# 99 questions/Solutions/10

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< 99 questions | Solutions(Difference between revisions)

Lyklykkkkkkk (Talk | contribs) (adding one more solution) |
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| x == y = encode' (n + 1) x ys |
| x == y = encode' (n + 1) x ys |
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| otherwise = (n, x) : encode' 1 y ys |
| otherwise = (n, x) : encode' 1 y ys |
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+ | </haskell> |
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+ | |||

+ | Or we can make use of zip and group: |
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+ | |||

+ | <haskell> |
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+ | import List |
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+ | encode :: String -> [(Int, Char)] |
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+ | encode xs=zip (map length l) h where |
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+ | l = (group xs) |
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+ | h = map head l |
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</haskell> |
</haskell> |

## Revision as of 06:34, 25 May 2012

(*) 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]

Or writing it Pointfree (Note that the type signature is essential here to avoid hitting the Monomorphism Restriction):

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 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 :: String -> [(Int, Char)] encode xs=zip (map length l) h where l = (group xs) h = map head l