# 99 questions/Solutions/18

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< 99 questions | Solutions(Difference between revisions)

(added a cleaner version of the final solution) |
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<haskell> | <haskell> | ||

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

+ | </haskell> | ||

+ | |||

+ | A solution using <hask>zip</hask>, <hask>filter</hask> then <hask>map</hask> seems straight-forward to me | ||

+ | |||

+ | <haskell> | ||

+ | slice xs i j = map snd | ||

+ | $ filter (\(x,_) -> x >= i && x <= j) | ||

+ | $ zip [1..] xs | ||

</haskell> | </haskell> |

## Revision as of 06:27, 20 November 2010

(**) 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)

splitAt

take

drop

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

splitAt

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

zip

filter

map

slice xs i j = map snd $ filter (\(x,_) -> x >= i && x <= j) $ zip [1..] xs