Difference between revisions of "99 questions/Solutions/18"

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(Take and drop can be applied in the opposite order too.)
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slice xs i k = [x | (x,j) <- zip xs [1..k], i <= j]
 
slice xs i k = [x | (x,j) <- zip xs [1..k], i <= j]
 
</haskell>
 
</haskell>
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Another simple solution using take and drop:
 
Another simple solution using take and drop:
 
<haskell>
 
<haskell>
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| i > k = []
 
| i > k = []
 
| otherwise = (take (k-i+1) (drop (i-1) l))
 
| otherwise = (take (k-i+1) (drop (i-1) l))
  +
</haskell>
  +
  +
Zip, filter, unzip:
  +
<haskell>
  +
slice :: [a] -> Int -> Int -> [a]
  +
slice xs a b = fst $ unzip $ filter ((>=a) . snd) $ zip xs [1..b]
  +
</haskell>
  +
  +
Take and drop can be applied in the opposite order too:
  +
<haskell>
  +
slice xs i k = drop (i-1) $ take k xs
 
</haskell>
 
</haskell>

Revision as of 01:54, 25 February 2012

(**) 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 k | i>0 = take (k-i+1) $ drop (i-1) 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)

A solution using zip, filter then map seems straight-forward to me (NB: this won't work for infinite lists):

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

A solution using list comprehension:

slice xs i k = [x | (x,j) <- zip xs [1..k], i <= j]

Another simple solution using take and drop:

slice :: [a] -> Int -> Int -> [a]
slice l i k 
  | i > k = []
  | otherwise = (take (k-i+1) (drop (i-1) l))

Zip, filter, unzip:

slice :: [a] -> Int -> Int -> [a]
slice xs a b = fst $ unzip $ filter ((>=a) . snd) $ zip xs [1..b]

Take and drop can be applied in the opposite order too:

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