99 questions/Solutions/3

(*) Find the K'th element of a list. The first element in the list is number 1.

This is (almost) the infix operator !! in Prelude, which is defined as:

(!!)                :: [a] -> Int -> a
(x:_)  !! 0         =  x
(_:xs) !! n         =  xs !! (n-1)

Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So:

elementAt :: [a] -> Int -> a
elementAt list i    = list !! (i-1)

Or without using the infix operator:

elementAt' :: [a] -> Int -> a
elementAt' (x:_) 1  = x
elementAt' [] _     = error "Index out of bounds"
elementAt' (_:xs) k
  | k < 1           = error "Index out of bounds"
  | otherwise       = elementAt' xs (k - 1)

Alternative version:

elementAt'' :: [a] -> Int -> a
elementAt'' (x:_) 1  = x
elementAt'' (_:xs) i = elementAt'' xs (i - 1)
elementAt'' _ _      = error "Index out of bounds"

This does not work correctly on invalid indexes and infinite lists, e.g.:

elementAt'' [1..] 0

A few more solutions using prelude functions:

elementAt'' xs n 
  | length xs < n = error "Index out of bounds"
  | otherwise = fst . last $ zip xs [1..n] 

elementAt''' xs n = head $ foldr ($) xs 
                         $ replicate (n - 1) tail
-- Negative indices not handled correctly:
-- Main> elementAt''' "haskell" (-1)
-- 'h'

elementAt'''' xs n
  | length xs < n = error "Index out of bounds"
  | otherwise = last $ take n xs

elementAt''''' xs n
  | length xs < n = error "Index out of bounds"
  | otherwise = head . reverse $ take n xs

elementAt'''''' xs n 
  | length xs < n = error "Index out of bounds"
  | otherwise = head $ drop (n - 1) xs

or elementAt_w' correctly in point-free style:

elementAt_w'pf = (last .) . take . (+ 1)

Pedantic note: the above definition of elementAt_w'pf does not conform to the order of arguments specified by the question, but the following does:

elementAt_w'pf' = flip $ (last .) . take . (+ 1)