| Safe Haskell | None |
|---|---|
| Language | Haskell2010 |
Data.Set.Util
Documentation
>>>import Data.Ord(comparing)>>>data S = S String deriving Show>>>cmpS = comparing (\(S s) -> length s)
splitOn :: Ord b => (a -> b) -> b -> Set a -> (Set a, Set a, Set a) Source #
Given a monotonic function f that maps a to b, split the sequence s depending on the b values. I.e. the result (l,m,r) is such that * all (< x) . fmap f $ l * all (== x) . fmap f $ m * all (> x) . fmap f $ r
running time: \(O(\log n)\)
fromListBy :: (a -> a -> Ordering) -> [a] -> Set a Source #
Constructs a Set using the given Order.
Note that this is dangerous as the resulting set may not abide the ordering expected of such sets.
running time: \(O(n\log n)\)
join :: Set a -> Set a -> Set a Source #
Given two sets l and r, such that all elements of l occur before r, join the two sets into a combined set.
running time: \(O(\log n)\)
insertBy :: (a -> a -> Ordering) -> a -> Set a -> Set a Source #
Inserts an element into the set, assuming that the set is ordered by the given order.
>>>insertBy cmpS (S "ccc") $ fromListBy cmpS [S "a" , S "bb" , S "dddd"]fromList [S "a",S "bb",S "ccc",S "dddd"]
When trying to insert an element that equals an element already in the set (according to the given comparator), this function replaces the old element by the new one:
>>>insertBy cmpS (S "cc") $ fromListBy cmpS [S "a" , S "bb" , S "dddd"]fromList [S "a",S "cc",S "dddd"]
running time: \(O(\log n)\)
deleteAllBy :: (a -> a -> Ordering) -> a -> Set a -> Set a Source #
Deletes an element from the set, assuming the set is ordered by the given ordering.
>>>deleteAllBy cmpS (S "bb") $ fromListBy cmpS [S "a" , S "bb" , S "dddd"]fromList [S "a",S "dddd"]>>>deleteAllBy cmpS (S "bb") $ fromListBy cmpS [S "a" , S "bb" , S "cc", S "dd", S "ee", S "ff", S "dddd"]fromList [S "a",S "dddd"]
running time: \(O(\log n)\)