99 questions/Solutions/18

(**) Extract a slice from a list.

Given two indices, i and k, the slice is the list containing the elements between the i'th and k'th element of the original list (both limits included). Start counting the elements with 1.

slice xs (i+1) k = take (k-i) $ drop i xs

The same solution as above, but the more paranoid (maybe too paranoid?) version of it (uses guards and Maybe):

slice :: [a] -> Int -> Int -> Maybe [a]
slice [] _ _ = Just []
slice xs k n 	| k == n = Just []
		| k > n || k > length xs || n > length xs || k < 0 || n < 0 = Nothing
		| k == 0 = Just (take n xs)
		| otherwise = Just (drop (k-1) $ take n xs)

Or, an iterative solution:

slice :: [a]->Int->Int->[a]
slice lst 1 m = slice' lst m []
        where
                slice' :: [a]->Int->[a]->[a]
                slice' _ 0 acc = reverse acc
                slice' (x:xs) n acc = slice' xs (n - 1) (x:acc)
slice (x:xs) n m = slice xs (n - 1) (m - 1)

Or:

slice :: [a] -> Int -> Int -> [a]
slice (x:xs) i k
 | i > 1	= slice xs (i - 1) (k - 1)
 | k < 1	= []
 | otherwise	= x:slice xs (i - 1) (k - 1)

Another way using splitAt, though not nearly as elegant as the take and drop version:

slice :: [a] -> Int -> Int -> [a]
slice xs i k = chunk
  where chop  = snd $ splitAt i' xs          -- Get the piece starting at i
        chunk = fst $ splitAt (k - i') chop  -- Remove the part after k
        i'    = i - 1

A little cleaner, using the previous problem's split (a.k.a. splitAt):

slice xs (i+1) k = snd (split (fst (split xs k)) i)