99 questions/Solutions/8
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(**) 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 fnomonomorphismrestriction
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