Difference between revisions of "99 questions/Solutions/10"
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h = map head l
h = map head l
Revision as of 19:31, 18 January 2014
(*) Run-length encoding of a list.
Use the result of problem P09 to implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.
encode xs = map (\x -> (length x,head x)) (group xs)
which can also be expressed as a list comprehension:
[(length x, head x) | x <- group xs]
encode :: Eq a => [a] -> [(Int, a)] encode = map (\x -> (length x, head x)) . group
Or (ab)using the "&&&" arrow operator for tuples:
encode :: Eq a => [a] -> [(Int, a)] encode xs = map (length &&& head) $ group xs
Or using the slightly more verbose (w.r.t.
(&&&)) Applicative combinators:
encode :: Eq a => [a] -> [(Int, a)] encode = map ((,) <$> length <*> head) . pack
Or with the help of foldr (pack is the resulting function from P09):
encode xs = (enc . pack) xs where enc = foldr (\x acc -> (length x, head x) : acc) 
Or using takeWhile and dropWhile:
encode  =  encode (x:xs) = (length $ x : takeWhile (==x) xs, x) : encode (dropWhile (==x) xs)
Or without higher order functions:
encode  =  encode (x:xs) = encode' 1 x xs where encode' n x  = [(n, x)] encode' n x (y:ys) | x == y = encode' (n + 1) x ys | otherwise = (n, x) : encode' 1 y ys
Or we can make use of zip and group:
import List encode :: Eq a => [a] -> [(Int, a)] encode xs=zip (map length l) h where l = (group xs) h = map head l