99 questions/Solutions/8

(**) Eliminate consecutive duplicates of list elements.

compress :: Eq a => [a] -> [a]
compress = map head . group

We simply group equal values together (using Data.List.group), then take the head of each.

An alternative solution is

compress (x:ys@(y:_))
| x == y    = compress ys
| otherwise = x : compress ys
compress ys = ys

A variation of the above using foldr (note that GHC erases the Maybes, producing efficient code):

compress xs = foldr f (const []) xs Nothing
where
f x r a@(Just q) | x == q = r a
f x r _ = x : r (Just x)

Another possibility using foldr (this one is not so efficient, because it pushes the whole input onto the "stack" before doing anything else):

compress :: (Eq a) => [a] -> [a]
compress = foldr skipDups []
where skipDups x [] = [x]
skipDups x acc
| x == head acc = acc
| otherwise = x : acc

A similar solution without using foldr.

compress :: (Eq a) => [a] -> [a]
compress list = compress_acc list []
where compress_acc [] acc = acc
compress_acc [x] acc = (acc ++ [x])
compress_acc (x:xs) acc
| x == (head xs)  = compress_acc xs acc
| otherwise       = compress_acc xs (acc ++ [x])

A very simple approach:

compress []     = []
compress (x:xs) = x : (compress \$ dropWhile (== x) xs)

Another approach, using foldr

compress :: Eq a => [a] -> [a]
compress x = foldr (\a b -> if a == (head b) then b else a:b) [last x] x

Wrong solution using foldr

compress :: Eq a => [a] -> [a]
compress xs = foldr (\x acc -> if x `elem` acc then acc else x:acc) [] xs
-- Main> compress [1, 1, 1, 2, 2, 1, 1]
-- [2,1] - must be [1,2,1]

and using foldl

compress :: (Eq a) => [a] -> [a]
compress x = foldl (\a b -> if (last a) == b then a else a ++ [b]) [head x] x
compress' x = reverse \$ foldl (\a b -> if (head a) == b then a else b:a) [head x] x

A crazy variation that acts as a good transformer for fold/build fusion

{-# INLINE compress #-}
compress :: Eq a => [a] -> [a]
compress xs = build (\c n ->
let
f x r a@(Just q) | x == q = r a
f x r _ = x `c` r (Just x)
in
foldr f (const n) xs Nothing)

A simple approach that pairs each element with its consecutive element. We ignore all pairs with the same element, and return the list of all 'firsts' of these pairs. The last element has to be appended at the end.

consecutivePairs a = zip (init a) (tail a)
compress a = [ fst x | x <- consecutivePairs a, (fst x /= snd x) ] ++ [last a]