rio-0.1.24.0: A standard library for Haskell

Safe Haskell | Safe-Inferred
Language | Haskell2010

RIO.Set.Partial

Contents

- Indexed
- Min/Max

Description

`Set` partial functions. Import as:

    import qualified RIO.Set.Partial as Set'

Synopsis

- findIndex :: Ord a => a -> Set a -> Int
- elemAt :: Int -> Set a -> a
- deleteAt :: Int -> Set a -> Set a
- findMin :: Set a -> a
- findMax :: Set a -> a
- deleteFindMin :: Set a -> (a, Set a)
- deleteFindMax :: Set a -> (a, Set a)

# Indexed

findIndex :: Ord a => a -> Set a -> Int Source #

\(O(\log n)\). Return the index of an element, which is its zero-based index in the sorted sequence of elements. The index is a number from 0 up to, but not including, the `size` of the set. Calls `error` when the element is not a `member` of the set.

    findIndex 2 (fromList [5,3])    Error: element is not in the set
    findIndex 3 (fromList [5,3]) == 0
    findIndex 5 (fromList [5,3]) == 1
    findIndex 6 (fromList [5,3])    Error: element is not in the set

Since: containers-0.5.4

elemAt :: Int -> Set a -> a Source #

\(O(\log n)\). Retrieve an element by its index, i.e. by its zero-based index in the sorted sequence of elements. If the index is out of range (less than zero, greater or equal to `size` of the set), `error` is called.

    elemAt 0 (fromList [5,3]) == 3
    elemAt 1 (fromList [5,3]) == 5
    elemAt 2 (fromList [5,3])    Error: index out of range

Since: containers-0.5.4

deleteAt :: Int -> Set a -> Set a Source #

\(O(\log n)\). Delete the element at index, i.e. by its zero-based index in the sorted sequence of elements. If the index is out of range (less than zero, greater or equal to `size` of the set), `error` is called.

    deleteAt 0    (fromList [5,3]) == singleton 5
    deleteAt 1    (fromList [5,3]) == singleton 3
    deleteAt 2    (fromList [5,3])    Error: index out of range
    deleteAt (-1) (fromList [5,3])    Error: index out of range

Since: containers-0.5.4

# Min/Max

findMin :: Set a -> a Source #

\(O(\log n)\). The minimal element of a set.

findMax :: Set a -> a Source #

\(O(\log n)\). The maximal element of a set.

deleteFindMin :: Set a -> (a, Set a) Source #

\(O(\log n)\). Delete and find the minimal element.

    deleteFindMin set = (findMin set, deleteMin set)

deleteFindMax :: Set a -> (a, Set a) Source #

\(O(\log n)\). Delete and find the maximal element.

    deleteFindMax set = (findMax set, deleteMax set)