# 99 questions/Solutions/18

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< 99 questions | Solutions(Difference between revisions)

(another solution using splitAt) |
(I've come up with a solution that actually is an extended version of the 1st solution) |
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Line 5: | Line 5: | ||

<haskell> | <haskell> | ||

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

+ | </haskell> | ||

+ | |||

+ | The same solution as above, but with guards: | ||

+ | |||

+ | <haskell> | ||

+ | slice [] _ _ = [] | ||

+ | slice xs k n | k == n = [] | ||

+ | | k > n = error "k > n" | ||

+ | | k == 0 = take n xs | ||

+ | | otherwise = drop (k-1) $ take n xs | ||

</haskell> | </haskell> | ||

## Revision as of 13:00, 18 August 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 with guards:

slice [] _ _ = [] slice xs k n | k == n = [] | k > n = error "k > n" | k == 0 = take n xs | otherwise = 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