An exercise in Haskell from First Principles says to implement filter using foldr and this is what I came up with but it feels and looks clunky. Is there a more natural way to implement it with a foldr?
import Data.Bool
myFilter :: (a -> Bool) -> [a] -> [a]
myFilter f = foldr (\x -> bool (++ []) ((:) x) (f x)) []
I would only use bool if it let me get rid of the lambda expression simply, by composing a call to bool with the predicate p: bool iffalse iftrue . p. However, p isn't the only function that needs to be called on a list element; (:) does as well. You could use the Applicative instance for functions, to write
myfilter p = foldr (bool id . (:) <*> p) [] -- yuck
but in this case I would just use a plain if expression, inside the lambda expression:
myfilter p = foldr (\x -> if p x then (x:) else id) [] -- much clearer!
Note that when specialized to functions, the Applicative's (<*>) operator is defined as f <*> g = \x -> f x (g x). I leave it as an exercise to use that definition to transform bool id . (:) <*> p into
\x -> bool id (x:) (p x).