# 99 questions/Solutions/8

### From HaskellWiki

(**) 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

Maybe

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

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)