# 99 questions/Solutions/8

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

Flowerking (Talk | contribs) m |
m (replace (++) with (:) as the element is consed to the list) |
||

Line 47: | Line 47: | ||

<haskell> |
<haskell> |
||

compress [] = [] |
compress [] = [] |
||

− | compress (x:xs) = [x] ++ (compress $ dropWhile (== x) xs) |
+ | compress (x:xs) = x : (compress $ dropWhile (== x) xs) |

</haskell> |
</haskell> |
||

## Revision as of 04:05, 1 December 2011

(**) 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