# 99 questions/Solutions/8

From HaskellWiki

< 99 questions | Solutions

(**) 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
```