# 99 questions/Solutions/18

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< 99 questions | Solutions(Difference between revisions)

(Take and drop can be applied in the opposite order too.) |
m (Fix for 'Non-exhaustive patterns in function slice' error) |
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(One intermediate revision by one user not shown) | |||

Line 22: | 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 35: | 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> | ||

## Revision as of 20:34, 8 April 2013

(**) 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