Apply a function multiple times in Haskell [duplicate]

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This question already has answers here:

How to call the same function 'n' times? [duplicate] (4 answers)

Library function to compose a function with itself n times (10 answers)

Closed 9 years ago.

Learn You a Haskell For Great Good (section "Higher Order Functions", subsection "Some higher-orderism is in order") describes an example function applyTwice that calls a function on an argument twice:

applyTwice :: (a -> a) -> a -> a  
applyTwice f x = f (f x)

But i need a function that applies some function over some argument an arbitrary amount of times. For example applyN 3 f x would be equivalent to f $ f $ f x. How would i write a function of repeated application in Haskell? Please post any possible solutions, using recursion, higher-order functions or anything else.

Answer 1

I always did something like iterate f x !! n.

Answer 2

You will have to do a recursive function. The obvious case will be be when you apply the function 0 time, it will be like you don't modify the input. The recursive will come from the fact that applyN n f x == f (applyN (n -1) f x or because composition of function is associative applyN n f x == apply (n - 1) f (f x). The second option lead to better performance because it will be tail recursive

applyN :: Int n => n -> (a -> a) -> a -> a
applyN 0 _ x = x
applyN n f x = applyN (n - 1) f (f x)

Answer 3

applyN = (foldr (.) id.) . replicate

-->>
applyN 0 f = id
applyN 1 f = f
applyN 2 f = (f.f)
-- ...

Or just use iterate, as was already said before. The only real difference is that using iterate you will get an exception if you use a negative n, whereas in this solution you'd get id. If that's a case that can plausibly happen for your use case, you should consider which is the behavior you like better.