# 99 questions/Solutions/9

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

Line 1: | Line 1: | ||

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

+ | |||

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

Line 9: | Line 10: | ||

A more verbose solution is |
A more verbose solution is |
||

+ | |||

<haskell> |
<haskell> |
||

pack :: Eq a => [a] -> [[a]] |
pack :: Eq a => [a] -> [[a]] |
||

Line 22: | Line 24: | ||

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

+ | |||

+ | Another solution using <hask>takeWhile</hask> and <hask>dropWhile</hask>: |
||

+ | |||

+ | <haskell> |
||

+ | pack :: (Eq a) => [a] -> [[a]] |
||

+ | pack [] = [] |
||

+ | pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs) |
||

+ | </haskell> |

## Revision as of 03:00, 14 July 2010

(**) 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 [] = []

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

group

Data.List

takeWhile

dropWhile

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