Ordset2 = ordset(T | E)

Returns a new ordered set formed from Ordset1 with Element inserted.

Examples

1> S0 = ordsets:new().
[]
2> S1 = ordsets:add_element(7, S0).
[7]
3> S2 = ordsets:add_element(42, S1).
[7,42]
4> ordsets:add_element(42, S2).
[7,42]

Ordset1 = Ordset2 = ordset(T)

Returns a copy of Ordset1 with Element removed.

Examples

1> S = ordsets:from_list([a,b,c]).
2> ordsets:del_element(c, S).
[a,b]
3> ordsets:del_element(x, S).
[a,b,c]

Pred = fun((Element :: T) -> boolean())

Ordset1 = Ordset2 = ordset(T)

Filters elements in Ordset1 using predicate function Pred.

Examples

1> S = ordsets:from_list([1,2,3,4,5,6,7]).
2> IsEven = fun(N) -> N rem 2 =:= 0 end.
3> ordsets:filter(IsEven, S).
[2,4,6]

Fun = fun((Element1 :: T1) -> boolean() | {true, Element2 :: T2})

Ordset1 = ordset(T1)

Ordset2 = ordset(T1 | T2)

Calls Fun(Elem) for each Elem of Ordset1 to update or remove elements from Ordset1.

Fun/1 must return either a Boolean or a tuple {true, Value}. The function returns the set of elements for which Fun returns a new value, with true being equivalent to {true, Elem}.

ordsets:filtermap/2 behaves as if it were defined as follows:

filtermap(Fun, Ordset1) ->
    ordsets:from_list(lists:filtermap(Fun, ordsets:to_list(Ordset1))).

Examples

1> S = ordsets:from_list([2,4,5,6,8,9]).
2> F = fun(X) ->
           case X rem 2 of
               0 -> {true, X div 2};
               1 -> false
           end
        end.
3> ordsets:filtermap(F, S).
[1,2,3,4]

Function = fun((Element :: T, AccIn :: term()) -> AccOut :: term())

Acc0 = Acc1 = term()

Ordset = ordset(T)

Folds Function over every element in Ordset and returns the final value of the accumulator.

Examples

1> S = ordsets:from_list([1,2,3,4]).
2> Plus = fun erlang:'+'/2.
3> ordsets:fold(Plus, 0, S).
10

Ordset = ordset(T)

Returns an ordered set of the elements in List.

Examples

1> ordsets:from_list([a,b,a,b,b,c]).
[a,b,c]

Ordset = ordset(_)

Returns the intersection of the non-empty list of sets.

The intersection of multiple sets is a new set that contains only the elements that are present in all sets.

Examples

1> S0 = ordsets:from_list([a,b,c,d]).
2> S1 = ordsets:from_list([d,e,f]).
3> S2 = ordsets:from_list([q,r]).
4> Sets = [S0, S1, S2].
5> ordsets:intersection([S0, S1, S2]).
[]
6> ordsets:intersection([S0, S1]).
[d]
7> ordsets:intersection([]).
** exception error: no function clause matching ordsets:intersection([])

Ordset1 = Ordset2 = Ordset3 = ordset(_)

Returns the intersection of Ordset1 and Ordset2.

The intersection of two sets is a new set that contains only the elements that are present in both sets.

Examples

1> S0 = ordsets:from_list([a,b,c,d]).
2> S1 = ordsets:from_list([c,d,e,f]).
3> S2 = ordsets:from_list([q,r]).
4> ordsets:intersection(S0, S1).
[c,d]
5> ordsets:intersection(S1, S2).
[]

Ordset1 = Ordset2 = ordset(_)

Returns true if Ordset1 and Ordset2 are disjoint; otherwise, returns false.

Two sets are disjoint if they have no elements in common.

This function is equivalent to ordsets:is_empty(ordsets:intersection(Ordset1, Ordset2)), but faster.

Examples

1> S0 = ordsets:from_list([a,b,c,d]).
2> S1 = ordsets:from_list([d,e,f]).
3> S2 = ordsets:from_list([q,r]).
4> ordsets:is_disjoint(S0, S1).
false
5> ordsets:is_disjoint(S1, S2).
true

Returns true if Element is an element of Ordset; otherwise, returns false.

Examples

1> S = ordsets:from_list([a,b,c]).
2> ordsets:is_element(42, S).
false
3> ordsets:is_element(b, S).
true

Returns true if Ordset is an empty set; otherwise, returns false.

Examples

1> ordsets:is_empty(ordsets:new()).
true
2> ordsets:is_empty(ordsets:from_list([1])).
false

Ordset1 = Ordset2 = ordset(_)

Returns true if Ordset1 and Ordset2 are equal, that is, if every element of one set is also a member of the other set; otherwise, returns false.

Examples

1> Empty = ordsets:new().
2> S = ordsets:from_list([a,b]).
3> ordsets:is_equal(S, S).
true
4> ordsets:is_equal(S, Empty).
false

Returns true if Ordset is an ordered set of elements; otherwise, returns false.

Examples

1> ordsets:is_set(ordsets:from_list([a,x,13,{p,q}])).
true
2> ordsets:is_set([a,b,c]).
true
3> ordsets:is_set([z,a]).
false
4> ordsets:is_set({a,b}).
false

Ordset1 = Ordset2 = ordset(_)

Returns true when every element of Ordset1 is also a member of Ordset2; otherwise, returns false.

Examples

1> S0 = ordsets:from_list([a,b,c,d]).
2> S1 = ordsets:from_list([c,d]).
3> ordsets:is_subset(S1, S0).
true
4> ordsets:is_subset(S0, S1).
false
5> ordsets:is_subset(S0, S0).
true

Fun = fun((Element1 :: T1) -> Element2 :: T2)

Ordset1 = ordset(T1)

Ordset2 = ordset(T2)

Maps elements in Ordset1 with mapping function Fun.

Examples

1> S = ordsets:from_list([1,2,3,4,5,6,7]).
2> F = fun(N) -> N div 2 end.
3> ordsets:map(F, S).
[0,1,2,3]

Returns a new empty ordered set.

Examples

1> ordsets:new().
[]

Returns the number of elements in Ordset.

Examples

1> ordsets:size(ordsets:new()).
0
2> ordsets:size(ordsets:from_list([4,5,6])).
3

Ordset1 = Ordset2 = Ordset3 = ordset(_)

Returns a new ordered set containing the elements of Ordset1 that are not elements in Ordset2.

Examples

1> S0 = ordsets:from_list([a,b,c,d]).
2> S1 = ordsets:from_list([c,d,e,f]).
3> ordsets:subtract(S0, S1).
[a,b]
4> ordsets:subtract(S1, S0).
[e,f]

List = [T]

Returns the elements of Ordset as a list.

Examples

1> S = ordsets:from_list([a,b]).
2> ordsets:to_list(S).
[a,b]

Ordset = ordset(T)

Returns the union of a list of sets.

The union of multiple sets is a new set that contains all the elements from all sets, without duplicates.

Examples

1> S0 = ordsets:from_list([a,b,c,d]).
2> S1 = ordsets:from_list([d,e,f]).
3> S2 = ordsets:from_list([q,r]).
4> Sets = [S0, S1, S2].
5> ordsets:union(Sets).
[a,b,c,d,e,f,q,r]

Ordset3 = ordset(T1 | T2)

Returns the union of Ordset1 and Ordset2.

The union of two sets is a new set that contains all the elements from both sets, without duplicates.

Examples

1> S0 = ordsets:from_list([a,b,c,d]).
2> S1 = ordsets:from_list([c,d,e,f]).
3> ordsets:union(S0, S1).
[a,b,c,d,e,f]