# 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.
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 : y : xs) = (if x == y then [] else [x]) ++ compress (y : xs)
compress x = x
```

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)
```