Instead of fmap, which applies a function to a value-in-a-functor:
fmap :: Functor f => (a -> b) -> f a -> f b
I needed a function where the functor has a function and the value is plain:
thing :: Functor f => f (a -> b) -> a -> f b
but I can't find one.
What is this pattern called, where I apply a function-in-a-functor (or in an applicative, or in a monad) to a plain value?
I've implemented it already, I just don't quite understand what I did and why there wasn't already such a function in the standard libraries.
You don't need Applicative for this; Functor will do just fine:
apply f x = fmap ($ x) f
-- or, expanded:
apply f x = fmap (\f' -> f' x) f
Interestingly, apply is actually a generalisation of flip; lambdabot replaces flip with this definition as one of its generalisations of standard Haskell, so that's a possible name, although a confusing one.
By the way, it's often worth trying Hayoo (which searches the entirety of Hackage, unlike Hoogle) to see what names a function is often given, and whether it's in any generic package. Searching for f (a -> b) -> a -> f b, it finds flip (in Data.Functor.Syntax, from the functors package) and ($#) (from the synthesizer package) as possible names. Still, I'd probably just use fmap ($ arg) f at the use site.
As Niklas says, this is application in some applicative functor to a lifted value.
\f a -> f <*> pure a
:: Applicative f => f (a -> b) -> a -> f b
or more generally (?), using Category (.)
\f a -> f . pure a
:: (Applicative (cat a), Category cat) => cat b c -> b -> cat a c
Related
I am studying for an exam and I'm confused at this function. Based on the output how do I know that the type declaration of the function is (a -> b -> c)? also, how I can evaluate my function?
zipWith' :: (a -> b -> c) -> [a] -> [b] -> [c]
zipWith' _ [] _ = []
zipWith' _ _ [] = []
zipWith' f (x:xs) (y:ys) = f x y : zipWith' f xs ys
What I understand is that high order functions in haskell mean that they take a function as a parameter and return a function as well, right? how can I call this particular function?
I did this:
zipWith' [1..5] ['a','z','r']
but I know is wrong because I am calling it as if it were the regular zip function that takes 2 lists and returns a tuple. I am just confused at the type declaration
zipWith' :: [a] -> [b] -> [(a,b)]
For this answer, we'll acknowledge that all functions are curried. That is, every function has a type a -> b, where a and b are some type.
A higher-order function is one whose type includes a function for either its argument or return type. Return values are easy: it's any function you ordinary think of as taking more than one argument:
take :: Int -> [a] -> [a]. It takes an Int and returns a (polymorphic) function that takes a list and returns a list.
map :: (a -> b) -> [a] -> [b]. It takes a function (literally any function) and returns a function from lists to lists. The types of the return value is determined by the type of the argument.
Higher-order functions that take an function and return something that isn't a function are actually somewhat rare. Perhaps a commenter will point out an obvious one I am overlooking, but until then, consider fix :: (a -> a) -> a.
Now, zipWith' is an example of a higher-order function whose argument itself must be a higher-order function. The general type a -> b can unify with an ordinary function like ord :: Char -> Int (a ~ Char and b ~ Int) as well as a higher-order function like (+) (with a ~ Num t => t and b ~ Num t => t -> t. The type a -> b -> c will only unify with higher-order functions. a may or may not be a function type, but b -> c is an explicit function type.
This means that once you apply zipWith' to some higher-order function, type inference gives you more information about what the types of [a], [b], and [c] must be in the resulting function.
zipWith' (+) tells you that a ~ b ~ c ~ Num t => [t].
zipWith' (,) tells you that a and b are still unrestricted, but c ~ (a, b)
zipWith' (:) tells you that a is unrestricted, but b ~ c ~ [a].
It may also help if you consider that zip :: [a] -> [b] -> [(a,b)] could be defined as zip = zipWith' (,).
I'm trying to become familiar with Haskell and I was wondering if the following was possible and if so, how?
Say I have a set of functions {f,g,..} for which I was to define a replacement function {f',g',..}. Now say I have a function c which uses these functions (and only these functions) inside itself e.g. c x = g (f x). Is there a way to automatically define c' x = g' (f' x) without explicitly defining it?
EDIT: By a replacement function f' I mean some function that is conceptually relates to f by is altered in some arbitrary way. For example, if f xs ys = (*) <$> xs <*> ys then f' (x:xs) (y:ys) = (x * y):(f' xs ys) etc.
Many thanks,
Ben
If, as seems to be the case with your example, f and f' have the same type etc., then you can easily pass them in as extra parameters. Like
cGen :: ([a] -> [a] -> [a]) -> (([a] -> [a]) -> b) -> [a] -> b
cGen f g x = g (f x)
...which BTW could also be written cGen = (.)...
If you want to group together specific “sets of functions”, you can do that with a “configuration type”
data CConfig a b = CConfig {
f :: [a] -> [a] -> [a]
, g :: ([a] -> [a]) -> b
}
cGen :: CConfig a b -> [a] -> b
cGen (CConfig f g) = f . g
The most concise and reliable way to do something like this would be with RecordWildCards
data Replacer ... = R {f :: ..., g :: ...}
c R{..} x = g (f x)
Your set of functions now is now pulled from the local scope from the record, rather than a global definition, and can be swapped out for a different set of functions at your discretion.
The only way to get closer to what you want would to be to use Template Haskell to parse the source and modify it. Regular Haskell code simply cannot inspect a function in any way - that would violate referential transparency.
Ok, about Monad, I am aware that there are enough questions having been asked. I am not trying to bother anyone to ask what is monad again.
Actually, I read What is a monad?, it is very helpful. And I feel I am very close to really understand it.
I create this question here is just to describe some of my thoughts on Monad and Function, and hope someone could correct me or confirm it correct.
Some answers in that post let me feel monad is a little bit like function.
Monad takes a type, return a wrapper type (return), also, it can take a type, doing some operations on it and returns a wrapper type (bind).
From my point of view, it is a little bit like function. A function takes something and do some operations and return something.
Then why we even need monad? I think one of the key reasons is that monad provides a better way or pattern for sequential operations on the initial data/type.
For example, we have an initial integer i. In our code, we need to apply 10 functions f1, f2, f3, f4, ..., f10 step by step, i.e., we apply f1 on i first, get a result, and then apply f2 on that result, then we get a new result, then apply f3...
We can do this by functions rawly, just like f1 i |> f2 |> f3.... However, the intermediate results during the steps are not consistent; Also if we have to handle possible failure somewhere in middle, things get ugly. And an Option anyway has to be constructed if we don't want the whole process fail on exceptions. So naturally, monad comes in.
Monad unifies and forces the return types in all steps. This largely simplifies the logic and readability of the code (this is also the purpose of those design patterns, isn't it). Also, it is more bullet proof against error or bug. For example, Option Monad forces every intermediate results to be options and it is very easy to implement the fast fail paradigm.
Like many posts about monad described, monad is a design pattern and a better way to combine functions / steps to build up a process.
Am I understanding it correctly?
It sounds to me like you're discovering the limits of learning by analogy. Monad is precisely defined both as a type class in Haskell and as a algebraic thing in category theory; any comparison using "... like ..." is going to be imprecise and therefore wrong.
So no, since Haskell's monads aren't like functions, since they are 1) implemented as type classes, and 2) intended to be used differently than functions.
This answer is probably unsatisfying; are you looking for intuition? If so, I'd suggest doing lots of examples, and especially reading through LYAH. It's very difficult to get an intuitive understanding of abstract things like monads without having a solid base of examples and experience to fall back on.
Why do we even need monads? This is a good question, and maybe there's more than one question here:
Why do we even need the Monad type class? For the same reason that we need any type class.
Why do we even need the monad concept? Because it's useful. Also, it's not a function, so it can't be replaced by a function. (Your example seems like it does not require a Monad (rather, it needs an Applicative)).
For example, you can implement context-free parser combinators using the Applicative type class. But try implementing a parser for the language consisting of the same string of symbols twice (separated by a space) without Monad, i.e.:
a a -> yes
a b -> no
ab ab -> yes
ab ba -> no
So that's one thing a monad provides: the ability to use previous results to "decide" what to do. Here's another example:
f :: Monad m => m Int -> m [Char]
f m =
m >>= \x ->
if x > 2
then return (replicate x 'a')
else return []
f (Just 1) -->> Just ""
f (Just 3) -->> Just "aaa"
f [1,2,3,4] -->> ["", "", "aaa", "aaaa"]
Monads (and Functors, and Applicative Functors) can be seen as being about "generalized function application": they all create functions of type f a ⟶ f b where not only the "values inside a context", like types a and b, are involved, but also the "context" -- the same type of context -- represented by f.
So "normal" function application involves functions of type (a ⟶ b), "generalized" function application is with functions of type (f a ⟶ f b). Such functions can too be composed under normal function composition, because of the more uniform types structure: f a ⟶ f b ; f b ⟶ f c ==> f a ⟶ f c.
Each of the three creates them in a different way though, starting from different things:
Functors: fmap :: Functor f => (a ⟶ b) ⟶ (f a ⟶ f b)
Applicative Functors: (<*>) :: Applicative f => f (a ⟶ b) ⟶ (f a ⟶ f b)
Monadic Functors: (=<<) :: Monad f => (a ⟶ f b) ⟶ (f a ⟶ f b)
In practice, the difference is in how do we get to use the resulting value-in-context type, seen as denoting some type of computations.
Writing in generalized do notation,
Functors: do { x <- a ; return (g x) } g <$> a -- fmap
Applicative do { x <- a ; y <- b ; return (g x y) } g <$> a <*> b
Functors: (\ x -> g x <$> b ) =<< a
Monadic do { x <- a ; y <- k x ; return (g x y) } (\ x -> g x <$> k x) =<< a
Functors:
And their types,
"liftF1" :: (Functor f) => (a ⟶ b) ⟶ f a ⟶ f b -- fmap
liftA2 :: (Applicative f) => (a ⟶ b ⟶ c) ⟶ f a ⟶ f b ⟶ f c
"liftBind" :: (Monad f) => (a ⟶ b ⟶ c) ⟶ f a ⟶ (a ⟶ f b) ⟶ f c
I'm trying to do find a way to do something like this:
(head, last) `someFunction` [1, 2, 3]
to produce the tuple (1, 3) as output.
It seems similar in theory to an applicative functor, but a little backwards. I'm guessing there's a similar function that does this (or some way to make one), but I can't seem to find it/figure it out.
I tried defining a function like this:
fmap' :: ((a -> b), (a -> b)) -> [a] -> (b, b)
fmap' (f1, f2) xs = (f1 xs, f2 xs)
but GHC won't actually compile this.
Any help would be great; thanks!
Edit (a whole year later!):
My fmap' wouldn't compile because the type signature was wrong. Obviously there are better ways to do what I was doing, but the type of my fmap' should instead be:
fmap' :: ((a -> b), (a -> b)) -> a -> (b, b)
In that case, it compiles and runs just fine.
I think you can do this with arrows.
head &&& last $ [1,2,3]
will return (1,3).
It seems similar in theory to an applicative functor, but a little backwards.
Actually, it's a boring old forwards applicative functor; specifically, the reader ((->) r).
Prelude Control.Applicative> liftA2 (,) head last [1,2,3]
(1,3)
Or, if you're into that kind of thing:
Prelude Control.Applicative> let sequenceA [] = pure []; sequenceA (x:xs) = (:) <$> x <*> sequenceA xs
Prelude Control.Applicative> [head, last] `sequenceA` [1,2,3]
[1,3]
The type of fmap' is wrong. It should be
fmap' :: ([a] -> b, [a] -> b) -> [a] -> (b, b)
or, it can be more generalized
fmap' :: (a -> b, a -> c) -> a -> (b, c)
It doesn't really resemble fmap :: (a -> b) -> f a -> f b.
Something to try in this situation is to omit the type signature and check what GHC infers.
Doing so and asking GHCi :t fmap' yields the signature
fmap' :: (t2 -> t, t2 -> t1) -> t2 -> (t, t1)
which is identical to KennyTM's generalized version, and will give you the behaviour you're looking for.
A while ago, I asked a question about $, and got useful answers -- in fact, I thought I understood how to use it.
It seems I was wrong :(
This example shows up in a tutorial:
instance Monad [] where
xs >>= f = concat . map f $ xs
I can't for the life of me see why $ was used there; ghci isn't helping me either, as even tests I do there seem to show equivalence with the version that would simply omit the $. Can someone clarify this for me?
The $ is used here because it has lower precedence than normal function application.
Another way to write this code is like so:
instance Monad [] where
xs >>= f = (concat . map f) xs
The idea here is to first construct a function (concat . map f) and then apply it to its argument (xs). As shown, this can also be done by simply putting parenthesis around the first part.
Note that omitting the $ in the original definition is not possible, it will result in a type error. This is because the function composition operator (the .) has a lower precedence than normal function application effectively turning the expression into:
instance Monad [] where
xs >>= f = concat . (map f xs)
Which doesn't make sense, because the second argument to the function composition operator isn't a function at all. Although the following definition does make sense:
instance Monad [] where
xs >>= f = concat (map f xs)
Incidentally, this is also the definition I would prefer, because it seems to me to be a lot clearer.
I'd like to explain why IMHO this is not the used style there:
instance Monad [] where
xs >>= f = concat (map f xs)
concat . map f is an example of so-called pointfree-style writing; where pointfree means "without the point of application". Remember that in maths, in the expression y=f(x), we say that f is applied on the point x. In most cases, you can actually do a final step, replacing:
f x = something $ x
with
f = something
like f = concat . map f, and this is actually pointfree style.
Which is clearer is arguable, but the pointfree style gives a different point of view which is also useful, so sometimes is used even when not exactly needed.
EDIT: I have replaced pointless with pointfree and fixed some examples, after the comment by Alasdair, whom I should thank.
The reason $ is used here is doe to the type signature of (.):
(.) :: (b -> c) -> (a -> c) -> a -> c
Here we have
map f :: [a] -> [[b]]
and
concat :: [[b]] -> [b]
So we end up with
concat . map f :: [a] -> [b]
and the type of (.) could be written as
(.) :: ([[b]] -> [b]) -> ([a] -> [[b]]) -> [a] -> [b]
If we were to use concat . map f xs, we'd see that
map f xs :: [[b]]
And so cannot be used with (.). (the type would have to be (.) :: (a -> b) -> a -> b