Haskell 99, Problems 1-10 (simple list processing)

I'm just learning Haskell and I wanted to know if I'm going in the right direction with my solving of the Haskell 99 problems. The file of interest is here which I've also reproduced below.

```
module OneToTen where

{-- Problem 1: Find the last element of a list. --}
myLast :: [a] -> a
myLast [] = error "Empty list."
myLast [x] = x
myLast (x:xs) = myLast xs

{-- Problem 2: Find the last but one element of a list. --}
myButLast :: [a] -> a
myButLast [] = error "Empty list."
myButLast [x] = error "One element list."
myButLast [x,y] = x
myButLast (x:xs) = myButLast xs

{-- Problem 3: Find the Kth element of a list. The first element in the list
-- is number 1 --}
elementAt :: [a] -> Int -> a
elementAt [] _ = error "Empty list."
elementAt (x:_) 1 = x
elementAt (_:xs) n = elementAt xs (n-1)

{-- Problem 4: Find the number of elements of a list. --}
myLength :: [a] -> Int
myLength xs = myLengthAux xs 0

myLengthAux :: [a] -> Int -> Int
myLengthAux [] acc = acc
myLengthAux (_:xs) acc = myLengthAux xs (acc+1)

{-- Problem 5: Reverse a list. --}
myReverse :: [a] -> [a]
myReverse xs = myReverseAux xs []

myReverseAux :: [a] -> [a] -> [a]
myReverseAux [] acc = acc
myReverseAux (x:xs) acc = myReverseAux xs (x:acc)

{-- Problem 7: Flatten a nested list structure. --}
data NestedList a = Elem a | List [NestedList a]

flatten :: NestedList a -> [a]
flatten xs = flattenAux xs []

flattenAux :: NestedList a -> [a] -> [a]
flattenAux (List []) acc = acc
flattenAux (Elem x) acc = x:acc
flattenAux (List (x:xs)) acc = flattenAux (List xs) (acc ++ (flattenAux x []))

{-- Problem 8: Eliminate consecutive duplicated of list elements. --}
compress :: (Eq a) => [a] -> [a]
compress [] = []
compress (x:xs) = compressAux xs x [x]

compressAux :: (Eq a) => [a] -> a -> [a] -> [a]
compressAux [] y acc = acc
compressAux (x:xs) y acc | x == y = compressAux xs y acc
| x /= y = compressAux xs x (acc ++ [x])

{-- Problem 9: Pack consecutive duplicates of list elements into s

Solutions 1 and 2 are fine.

Solution 3: Find the Kth element of a list

The error message for 3 is a bit misleading: you can also get the error if the requested index is beyond the end of the list. Strictly speaking, it should be an error to request a non-positive index as well.

Solution 4: Find the number of elements of a list

You should not need a helper function.

myLength :: [a] -> Int
myLength [] = 0
myLength (x:xs) = 1 + myLength xs

However, as @Franky points out, using a helper function can make it tail-recursive.

Solution 5: Reverse a list

You could write it as myReverse (x:xs) = (myReverse xs) ++ x, but it is much more efficient the way you wrote it using a helper function. The helper should be scoped using a where clause (or let … in, which does the same thing). It is customary to name trivial helper functions using the myReverse' convention.

myReverse :: [a] -> [a]
myReverse xs = myReverse' xs []
  where
    myReverse' []     rev = rev
    myReverse' (x:xs) rev = myReverse' xs (x:rev)

Solution 7: Flatten a nested list structure.

No helper is necessary.

data NestedList a = Elem a | List [NestedList a]

flatten :: NestedList a -> [a]
flatten (Elem x)      = [x]
flatten (List [])     = []
flatten (List (x:xs)) = flatten x ++ flatten (List xs)

Solution 8: Eliminate consecutive duplicated list elements

Again, no helper or accumulator is necessary. Furthermore, you should avoid ++, as it works by inefficiently walking to the end of its LHS argument.

compress :: Eq a => [a] -> [a]
compress []     = []
compress (x:[]) = [x]
compress (x:y:zs)
  | x == y      = compress (y:zs)
  | otherwise   = x : compress (y:zs)

Problem 9: Pack consecutive duplicates of list elements into sublists

I don't think that packing an empty list should be an error. Rather it should just return an empty list.

My advice is similar to the above. I think that a helper function is needed, but not in that way.

pack :: Eq a => [a] -> [[a]]
pack [] = []
pack (x:xs) = pack' [x] xs
  where
    pack' run []  = [run]
    pack' run@(r:_) (x:xs)
      | x == r    = pack' (x:run) xs
      | otherwise = run : pack' [x] xs

Problem 10: Run-length encoding of a list

Again, I don't think that encoding an empty list should be an error.

The implementation would read more nicely, in my opinion, using a list comprehension and better naming.

encode :: Eq a => [a] -> [(Int, a)]
encode xs = [(length run, head run) | run <- pack xs]

In summary, accumulators are sometimes the right approach (as in myReverse), but often they are harmful. Before using one, think hard about whether it is necessary and efficient.

StackExchange Code Review Q#86202, answer score: 3