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99 questions/Solutions/18

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(adding a remark; adding   after hask-tagged words, as theres no trailing space after hask tagged word as rendered by IE6 (a bug? someone knows how to fix it?))
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$ filter (\(x,_) -> x >= i && x <= j)
$ filter (\(x,_) -> x >= i && x <= j)
$ zip [1..] xs
$ zip [1..] xs
 +
</haskell>
 +
A solution using list comprehension:
 +
<haskell>
 +
slice xs i k = [x | (x,j) <- zip xs [1..k], i <= j]
</haskell>
</haskell>

Revision as of 23:06, 5 August 2011

(**) 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]
Retrieved from "http://www.haskell.org/haskellwiki/99_questions/Solutions/18"