# 99 questions/Solutions/9

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< 99 questions | Solutions(Difference between revisions)

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− | This is implemented as <hask>group</hask> in <hask>Data.List</hask>. |
+ | This is implemented as <hask>group</hask> in <hask>Data.List</hask>. |

A more verbose solution is |
A more verbose solution is |
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− | Similarly, using <hask>splitAt</hask> and <hask>findIndex</hask>: |
+ | Similarly, using <hask>splitAt</hask> and <hask>findIndex</hask>: |

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pack (x:xs) = (x:reps) : (pack rest) |
pack (x:xs) = (x:reps) : (pack rest) |
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where |
where |
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− | (reps, rest) = maybe (xs,[]) (\i -> splitAt i xs) (findIndex (/=x) xs) |
+ | (reps, rest) = maybe (xs,[]) (\i -> splitAt i xs) |

+ | (findIndex (/=x) xs) |
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− | Another solution using <hask>takeWhile</hask> and <hask>dropWhile</hask>: |
+ | Another solution using <hask>takeWhile</hask> and <hask>dropWhile</hask>: |

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− | Or we can use <hask>foldr</hask> to implement this: |
+ | Or we can use <hask>foldr</hask> to implement this: |

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## Revision as of 09:53, 1 June 2011

(**) Pack consecutive duplicates of list elements into sublists.

If a list contains repeated elements they should be placed in separate sublists.

pack (x:xs) = let (first,rest) = span (==x) xs in (x:first) : pack rest pack [] = []

group

Data.List

A more verbose solution is

pack :: Eq a => [a] -> [[a]] pack [] = [] pack (x:xs) = (x:first) : pack rest where getReps [] = ([], []) getReps (y:ys) | y == x = let (f,r) = getReps ys in (y:f, r) | otherwise = ([], (y:ys)) (first,rest) = getReps xs

splitAt

findIndex

pack :: Eq a => [a] -> [[a]] pack [] = [] pack (x:xs) = (x:reps) : (pack rest) where (reps, rest) = maybe (xs,[]) (\i -> splitAt i xs) (findIndex (/=x) xs)

takeWhile

dropWhile

pack :: (Eq a) => [a] -> [[a]] pack [] = [] pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs)

foldr

pack :: (Eq a) => [a] -> [[a]] pack = foldr func [] where func x [] = [[x]] func x (y:xs) = if x == (head y) then ((x:y):xs) else ([x]:y:xs)

A simple solution:

pack :: (Eq a) => [a] -> [[a]] pack [] = [] pack [x] = [[x]] pack (x:xs) = if x `elem` (head (pack xs)) then (x:(head (pack xs))):(tail (pack xs)) else [x]:(pack xs)