# ordsets

## Functions for manipulating sets as ordered lists.

Sets are collections of elements with no duplicate elements. An ordset is a representation of a set, where an ordered list is used to store the elements of the set. An ordered list is more efficient than an unordered list. Elements are ordered according to the Erlang term order.

This module provides the same interface as the sets(3) module but with a defined representation. One difference is that while sets considers two elements as different if they do not match (=:=), this module considers two elements as different if and only if they do not compare equal (==).

### ordset(T) = [T]

As returned by new/0.

#### Functions

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

### del_element(Element, Ordset1) -> Ordset2

• Element = term()
• Ordset1 = Ordset2 = ordset(T)

Returns Ordset1, but with Element removed.

### filter(Pred, Ordset1) -> Ordset2

• Pred = fun((Element :: T) -> boolean())
• Ordset1 = Ordset2 = ordset(T)

Filters elements in Ordset1 with boolean function Pred.

### fold(Function, Acc0, Ordset) -> Acc1

• Function =
fun((Element :: T, AccIn :: term()) -> AccOut :: term())
• Ordset = ordset(T)
• Acc0 = Acc1 = term()

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

### from_list(List) -> Ordset

Returns an ordered set of the elements in List.

### intersection(OrdsetList) -> Ordset

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

### intersection(Ordset1, Ordset2) -> Ordset3

• Ordset1 = Ordset2 = Ordset3 = ordset(term())

Returns the intersection of Ordset1 and Ordset2.

### is_disjoint(Ordset1, Ordset2) -> boolean()

• Ordset1 = Ordset2 = ordset(term())

Returns true if Ordset1 and Ordset2 are disjoint (have no elements in common), otherwise false.

### is_element(Element, Ordset) -> boolean()

• Element = term()
• Ordset = ordset(term())

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

### is_empty(Ordset) -> boolean()

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

### is_set(Ordset) -> boolean()

• Ordset = term()

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

### is_subset(Ordset1, Ordset2) -> boolean()

• Ordset1 = Ordset2 = ordset(term())

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

### new() -> []

Returns a new empty ordered set.

### size(Ordset) -> integer() >= 0

Returns the number of elements in Ordset.

### subtract(Ordset1, Ordset2) -> Ordset3

• Ordset1 = Ordset2 = Ordset3 = ordset(term())

Returns only the elements of Ordset1 that are not also elements of Ordset2.

### to_list(Ordset) -> List

Returns the elements of Ordset as a list.

### union(OrdsetList) -> Ordset

Returns the merged (union) set of the list of sets.

### union(Ordset1, Ordset2) -> Ordset3

Returns the merged (union) set of Ordset1 and Ordset2.