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(Added solution using fold)
 
(10 intermediate revisions by 7 users not shown)
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<haskell>
 
<haskell>
slice xs (i+1) k = take (k-i) $ drop i xs
+
slice xs i k | i>0 = take (k-i+1) $ drop (i-1) xs
 
</haskell>
 
</haskell>
   
Line 13: Line 13:
 
slice [] _ _ = Just []
 
slice [] _ _ = Just []
 
slice xs k n | k == n = Just []
 
slice xs k n | k == n = Just []
| k > n || k > length xs || n > length xs || k < 0 || n < 0 = Nothing
+
| k > n || k > length xs ||
  +
n > length xs || k < 0 || n < 0 = Nothing
 
| 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)
Line 21: Line 21:
   
 
<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 34: Line 36:
 
<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>
   
Another way using <hask>splitAt</hask>, though not nearly as elegant as the <hask>take</hask> and <hask>drop</hask> version:
+
Another way using <hask>splitAt</hask>, though not nearly as elegant as the <hask>take</hask>&nbsp;and <hask>drop</hask>&nbsp;version:
   
 
<haskell>
 
<haskell>
Line 54: Line 57:
 
</haskell>
 
</haskell>
   
A solution using <hask>zip</hask>, <hask>filter</hask> then <hask>map</hask> seems straight-forward to me
+
A solution using <hask>zip</hask>, <hask>filter</hask>&nbsp;then <hask>map</hask>&nbsp;seems straight-forward to me (''NB: this won't work for infinite lists''):
   
 
<haskell>
 
<haskell>
Line 61: Line 64:
 
$ zip [1..] xs
 
$ zip [1..] xs
 
</haskell>
 
</haskell>
  +
A solution using list comprehension:
  +
<haskell>
  +
slice xs i k = [x | (x,j) <- zip xs [1..k], i <= j]
  +
</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>
  +
  +
[[Category:Programming exercise spoilers]]

Latest revision as of 15:44, 18 May 2014

(**) 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' [] _ _ = []
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)