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