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

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(Updated incorrect solution)
 
(11 intermediate revisions by 9 users not shown)
Line 17: Line 17:
 
| k == 0 = Just (take n xs)
 
| k == 0 = Just (take n xs)
 
| otherwise = Just (drop (k-1) $ take n xs)
 
| otherwise = Just (drop (k-1) $ take n xs)
  +
</haskell>
  +
  +
Or, a concise but inefficient for lists solution using list comprehension:
  +
  +
<haskell>
  +
-- Incorrect - only works on sequential data.
  +
--slice :: (Enum a) => [a] -> Int -> Int -> [a]
  +
--slice [] _ _ = []
  +
--slice xs m n = [(xs !! (m-1)) .. (xs !! (n-1))]
  +
  +
slice xs i k = map (xs!!) [i-1 .. k-1]
 
</haskell>
 
</haskell>
   
Line 22: Line 33:
   
 
<haskell>
 
<haskell>
slice :: [a]->Int->Int->[a]
+
slice :: [a] -> Int -> Int -> [a]
slice lst 1 m = slice' lst m []
+
slice lst 1 m = slice' lst m []
 
where
 
where
 
slice' :: [a]->Int->[a]->[a]
 
slice' :: [a]->Int->[a]->[a]
 
slice' _ 0 acc = reverse acc
 
slice' _ 0 acc = reverse acc
 
slice' (x:xs) n acc = slice' xs (n - 1) (x:acc)
 
slice' (x:xs) n acc = slice' xs (n - 1) (x:acc)
  +
slice' [] _ _ = []
 
slice (x:xs) n m = slice xs (n - 1) (m - 1)
 
slice (x:xs) n m = slice xs (n - 1) (m - 1)
  +
slice [] _ _ = []
 
</haskell>
 
</haskell>
   
Line 35: Line 48:
 
<haskell>
 
<haskell>
 
slice :: [a] -> Int -> Int -> [a]
 
slice :: [a] -> Int -> Int -> [a]
  +
slice [] _ _ = []
 
slice (x:xs) i k
 
slice (x:xs) i k
| i > 1 = slice xs (i - 1) (k - 1)
+
| i > 1 = slice xs (i - 1) (k - 1)
| k < 1 = []
+
| k < 1 = []
| otherwise = x:slice xs (i - 1) (k - 1)
+
| otherwise = x:slice xs (i - 1) (k - 1)
 
</haskell>
 
</haskell>
   
Line 66: Line 80:
 
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>
  +
  +
Another simple solution using take and drop:
  +
<haskell>
  +
slice :: [a] -> Int -> Int -> [a]
  +
slice l i k
  +
| i > k = []
  +
| 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>
  +
  +
Using a fold:
  +
<haskell>
  +
slice :: [a] -> Int -> Int -> [a]
  +
slice (x:xs) begin end = snd $ foldl helper (1, []) (x:xs)
  +
where helper (i, acc) x = if (i >= begin) && (i <= end) then (i+1, acc ++ [x]) else (i+1, acc)
  +
</haskell>
  +
  +
Another simple solution from first principles:
  +
<haskell>
  +
slice :: [a] -> Int -> Int -> [a]
  +
slice [] _ _ = []
  +
slice _ _ 0 = []
  +
slice (x:xs) 1 n = x:slice xs 1 (n-1)
  +
slice (_:xs) m n = slice xs (m-1) (n-1)
  +
</haskell>
  +
  +
[[Category:Programming exercise spoilers]]

Latest revision as of 22:51, 3 December 2020

(**) 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, a concise but inefficient for lists solution using list comprehension:

 
-- Incorrect - only works on sequential data.
--slice :: (Enum a) => [a] -> Int -> Int -> [a]
--slice [] _ _ = []
--slice xs m n = [(xs !! (m-1)) .. (xs !! (n-1))]

slice xs i k = map (xs!!) [i-1 .. k-1]

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' [] _ _ = []
slice (x:xs) n m = slice xs (n - 1) (m - 1)
slice []     _ _ = []

Or:

slice :: [a] -> Int -> Int -> [a]
slice [] _ _  = []
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

Using a fold:

slice :: [a] -> Int -> Int -> [a]
slice (x:xs) begin end = snd $ foldl helper (1, []) (x:xs)
    where helper (i, acc) x = if (i >= begin) && (i <= end) then (i+1, acc ++ [x]) else (i+1, acc)

Another simple solution from first principles:

slice :: [a] -> Int -> Int -> [a]
slice [] _ _ = []
slice _ _ 0 = []
slice (x:xs) 1 n = x:slice xs 1 (n-1)
slice (_:xs) m n = slice xs (m-1) (n-1)