value of a list) but to be honest I'm not entirely sure.
-- Can't do this :(
-- listApply (**) $ foldr (:$) End [2, 32]
If you feel uncomfortable about using a heterogeneous list, all you have to do is add an extra parameter to the LApp type, e.g:
-- Alternative definition
-- data FList n o a where
-- Nil :: FList o o a
-- Cons :: a -> FList f o a -> FList (a -> f) o a
Here a represents the argument type, where the function which is applied to will also have to accept arguments of all the same type.
This isn't precisely an answer to your original question, but I think it might be an answer to your use-case.
pure f <*> [arg] <*> [arg2] ...
-- example
λ>pure (\a b c -> (a*b)+c) <*> [2,4] <*> [3] <*> [1]
[7,13]
λ>pure (+) <*> [1] <*> [2]
[3]
The applicative instance of list is a lot broader than this super narrow use-case though...
λ>pure (+1) <*> [1..10]
[2,3,4,5,6,7,8,9,10,11]
-- Or, apply (+1) to items 1 through 10 and collect the results in a list
λ>pure (+) <*> [1..5] <*> [1..5]
[2,3,4,5,6,3,4,5,6,7,4,5,6,7,8,5,6,7,8,9,6,7,8,9,10]
{- The applicative instance of list gives you every possible combination of
elements from the lists provided, so that is every possible sum of pairs
between one and five -}
λ>pure (\a b c -> (a*b)+c) <*> [2,4] <*> [4,3] <*> [1]
[9,7,17,13]
{- that's - 2*4+1, 2*3+1, 4*4+1, 4*3+1
Or, I am repeating argC when I call this function twice, but a and b are
different -}
λ>pure (\a b c -> show (a*b) ++ c) <*> [1,2] <*> [3,4] <*> [" look mah, other types"]
["3 look mah, other types","4 look mah, other types","6 look mah, other types","8 look mah, other types"]
So it's not the same concept, precisely, but it a lot of those compositional use-cases, and adds a few more.
I asked a similar question before (how to implement mapAccumM?).
I need the one which folds from the right as well (mapAccumR):
mapAccumRM :: (Monad m, Traversable t)
=> (a -> b -> m (a, c)) -> a -> t b -> m (a, t c)
Is there a simple implementation for this?
One approach is to define new instances of Traversable that use the ordering you like. For example, for lists, one might simply define a new type:
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
import Control.Applicative
import Data.Traversable
newtype BackwardsList a = BackwardsList [a]
deriving (Eq, Ord, Read, Show, Functor, Foldable)
instance Traversable BackwardsList where
traverse f (BackwardsList xs) = BackwardsList <$> go xs where
go [] = pure []
go (x:xs) = liftA2 (flip (:)) (go xs) (f x)
In ghci, we can see the difference between this and the standard instance:
> runState (traverse (\_ -> modify (+1) >> get) "hello, world!") 0
([1,2,3,4,5,6,7,8,9,10,11,12,13],13)
> runState (traverse (\_ -> modify (+1) >> get) (BackwardsList "hello, world!")) 0
(BackwardsList [13,12,11,10,9,8,7,6,5,4,3,2,1],13)
This approach is fairly simple; however, it requires a new type (and the associated newtype wrapping/unwrapping garbage) for every new traversal order that you are interested in.
One could consider defining a new type class for ordered traversals. Let's see one way this might be done. We'll need a little prelude:
{-# LANGUAGE Rank2Types, TypeFamilies #-}
import Control.Applicative
import Data.Foldable
import Data.Traversable
import Data.Tree
Most Haskell data types can be viewed as fixed-points of polynomial functors; and the functors that they are fixed-points of are good descriptions of the "spine" of the data structure. We'll abuse this idea to give a concrete way to encode the ordering that should be used during traversal. The class itself looks like this:
type Order t = forall f a. Applicative f => Spine t (f a) (f (t a)) -> f (t a)
class OrderedTraversable t where
data Spine t :: * -> * -> *
otraverse :: Applicative f => Order t -> (a -> f b) -> t a -> f (t b)
Notice that the type of otraverse looks just like the type of traverse, except that it now takes an extra ordering argument. The ordering argument is, in a sense, variadic; since different data types have different numbers of values/children at various places in their structure, and the ordering may care about all of them. (Of special interest is here is the technique of using rank-2 types to prevent an ordering from observing "too much" about a data structure: it can't use special facts about a given instance of Applicative or given kind of element to decide how to traverse a spine, only decisions based on the shape of the spine are allowed.) Let's see a simple example, for lists:
instance OrderedTraversable [] where
-- Cute hack: the normal presentation for the spine of a list uses both
-- a `Cons` and a `Nil`; but parametricity says the only thing an
-- `Order []` can do with a `Nil` is `pure []` anyway. So let's just
-- bake that into `otraverse`.
data Spine [] a r = Cons a r
otraverse order f = go where
go [] = pure []
go (x:xs) = order (Cons (f x) (go xs))
Compare with the implementation of Traversable for lists in the standard library (I have taken the liberty of expanding the definition of foldr to make it more closely match the code above):
instance Traversable [] where
traverse f = go where
go [] = pure []
go (x:xs) = (:) <$> f x <*> go xs
As you can see, the primary difference is that we have abstracted which function to use to combine f x and go xs. We can recover the standard Traversable instance with a "head-first" order. There is also a "last-first" order; and these are basically the only two orders that make sense for lists.
headFirst, lastFirst :: Order []
headFirst (Cons fx fxs) = liftA2 (:) fx fxs
lastFirst (Cons fx fxs) = liftA2 (flip (:)) fxs fx
In ghci, we can now see how they differ:
> runState (traverse (\_ -> modify (+1) >> get) "hello, world!") 0
([1,2,3,4,5,6,7,8,9,10,11,12,13],13)
> runState (otraverse headFirst (\_ -> modify (+1) >> get) "hello, world!") 0
([1,2,3,4,5,6,7,8,9,10,11,12,13],13)
> runState (otraverse lastFirst (\_ -> modify (+1) >> get) "hello, world!") 0
([13,12,11,10,9,8,7,6,5,4,3,2,1],13)
To give another example, here is how you might use this class with rose trees:
instance OrderedTraversable Tree where
data Spine Tree a r = SNode a [r]
otraverse order f = go where
go (Node x ts) = order (SNode (f x) (map go ts))
-- two example orders for trees
prefix, postfix :: Order [] -> Order Tree
prefix list (SNode fx fts) = liftA2 Node fx (otraverse list id fts)
postfix list (SNode fx fts) = liftA2 (flip Node) (otraverse list id fts) fx
Note that there are actually infinitely many "good" ordering functions for rose trees; two that are particularly likely to be what you want are included above.
Learn You a Haskell offers the following exercise:
Let's try implementing a function that takes a list of applicatives and
returns an applicative that has a list as its result value.
LYAH gives the type signature sequenceA' :: (Applicative f) => [f a] -> f [a]
I started with Applicative, but wasn't sure how to extract a from f a in a general way for all Applicative's.
So, I implemented it as a Maybe. Of course this is unacceptable for all `Applicative's.
import Control.Applicative
sequenceA' :: (Num a) => [Maybe a] -> Maybe [a]
sequenceA' [] = pure []
sequenceA' xs = pure $ [extract' x | x <- xs ]
extract' :: (Num a) => Maybe a -> a
extract' x = case x of
Just y -> y
Nothing -> 0
How can I extract a from f a where f is an Applicative?
You can't in general. In fact, the Maybe example you gave is a good example of why since it requires it to be a Num instance. It wouldn't make sense for the Maybe Applicative in general, so that would be a counterexample to a general solution. Another counterexample would be IO. There is no valid extract implementation for IO.
To make a function that is general enough to work with all Applicative instances, you must be able to construct the function using only methods from Applicative and its super-class, Functor. There is no way to make extract using only fmap, pure and (<*>).
It's not necessary to take things out of the applicative functor to achieve this.
The great thing about applicative functors is they allow you to use ordinary functions on the results of each applicative computation, so
if you have applicatives c1, c2 and c3 of types f a, f b, f c
that produce values v1, v2 and v3 of types a, b and c,
but you actually want to use a function g :: a -> b -> c -> d on the values
to produce g v1 v2 v3 :: d, then you can do
g <$> c1 <*> c2 <*> c3
which has type f d.
So we can use the function (:) to join the first value out of our list of applicatives with the rest of them, so you can do someytthing like (:) <$> thingThatGivesFirstValue <*> thing that gives the rest of the list. So it'll be a nice recursion if you pattern match on the list of applicatives.
sequenceA' :: (Applicative f) => [f a] -> f [a]
sequenceA' [] = -- just give the empty list
sequenceA' (c:cs) = -- hmm. What could go here then?
so for example you should get
ghci> sequenceA' [getLine, getLine, getLine]
Hello
there
everyone
["Hello","there","everyone"]
ghci> sequenceA' [Just 3, Just 4, Just 5]
Just [3,4,5]
Here's an example function to help you along with the recursice case:
nth :: Applicative f => f Int -> f [a] -> f a
nth wrappedInt wrappedList = (!!) <$> wrappedInt <*> wrappedList
So you don't need to unwrap anything or get values out, the operators<$> and <*> let you do what you like inside.
nth (Just 3) (Just "Hello") == 'l'
Here's a hint:
foo :: Applicative f => f Int -> f Int -> f Int
foo fx fy = (+) <$> fx <*> fy -- apply + "under" the functor
bar :: Applicative f => f a -> f [a] -> f [a]
bar fx fxs = ??? <$> fx <*> fxs -- apply ??? "under" the functor
sequenceA' :: Applicative f => [f a] -> f [a]
sequenceA' [] = pure [] -- as in your solution
sequenceA' (x:xs) = let y = x -- :: f a
ys = sequenceA' xs -- :: f [a]
in ???
I use let in the last function to clarify the types which are involved. After you fill in the ??? you can of course remove the let.
You can extract a from f a with pattern matching or evaluation, if f is not IO
import Control.Applicative
import System.IO
import Control.Monad.ST
-- with AndrewC sequenceA' definition
sequenceA' :: (Applicative f) => [f a] -> f [a]
sequenceA' [] = pure []
sequenceA' (c:cs) = (:) <$> c <*> sequenceA' cs
seqOfMaybes :: [Maybe a] -> [a]
seqOfMaybes listOfMb = case sequenceA' listOfMb of
Nothing -> []
Just list -> list
-- sequencing ST computations
compute :: a -> ST s a
compute x = return x
seqOfSTData :: [Int] -> [Int]
seqOfSTData vals = runST $ (sequenceA' (map compute vals) :: ST s [Int])
-- but you cannot escape the applicative f if f is IO
readAnInt :: IO Int
readAnInt = putStrLn "give me an Int>" >> hFlush stdout >> getLine >>= (return . read)
seqOfIO :: [IO a] -> IO [a]
seqOfIO listOfIO = sequenceA' listOfIO
main :: IO ()
main = do
print $ seqOfMaybes [Just 3, Just 4, Just 5]
print $ seqOfSTData [1,2,3]
seqOfIO [readAnInt, readAnInt] >>= print
Take a quick peek at the following interactive session in GHCi:
Prelude> import Control.Applicative
Prelude Control.Applicative> (+1) <$> [1,2]
[2,3]
Prelude Control.Applicative> (+1) <$> (1,2)
(1,3)
I guess there is a good reason for the behavior of <$> regarding pairs, but I wasn't able to find one so far, so:
Why is <$> (or fmap) defined to act only on the second member of a pair and not on both values?
<$> (aka fmap) is a member of the Functor class like so:
class Functor f where
fmap :: (a -> b) -> f a -> f b
So whatever f is must be a parameterised type with one type argument. Lists are one such type, when written in their prefix form [] ([] a is the same as [a]). So the instance for lists is:
instance Functor [] where
-- fmap :: (a -> b) -> [] a -> [] b
fmap = map
Pairs can also be written in prefix form: (,) a b is the same as (a, b). So let's consider what we do if we want a Functor instance involving pairs. We can't declare an instance Functor (,) because the pair constructor (,) takes two types -- and they can be different types! What we can do is declare an instance for (,) a -- that's a type that only needs one more type:
instance Functor ( (,) a ) where
-- fmap :: (b -> c) -> (,) a b -> (,) a c
fmap f (x, y) = (x, f y)
Hopefully you can see that the definition of fmap is the only sensible one we can give. The answer as to why the functor instance operates on the second item in a pair is that the type for the second item comes last in the list! We can't easily declare a functor instance that operates on the first item in a pair. Incidentally, this generalises to larger tuples, e.g. the quadruple (,,,) a b c d (aka (a, b, c, d)) can also have a Functor instance on the last item:
instance Functor ( (,,,) a b c) where
-- fmap :: (d -> e) -> (,,,) a b c d -> (,,,) a b c e
fmap f (p, q, r, s) = (p, q, r, f s)
Hope that helps explain it all!
Consider the definition of the Functor typeclass:
class Functor f where
fmap :: (a -> b) -> f a -> f b
Obviously, f has kind * -> *. So you can only declare instances for a datatype, which has kind * -> *. What you can do, is doing some stuff like this:
instance Functor (,) where
fmap :: (a -> b) -> (,) a -> (,) b
This would work on partitial applicated tuples and is really unhandy. So one defined the instance like this:
instance Functor ((,) a) where
fmap :: (b -> c) -> (,) a b -> (,) a c
fmap f (x,y) = (x,f y)
In a nutshell: It is not possible in plain Haskell 98 (although I believe tht there's a syntax extension for this) to define an instance as
What you can do, is defining your own tuple:
data T a = T a a
instance Functor T where
fmap f (T a b) = T (f a) (f b)
Then you can do whatever you like. You see, because the kind is * -> * instead of * -> * -> *, anything's okay.
I guess, a tuple doesn't need to be homogeneous, I mean both types can be different. If you want a homogeneous tuple you can use a list and then fmap will work.
How would you expect (+1) ("Hello", 2) to work ?
Prelude> import Control.Applicative
Prelude Control.Applicative> (+1) <$> ("hello",2)
("hello",3)
That's just work, but there is not special behavior when both type are same.
By the way I don't know why the second value is not used rather than the first one, but anyway you can only use one value.
I'm not a Haskell programmer, but I'm curious about the following questions.
Informal function specification:
Let MapProduct be a function that takes a function called F and multiple lists. It returns a list containing the results of calling F with one argument from each list in each possible combination.
Example:
Call MapProduct with F being a function that simply returns a list of its arguments, and two lists. One of the lists contains the integers 1 and 2, the other one contains the strings "a" and "b". It should return a list that contains the lists: 1 and "a", 1 and "b", 2 and "a", 2 and "b".
Questions:
How is MapProduct implemented?
What is the function's type? What is F's type?
Can one guess what the function does just by looking at its type?
Can you handle inhomogeneous lists as input? (e.g. 1 and "a" in one of the input lists)
What extra limitation (if any) do you need to introduce to implement MapProduct?
How is MapProduct implemented?
.
Prelude> :m + Control.Applicative
Prelude Control.Applicative> (,,) <$> [1,2,3] <*> ["a","b","c"] <*> [0.8, 1.2, 4.4]
[(1,"a",0.8),(1,"a",1.2),...,(3,"c",4.4)]
What is the function's type? What is F's type?
The type of F depends on the list you want to apply. <$> here is fmap, and (<*>) :: f(a->b) -> f a -> f b where f = [] here.
Can you handle inhomogeneous lists as input? (e.g. 1 and "a" in one of the input lists)
There is no such thing as an heterogeneous list. But you can simulate a heterogeneous list for a specific context with existential types. And then you can just use the method above to do the MapProduct.
*Main Control.Applicative> :i SB
data ShowBox where
SB :: forall s. (Show s) => s -> ShowBox
-- Defined at v.hs:1:35-36
*Main Control.Applicative> [SB 2, SB "a", SB 6.4]
[2,"a",6.4]
*Main Control.Applicative> (,) <$> [SB 2, SB "a", SB 6.4] <*> [SB 'z', SB 44]
[(2,'z'),(2,44),("a",'z'),("a",44),(6.4,'z'),(6.4,44)]
It is possible to define a function mapProduct that works for any arity of function:
{-# LANGUAGE FlexibleInstances, TypeFamilies #-}
module MapProduct (
mapProduct
) where
import Control.Monad
newtype ProdFuncList a b = ProdFuncList [ a -> b ]
class MapProdResult p where
type MapProdArg p
apNext :: ProdFuncList x (MapProdArg p) -> [x] -> p
instance (MapProdResult b) => MapProdResult ([a] -> b) where
type MapProdArg ([a] -> b) = (a -> MapProdArg b)
apNext (ProdFuncList fs) = apNext . ProdFuncList . ap fs
instance MapProdResult [b] where
type MapProdArg [b] = b
apNext (ProdFuncList fs) = ap fs
mapProduct :: (MapProdResult q) => (a -> MapProdArg q) -> [a] -> q
mapProduct f = apNext (ProdFuncList [f])
Here it is in action:
> :l MapProduct.hs
[1 of 1] Compiling MapProduct ( MapProduct.hs, interpreted )
Ok, modules loaded: MapProduct.
> mapProduct (+10) [1..4] :: [Int]
[11,12,13,14]
> mapProduct (*) [1..4] [10..12] :: [Int]
[10,11,12,20,22,24,30,33,36,40,44,48]
> mapProduct (\a b c -> a:b:c:[]) "bcs" "ao" "dnt" :: [String]
["bad","ban","bat","bod","bon","bot","cad","can","cat","cod","con","cot","sad","san","sat","sod","son","sot"]
The downside of this approach is that you'll most likely have to type annotate the result (as shown in the examples above). It would be much more idiomatic to simply use fmap and ap directly:
> :m + Control.Monad
> (+10) `fmap` [1..4]
[11,12,13,14]
> (*) `fmap` [1..4] `ap` [10..12]
[10,11,12,20,22,24,30,33,36,40,44,48]
> (\a b c -> a:b:c:[]) `fmap` "bcs" `ap` "ao" `ap` "dnt"
["bad","ban","bat","bod","bon","bot","cad","can","cat","cod","con","cot","sad","san","sat","sod","son","sot"]
This requires no type annotations, and is fully general over all monads, not just [].
(The MapProduct module above could easily be generalized over all monads as well. I didn't so as to make it clearly solve the original question.)
The function that you describe is closely related to the zipWithN functions. It will have the same type - it will just result in bigger result lists. Now the problem is that there is no way to express "a function that takes N arguments of the types t_1, ..., t_n" or "n lists of the types [t_1],...,[t_n]" (or "an n-tuple of type ([t_1], ..., [t_n]") ) in haskell's type system (without extensions like template haskell). This is why there is not one zipWith function, but one for each number of arguments lists that is supported.
So to answer your questions:
It is implemented by defining a function mapProductN for every number N that you want to support. For N=2 it would look like this:
mapProduct f l1 l2 = [f x1 x2 | x1 <- l1, x2 <- x2]
Or as a general blueprint (i.e. pseudo-code) how to define the functions for any N:
mapProduct f l1 ... ln = [f x1 ... xn | x1 <- l1, ..., xn <- ln]
As I said it's the same as the types of the zipWith functions, i.e.:
zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
zipWith3 :: (a -> b -> c -> d) -> [a] -> [b] -> [c] -> [d]
zipWith4 :: (a -> b -> c -> d -> e) -> [a] -> [b] -> [c] -> [d] -> [e]
Since f is the first argument to the function, the type of the first argument is f's type (so for n=2 it'd be a -> b -> c)
Well since it has the same type as zipWith and zipWith does something else, that'd be a no.
Haskell doesn't allow inhomogeneous lists without an extension.
There's an upper limit on the number of lists unless you're willing to spend the time writing infinite versions of mapProduct.