# 99 questions/Solutions/8

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

m |
Jargonjustin (Talk | contribs) |
||

Line 22: | Line 22: | ||

<haskell>compress xs = map head $ group xs</haskell> | <haskell>compress xs = map head $ group xs</haskell> | ||

+ | |||

An alternative solution is | An alternative solution is | ||

<haskell> | <haskell> | ||

− | compress (x : y : | + | compress (x:ys@(y:_)) |

− | compress | + | | x == y = compress ys |

+ | | otherwise = x : compress ys | ||

+ | compress ys = ys | ||

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

## Revision as of 04:56, 11 May 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