# 99 questions/Solutions/20

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< 99 questions | Solutions(Difference between revisions)

(Adding a third solution.) |
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<haskell> | <haskell> | ||

removeAt n xs = (xs!!n,take n xs ++ drop (n+1) xs) | removeAt n xs = (xs!!n,take n xs ++ drop (n+1) xs) | ||

+ | </haskell> | ||

+ | |||

+ | Another solution that avoids throwing an error and using ++ operators. Treats 1 as the first element in the list. | ||

+ | |||

+ | <haskell> | ||

+ | removeAt :: Int -> [a] -> (Maybe a, [a]) | ||

+ | removeAt _ [] = (Nothing, []) | ||

+ | removeAt 1 (x:xs) = (Just x, xs) | ||

+ | removeAt k (x:xs) = let (a, r) = removeAt (k - 1) xs in (a, x:r) | ||

</haskell> | </haskell> |

## Revision as of 08:10, 29 July 2010

(*) Remove the K'th element from a list.

removeAt :: Int -> [a] -> (a, [a]) removeAt k xs = case back of [] -> error "removeAt: index too large" x:rest -> (x, front ++ rest) where (front, back) = splitAt k xs

splitAt

If the original list has fewer than k+1 elements, the second list will be empty, and there will be no element to extract. Note that the Prolog and Lisp versions treat 1 as the first element in the list, and the Lisp version appends NIL elements to the end of the list if k is greater than the list length.

or

removeAt n xs = (xs!!n,take n xs ++ drop (n+1) xs)

Another solution that avoids throwing an error and using ++ operators. Treats 1 as the first element in the list.

removeAt :: Int -> [a] -> (Maybe a, [a]) removeAt _ [] = (Nothing, []) removeAt 1 (x:xs) = (Just x, xs) removeAt k (x:xs) = let (a, r) = removeAt (k - 1) xs in (a, x:r)