Difference between revisions of "99 questions/Solutions/10"
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m (Linked to MR) |
(Added a solution that makes the usage of foldr) |
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encode :: Eq a => [a] -> [(Int, a)] |
encode :: Eq a => [a] -> [(Int, a)] |
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encode xs = map (length &&& head) $ group xs |
encode xs = map (length &&& head) $ group xs |
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+ | </haskell> |
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+ | |||
+ | Or with the help of foldr: |
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+ | |||
+ | <haskell> |
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+ | encode xs = (enc . pack) xs |
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+ | where enc = foldr (\x acc -> (length x, head x) : acc) [] |
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</haskell> |
</haskell> |
Revision as of 14:46, 16 August 2010
(*) 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:
encode xs = (enc . pack) xs
where enc = foldr (\x acc -> (length x, head x) : acc) []