Say I have functions which accept the same parameters and I want to test if their outputs are equivalent for the same input.
f :: a -> b -> c
g :: a -> b -> c
f a b == g a b
How can I package the parameters a and b in x so I can write the following instead.
f x == g x
What are the best ways to accomplish this without needing to wrap the functions themselves?
The only way to do exactly what you’re asking is to use uncurry:
let
x = (a, b)
in uncurry f x == uncurry g x
(Or uncurryN for N arguments.)
However, instead of packaging the arguments in a tuple, you could use the (->) x instance of Applicative (i.e., functions taking x as input) to implicitly “spread” the arguments to the parameters of both functions, so at least you only have to mention them once. This instance is commonly used in point-free code.
For example, using liftA2 specialised to this instance:
-- General type:
liftA2 :: Applicative f => (a -> b -> c) -> f a -> f b -> f c
-- Specialised to ‘(->) x’ (using TypeApplications syntax):
liftA2 #((->) _) :: (a -> b -> c) -> (x -> a) -> (x -> b) -> (x -> c)
You get this pattern:
liftA2 h f g x
-- =
(h <$> f <*> g) x
-- =
h (f x) (g x)
To lift more arguments, you add another liftA2 or … <$> … <*> …:
liftA2 (liftA2 h) f g x y
-- =
(liftA2 h <$> f <*> g) x y
-- =
h (f x y) (g x y)
So in a case like yours:
f, g :: Int -> Char -> Bool
f i c = chr i == c
g i c = i == ord c
(liftA2 . liftA2) (==) f g :: Int -> Char -> Bool
-- =
liftA2 (liftA2 (==)) f g
-- =
(\ x y -> f x y == g x y)
The N in liftAN corresponds to the number of functions; the number of liftAN calls corresponds to the number of arguments.
I am trying to understand the Interchange law of applicative functor:
u <*> pure y = pure ($ y) <*> u
What make me confuse is, the function application $ y, consider following example:
($ 2) :: (a -> b) -> b
Why does the second argument get applied not the first?
That's an operator section. A few simple examples:
Prelude> (/2) <$> [1..8]
[0.5,1.0,1.5,2.0,2.5,3.0,3.5,4.0]
Prelude> (:"!") <$> ['a'..'e']
["a!","b!","c!","d!","e!"]
The section (:"!") is syntactic sugar for \c -> c:"!", i.e. it takes a character c and prepends it to the string "!".
Likewise, the section ($ 2) takes a function f and simply applies it to the number 2.
Note that this is different from ordinary partial application:
Prelude> ((/) 2) <$> [1..8]
[2.0,1.0,0.6666666666666666,0.5,0.4,0.3333333333333333,0.2857142857142857,0.25]
Here, I've simply applied the function (/) to one fixed argument 2, the dividend. This can also be written as a left section (2/). But the right section (/2) applies 2 as the divisor instead.
You can do that with operator sections. For example:
(5+ ) -- Same as \ x -> 5+x
( +5) -- Same as \ x -> x+5
It's only operators you can do this with; normal named functions can only be curried from left to right.
Haskell cheat sheet operator sections entry could be:
(a `op` b) = (a `op`) b = (`op` b) a = (op) a b
When op is an actual operator (not an alpha-numerical name), backticks aren't needed.
The above can be seen as partially applying implicitly defined lambda expressions:
(a `op`) b = (a `op` b) = (\y -> a `op` y) b = (\x y -> x `op` y) a b = op a b
(`op` b) a = (a `op` b) = (\x -> x `op` b) a = (\y x -> x `op` y) b a = flip op b a
If a function f expects more than two arguments eventually, we can similarly create its curried version by partially applying an explicit lambda expression:
(\y z x -> f x y z) b c -- = (\x -> f x b c)
(\x z y -> f x y z) a c -- = (\y -> f a y c)
(\x y z -> f x y z) a b -- = (\z -> f a b z)
The last case is equivalent to just f a b, and the second to (flip . f) a c:
g b c a = f a b c = flip f b a c = flip (flip f b) c a = (flip . flip f) b c a
g a c b = f a b c = flip (f a) c b = (flip . f) a c b
g a b c = f a b c
While musing what more useful standard class to suggest to this one
class Coordinate c where
createCoordinate :: x -> y -> c x y
getFirst :: c x y -> x
getSecond :: c x y -> y
addCoordinates :: (Num x, Num y) => c x y -> c x y -> c x y
it occured me that instead of something VectorSpace-y or R2, a rather more general beast might lurk here: a Type -> Type -> Type whose two contained types can both be extracted. Hm, perhaps they can be extracted?
Turns out neither the comonad nor bifunctors package contains something called Bicomonad. Question is, would such a class even make sense, category-theoretically? Unlike Bimonad (which also isn't defined, and I couldn't really see how might look), a naïve definition seems plausible:
class Bifunctor c => Bicomonad c where
fst :: c x y -> x
snd :: c x y -> y
bidup :: c x y -> c (c x y) (c x y)
probably with the laws
fst . bidup ≡ id
snd . bidup ≡ id
bimap fst snd . bidup ≡ id
bimap bidup bidup . bidup ≡ bidup . bidup
but I find it disquieting that both fields of the result of bidup contain the same type, and there are quite a number of other, perhaps “better” conceivable signatures.
Any thoughts?
This is not an answer, but for Bimonad, how about this?
class Biapplicative p => Bimonad p where
(>>==) :: p a b -> (a -> b -> p c d) -> p c d
biap :: Bimonad p => p (a -> b) (c -> d) -> p a c -> p b d
biap p q = p >>== \ab cd -> q >>== \a c -> bipure (ab a) (cd c)
instance Bimonad (,) where
(a,b) >>== f = f a b
I don't know if this is categorically right/interesting, or even remotely useful, but it smells right from a Haskell perspective. Does it match your Bicomonad or something similar?
I'm going through the new book Haskell Programming from First Principles. It seems decent, but I feel that there are some confusing holes in the explanations. I apologize if I'm missing something basic.
The last problem in chapter 5 is to fill in the ??? below so that things make sense:
munge :: (x -> y) -> (y -> (w, z)) -> x -> w
munge = ???
The solution which was explained to me (after much head-scratching) goes:
g :: y -> (w, z)
g = undefined
f :: x -> y
f = undefined
munge :: (x -> y) -> (y -> (w, z)) -> x -> w
munge g f v = fst (g (f v))
I'm getting hung up on this example in two ways.
First, it seems like the munge function ought to take a function as input which takes x -> y. But the way munge is defined, it seems like we supply an additional argument v to the function f first. But if f :: x -> y, then won't the expression f v be of type just y instead of x -> y?
Second, I'm struggling to understand why the x appears in the second-to-last position in the type declaration. At that point I feel like the logical next piece after the (y -> (w,x)) step should just be w, since at that stage the function g is being applied to fst and w ought to be the type of what fst returns. I can feel that I'm close, but can't quite close the gap.
Clearly I'm not understanding the notation correctly. Can anyone set me straight?
EDIT: Ok, here is a clarifying question to the second part. Is it possible to revise the munge function so that it has the following type (i.e. original type with second-to-last x application omitted)? If so what would it look like?
munge :: (x -> y) -> (y -> (w, z)) -> w
The answer is incorrect and ill-typed. f and g should be swapped:
munge :: (x -> y) -> (y -> (w, z)) -> x -> w
munge g f v = fst (f (g v))
I'm not sure if that clears up your confusion.
EDIT In case it's interesting, here are more equivalent ways of writing this function and its type:
-- notice parens in type signature; `->` associates right
munge :: (x -> y) -> ((y -> (w, z)) -> (x -> w))
munge g f v = -- omitted
-- type signature omitted
munge boop _plort zOWY = fst (_plort (boop zOWY))
munge g f = fst . f . g
munge g = \f v -> fst . f . g $ v
-- don't do this please
munge = ((fst .) .) . (.)
EDIT2 It might be helpful to play around with this in GHCi, asking the inferred type of different expressions:
Prelude> let munge g f v = fst (f (g v))
Prelude> :t munge
munge :: (t1 -> t) -> (t -> (a, b)) -> t1 -> a
Prelude> :t munge head
munge head :: (t -> (a, b)) -> [t] -> a
Prelude> :t munge head (\x-> (x, not x))
munge head (\x-> (x, not x)) :: [Bool] -> Bool
Prelude> :t munge ((+1) . fst . snd . head)
munge ((+1) . fst . snd . head)
:: Num t => (t -> (a, b)) -> [(a1, (t, b1))] -> a
The solution, confusingly, is using the same variables f and g for two different things: as global names for two functions, and as parameter names in defining munge. Making a change of variable should make it clearer:
g :: y -> (w, z)
g = undefined
f :: x -> y
f = undefined
munge :: (x -> y) -> (y -> (w, z)) -> x -> w
munge f1 f2 v = fst (f2 (f1 v)) -- fst . f2 . f1 $ v
Then you would call munge on f and g will thing like
munge f g someArgumentForF
Inside munge, f (called f1) is first applied to someArgumentForF (called v) to get a value that can be passed to g (called f2). This produces a tuple, and applying fst to the tuple returns the value of type w needed as the final result.
Mind the pure function below, in an imperative language:
def foo(x,y):
x = f(x) if a(x)
if c(x):
x = g(x)
else:
x = h(x)
x = f(x)
y = f(y) if a(y)
x = g(x) if b(y)
return [x,y]
That function represents a style where you have to incrementally update variables. It can be avoided in most cases, but there are situations where that pattern is unavoidable - for example, writing a cooking procedure for a robot, which inherently requires a series of steps and decisions. Now, imagine we were trying to represent foo in Haskell.
foo x0 y0 =
let x1 = if a x0 then f x0 else x0 in
let x2 = if c x1 then g x1 else h x1 in
let x3 = f x2 in
let y1 = if a y0 then f y0 else y0 in
let x4 = if b y1 then g x3 else x3 in
[x4,y1]
That code works, but it is too complicated and error prone due to the need for manually managing the numeric tags. Notice that, after x1 is set, x0's value should never be used again, but it still can. If you accidentally use it, that will be an undetected error.
I've managed to solve this problem using the State monad:
fooSt x y = execState (do
(x,y) <- get
when (a x) (put (f x, y))
(x,y) <- get
if c x
then put (g x, y)
else put (h x, y)
(x,y) <- get
put (f x, y)
(x,y) <- get
when (a y) (put (x, f y))
(x,y) <- get
when (b y) (put (g x, x))) (x,y)
This way, need for tag-tracking goes away, as well as the risk of accidentally using an outdated variable. But now the code is verbose and much harder to understand, mainly due to the repetition of (x,y) <- get.
So: what is a more readable, elegant and safe way to express this pattern?
Full code for testing.
Your goals
While the direct transformation of imperative code would usually lead to the ST monad and STRef, lets think about what you actually want to do:
You want to manipulate values conditionally.
You want to return that value.
You want to sequence the steps of your manipulation.
Requirements
Now this indeed looks first like the ST monad. However, if we follow the simple monad laws, together with do notation, we see that
do
x <- return $ if somePredicate x then g x
else h x
x <- return $ if someOtherPredicate x then a x
else b x
is exactly what you want. Since you need only the most basic functions of a monad (return and >>=), you can use the simplest:
The Identity monad
foo x y = runIdentity $ do
x <- return $ if a x then f x
else x
x <- return $ if c x then g x
else h x
x <- return $ f x
y <- return $ if a x then f y
else y
x <- return $ if b y then g x
else y
return (x,y)
Note that you cannot use let x = if a x then f x else x, because in this case the x would be the same on both sides, whereas
x <- return $ if a x then f x
else x
is the same as
(return $ if a x then (f x) else x) >>= \x -> ...
and the x in the if expression is clearly not the same as the resulting one, which is going to be used in the lambda on the right hand side.
Helpers
In order to make this more clear, you can add helpers like
condM :: Monad m => Bool -> a -> a -> m a
condM p a b = return $ if p then a else b
to get an even more concise version:
foo x y = runIdentity $ do
x <- condM (a x) (f x) x
x <- fmap f $ condM (c x) (g x) (h x)
y <- condM (a y) (f y) y
x <- condM (b y) (g x) x
return (x , y)
Ternary craziness
And while we're up to it, lets crank up the craziness and introduce a ternary operator:
(?) :: Bool -> (a, a) -> a
b ? ie = if b then fst ie else snd ie
(??) :: Monad m => Bool -> (a, a) -> m a
(??) p = return . (?) p
(#) :: a -> a -> (a, a)
(#) = (,)
infixr 2 ??
infixr 2 #
infixr 2 ?
foo x y = runIdentity $ do
x <- a x ?? f x # x
x <- fmap f $ c x ?? g x # h x
y <- a y ?? f y # y
x <- b y ?? g x # x
return (x , y)
But the bottomline is, that the Identity monad has everything you need for this task.
Imperative or non-imperative
One might argue whether this style is imperative. It's definitely a sequence of actions. But there's no state, unless you count the bound variables. However, then a pack of let … in … declarations also gives an implicit sequence: you expect the first let to bind first.
Using Identity is purely functional
Either way, the code above doesn't introduce mutability. x doesn't get modified, instead you have a new x or y shadowing the last one. This gets clear if you desugar the do expression as noted above:
foo x y = runIdentity $
a x ?? f x # x >>= \x ->
c x ?? g x # h x >>= \x ->
return (f x) >>= \x ->
a y ?? f y # y >>= \y ->
b y ?? g x # x >>= \x ->
return (x , y)
Getting rid of the simplest monad
However, if we would use (?) on the left hand side and remove the returns, we could replace (>>=) :: m a -> (a -> m b) -> m b) by something with type a -> (a -> b) -> b. This just happens to be flip ($). We end up with:
($>) :: a -> (a -> b) -> b
($>) = flip ($)
infixr 0 $> -- same infix as ($)
foo x y = a x ? f x # x $> \x ->
c x ? g x # h x $> \x ->
f x $> \x ->
a y ? f y # y $> \y ->
b y ? g x # x $> \x ->
(x, y)
This is very similar to the desugared do expression above. Note that any usage of Identity can be transformed into this style, and vice-versa.
The problem you state looks like a nice application for arrows:
import Control.Arrow
if' :: (a -> Bool) -> (a -> a) -> (a -> a) -> a -> a
if' p f g x = if p x then f x else g x
foo2 :: (Int,Int) -> (Int,Int)
foo2 = first (if' c g h . if' a f id) >>>
first f >>>
second (if' a f id) >>>
(\(x,y) -> (if b y then g x else x , y))
in particular, first lifts a function a -> b to (a,c) -> (b,c), which is more idiomatic.
Edit: if' allows a lift
import Control.Applicative (liftA3)
-- a functional if for lifting
if'' b x y = if b then x else y
if' :: (a -> Bool) -> (a -> a) -> (a -> a) -> a -> a
if' = liftA3 if''
I'd probably do something like this:
foo x y = ( x', y' )
where x' = bgf y' . cgh . af $ x
y' = af y
af z = (if a z then f else id) z
cgh z = (if c z then g else h) z
bg y x = (if b y then g else id) x
For something more complicated, you may want to consider using lens:
whenM :: Monad m => m Bool -> m () -> m ()
whenM c a = c >>= \res -> when res a
ifM :: Monad m => m Bool -> m a -> m a -> m a
ifM mb ml mr = mb >>= \b -> if b then ml else mr
foo :: Int -> Int -> (Int, Int)
foo = curry . execState $ do
whenM (uses _1 a) $
_1 %= f
ifM (uses _1 c)
(_1 %= g)
(_1 %= h)
_1 %= f
whenM (uses _2 a) $
_2 %= f
whenM (uses _2 b) $ do
_1 %= g
And there's nothing stopping you from using more descriptive variable names:
foo :: Int -> Int -> (Int, Int)
foo = curry . execState $ do
let x :: Lens (a, c) (b, c) a b
x = _1
y :: Lens (c, a) (c, b) a b
y = _2
whenM (uses x a) $
x %= f
ifM (uses x c)
(x %= g)
(x %= h)
x %= f
whenM (uses y a) $
y %= f
whenM (uses y b) $ do
x %= g
This is a job for the ST (state transformer) library.
ST provides:
Stateful computations in the form of the ST type. These look like ST s a for a computation that results in a value of type a, and may be run with runST to obtain a pure a value.
First-class mutable references in the form of the STRef type. The newSTRef a action creates a new STRef s a reference with an initial value of a, and which can be read with readSTRef ref and written with writeSTRef ref a. A single ST computation can use any number of STRef references internally.
Together, these let you express the same mutable variable functionality as in your imperative example.
To use ST and STRef, we need to import:
{-# LANGUAGE NoMonomorphismRestriction #-}
import Control.Monad.ST.Safe
import Data.STRef
Instead of using the low-level readSTRef and writeSTRef all over the place, we can define the following helpers to match the imperative operations that the Python-style foo example uses:
-- STRef assignment.
(=:) :: STRef s a -> ST s a -> ST s ()
ref =: x = writeSTRef ref =<< x
-- STRef function application.
($:) :: (a -> b) -> STRef s a -> ST s b
f $: ref = f `fmap` readSTRef ref
-- Postfix guard syntax.
if_ :: Monad m => m () -> m Bool -> m ()
action `if_` guard = act' =<< guard
where act' b = if b then action
else return ()
This lets us write:
ref =: x to assign the value of ST computation x to the STRef ref.
(f $: ref) to apply a pure function f to the STRef ref.
action `if_` guard to execute action only if guard results in True.
With these helpers in place, we can faithfully translate the original imperative definition of foo into Haskell:
a = (< 10)
b = even
c = odd
f x = x + 3
g x = x * 2
h x = x - 1
f3 x = x + 2
-- A stateful computation that takes two integer STRefs and result in a final [x,y].
fooST :: Integral n => STRef s n -> STRef s n -> ST s [n]
fooST x y = do
x =: (f $: x) `if_` (a $: x)
x' <- readSTRef x
if c x' then
x =: (g $: x)
else
x =: (h $: x)
x =: (f $: x)
y =: (f $: y) `if_` (a $: y)
x =: (g $: x) `if_` (b $: y)
sequence [readSTRef x, readSTRef y]
-- Pure wrapper: simply call fooST with two fresh references, and run it.
foo :: Integral n => n -> n -> [n]
foo x y = runST $ do
x' <- newSTRef x
y' <- newSTRef y
fooST x' y'
-- This will print "[9,3]".
main = print (foo 0 0)
Points to note:
Although we first had to define some syntactical helpers (=:, $:, if_) before translating foo, this demonstrates how you can use ST and STRef as a foundation to grow your own little imperative language that's directly suited to the problem at hand.
Syntax aside, this matches the structure of the original imperative definition exactly, without any error-prone restructuring. Any minor changes to the original example can be mirrored directly to Haskell. (The addition of the temporary x' <- readSTRef x binding in the Haskell code is only in order to use it with the native if/else syntax: if desired, this can be replaced with an appropriate ST-based if/else construct.)
The above code demonstrates giving both pure and stateful interfaces to the same computation: pure callers can use foo without knowing that it uses mutable state internally, while ST callers can directly use fooST (and for example provide it with existing STRefs to modify).
#Sibi said it best in his comment:
I would suggest you to stop thinking imperatively and rather think in a functional way. I agree that it will take some time to getting used to the new pattern, but try to translate imperative ideas to functional languages isn't a great approach.
Practically speaking, your chain of let can be a good starting point:
foo x0 y0 =
let x1 = if a x0 then f x0 else x0 in
let x2 = if c x1 then g x1 else h x1 in
let x3 = f x2 in
let y1 = if a y0 then f y0 else y0 in
let x4 = if b y1 then g x3 else x3 in
[x4,y1]
But I would suggest using a single let and giving descriptive names to the intermediate stages.
In this example unfortunately I don't have a clue what the various x's and y's do, so I cannot suggest meaningful names. In real code you would use names such as x_normalized, x_translated, or such, instead of x1 and x2, to describe what those values really are.
In fact, in a let or where you don't really have variables: they're just shorthand names you give to intermediate results, to make it easy to compose the final expression (the one after in or before the where.)
This is the spirit behind the x_bar and x_baz below. Try to come up with names that are reasonably descriptive, given the context of your code.
foo x y =
let x_bar = if a x then f x else x
x_baz = f if c x_bar then g x_bar else h x_bar
y_bar = if a y then f y else y
x_there = if b y_bar then g x_baz else x_baz
in [x_there, y_bar]
Then you can start recognizing patterns that were hidden in the imperative code. For example, x_bar and y_bar are basically the same transformation, applied respectively to x and y: that's why they have the same suffix "_bar" in this nonsensical example; then your x2 probably doesn't need an intermediate name , since you can just apply f to the result of the entire "if c then g else h".
Going on with the pattern recognition, you should factor out the transformations that you are applying to variables into sub-lambdas (or whatever you call the auxiliary functions defined in a where clause.)
Again, I don't have a clue what the original code did, so I cannot suggest meaningful names for the auxiliary functions. In a real application, f_if_a would be called normalize_if_needed or thaw_if_frozen or mow_if_overgrown... you get the idea:
foo x y =
let x_bar = f_if_a x
y_bar = f_if_a y
x_baz = f (g_if_c_else_h x_bar)
x_there = g_if_b x_baz y_bar
in [x_there, y_bar]
where
f_if_a x
| a x = f x
| otherwise = x
g_if_c_else_h x
| c x = g x
| otherwise = h x
g_if_b x y
| b y = g x
| otherwise = x
Don't disregard this naming business.
The whole point of Haskell and other pure functional languages is to express algorithms without the assignment operator, meaning the tool that can modify the value of an existing variable.
The names you give to things inside a function definition, whether introduced as arguments, let, or where, can only refer to one value (or auxiliary function) throughout the entire definition, so that your code can be more easily reasoned about and proven correct.
If you don't give them meaningful names (and conversely giving your code a meaningful structure) then you're missing out on the entire purpose of Haskell.
(IMHO the other answers so far, citing monads and other shenanigans, are barking up the wrong tree.)
I always prefer layering state transformers to using a single state over a tuple: it definitely declutters things by letting you "focus" on a specific layer (representations of the x and y variables in our case):
import Control.Monad.Trans.Class
import Control.Monad.Trans.State
foo :: x -> y -> (x, y)
foo x y =
(flip runState) y $ (flip execStateT) x $ do
get >>= \v -> when (a v) (put (f v))
get >>= \v -> put ((if c v then g else h) v)
modify f
lift $ get >>= \v -> when (a v) (put (f v))
lift get >>= \v -> when (b v) (modify g)
The lift function allows us to focus on the inner state layer, which is y.