# 99 questions/Solutions/8

(Difference between revisions)

(**) 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. Note that (with GHC) we must give an explicit type to compress otherwise we get:

```Ambiguous type variable `a' in the constraint:
`Eq a'
arising from use of `group'
Possible cause: the monomorphism restriction applied to the following:
compress :: [a] -> [a]
Probable fix: give these definition(s) an explicit type signature
or use -fno-monomorphism-restriction```

We can circumvent the monomorphism restriction by writing compress this way (See: section 4.5.4 of the report):

`compress xs = map head \$ group xs`

An alternative solution is

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

Another possibility using foldr

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

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```