Swift does not have a native ordered set type. If you use Foundation, you can use NSOrderedSet in Swift. If not, you have the opportunity to write your own ordered set data structure. Update: Swift Package Manager includes an OrderedSet implementation that may be useful. It wraps both an array and a set and manages … Read more
You’re after the symmetric difference. This is discussed in the Java tutorial. Set<Type> symmetricDiff = new HashSet<Type>(set1); symmetricDiff.addAll(set2); // symmetricDiff now contains the union Set<Type> tmp = new HashSet<Type>(set1); tmp.retainAll(set2); // tmp now contains the intersection symmetricDiff.removeAll(tmp); // union minus intersection equals symmetric-difference
Use frozenset, >>> set([frozenset([1,2]), frozenset([2,3])]) set([frozenset([1, 2]), frozenset([2, 3])]) To represent a set of sets, the inner sets must be frozenset objects for the reason that the elements of a set must be hashable (all of Python’s immutable built-in objects are hashable). frozenset is immutable and set is mutable.
They are semantically different. Consider: #include <set> #include <map> #include <utility> #include <iostream> using namespace std; int main() { pair<int, int> p1(1, 1); pair<int, int> p2(1, 2); set< pair<int, int> > s; s.insert(p1); s.insert(p2); map<int, int> m; m.insert(p1); m.insert(p2); cout << “Set size = ” << s.size() << endl; cout << “Map size = ” … Read more
As per the docs: public boolean add(E e) Adds the specified element to this set if it is not already present. More formally, adds the specified element e to this set if this set contains no element e2 such that (e==null ? e2==null : e.equals(e2)). If this set already contains the element, the call leaves … Read more
Better than extending a particular implementation, you can easily write a proxy implementation of Set that checks for nulls. This analogous to Collections.checkedSet. Other than being applicable to any implementation, you can also be sure that you have overridden all applicable methods. Many flaws have been found by extending concrete collections which then have additional … Read more
From relational databases perspective this is a set. Databases do not preserve order and using a List is meaningless, the order in them is unspecified (unless using so called indexed collections). Using a Set also has great performance implications. When List is used, Hibernate uses PersistentBag collection underneath which has some terrible characteristics. I.e.: if … Read more
They are very different. One set changes the set in place, while the other leaves the original set alone, and returns a copy instead. >>> s = {1, 2, 3} >>> news = s | {4} >>> s set([1, 2, 3]) >>> news set([1, 2, 3, 4]) Note how s has remained unchanged. >>> s.update({4}) … Read more
Just use the .union() method. >>> l = [set([1,2,3]), set([4,5,6]), set([1,4,9])] >>> frozenset().union(*l) frozenset([1, 2, 3, 4, 5, 6, 9]) This works for any iterable of iterables.
You can do something like: std::set<int> s1; std::set<int> s2; // fill your sets s1.insert(s2.begin(), s2.end());