使Haskell功能无点

Question

I'm a Haskell beginner and I've been playing around with point-free functions. I have problems with two functions - lambdabot's solutions were absolutely unreadable and made the code obfuscated, so I'm asking here in case there is a way to simplify the functions.

The first function removes duplicates from a list.

func1 :: Eq a => [a] -> [a]
func1 [] = []
func1 (x:xs) = x : (func1 . filter (/=x) $ xs)

I tried making a point-free version of this function with foldr and >>= but did not succeed.

The second function maps a list to a list of tuples containing the original elements and how often they occurred in the list.

func2 :: Eq a => [a] -> [(a, Int)]
func2 xs = map ( \f -> (f, count f xs) ) xs

where count a = length.filter(==a) . I am not sure if making a point-free version of this function while maintaining readability is even possible, but I'd like to make sure.

Any help with making the two functions point-free will be appreciated.

Answer 1

Well, func1 can be written as a fold: func1 = foldr (\\x xs -> x : filter (/= x) xs) [] . ¹ However, you don't have to, as it's identical to the standard function nub .

And you can remove some points from func2 using the (&&&) :: (a -> b) -> (a -> c) -> a -> (b,c) ² combinator from Control.Arrow :

func2 xs = map (id &&& (`count` xs)) xs

Which can then be made fully point-free:

func2 = (id &&&) . flip count >>= map

But, frankly, I had to use lambdabot to do that last step; I'd suggest keeping that function in its original form. Point-free style is only helpful when it aids comprehension; if you're having difficulty making a function point-free, then it's probably counterproductive.

¹ Which can then be made fully point-free as foldr (liftM2 (.) (:) (filter . (/=))) [] (thanks again, lambdabot!) but, again, I really don't recommend this. There's not a point-free combinator for every situation.

² (&&&) actually has a more general type; it works with any Arrow , rather than just (->) . But that's not relevant here.