99 questions/Solutions/20
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(*) 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
Simply use the splitAt to split after k elements.
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)
Another solution that also uses Maybe to indicate failure:
removeAt :: Int -> [a] -> (Maybe a, [a])
removeAt _ [] = (Nothing, [])
removeAt 0 xs = (Nothing, xs)
removeAt nr xs | nr > length xs = (Nothing, xs)
| otherwise = (Just (xs !! nr), fst splitted ++ (tail . snd) splitted)
where splitted = splitAt nr xs