# 99 questions/Solutions/18

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< 99 questions | Solutions(Difference between revisions)

m (Updated my solution to a more paranoid one) |
(Added solution using fold) |
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(12 intermediate revisions by 9 users not shown) | |||

Line 4: | Line 4: | ||

<haskell> | <haskell> | ||

− | slice 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 22: | ||

<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 37: | ||

<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> and <hask>drop</hask> version: |

<haskell> | <haskell> | ||

Line 49: | Line 53: | ||

i' = i - 1 | i' = i - 1 | ||

</haskell> | </haskell> | ||

+ | A little cleaner, using the previous problem's split (a.k.a. <hask>splitAt</hask>): | ||

+ | <haskell> | ||

+ | slice xs (i+1) k = snd (split (fst (split xs k)) i) | ||

+ | </haskell> | ||

+ | |||

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

+ | |||

+ | <haskell> | ||

+ | slice xs i j = map snd | ||

+ | $ filter (\(x,_) -> x >= i && x <= j) | ||

+ | $ zip [1..] xs | ||

+ | </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]] |

## 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)

splitAt

take

drop

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

splitAt

slice xs (i+1) k = snd (split (fst (split xs k)) i)

zip

filter

map

*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)