I want to create different methods for a class called Multiset. I have all the required methods, but I’m unsure of how to write intersection, union, and subset methods. For intersection and union, my code starts like this: def intersect(var) x = Multiset.new end Here is an example: X = [1, 1, 2, 4] Y … Read more
This question already has answers here: Efficiently finding the intersection of a variable number of sets of strings (8 answers) Closed 1 year ago. Possible Duplicate: Efficiently finding the intersection of a variable number of sets of strings Say, have two Hashset, how to calculate the intersection of them? Set<String> s1 = new HashSet<String>(); Set<String> … Read more
a = [1,2,3,4,5] b = [1,3,5,6] c = a and b print c actual output: [1,3,5,6] expected output: [1,3,5] How can we achieve a boolean AND operation (list intersection) on two lists? 16 Answers 16
I know how to get an intersection of two flat lists: b1 = [1,2,3,4,5,9,11,15] b2 = [4,5,6,7,8] b3 = [val for val in b1 if val in b2] or def intersect(a, b): return list(set(a) & set(b)) print intersect(b1, b2) But when I have to find intersection for nested lists then my problems starts: c1 = … Read more
What’s the simplest, library-free code for implementing array intersections in javascript? I want to write intersection([1,2,3], [2,3,4,5]) and get [2, 3] 37 s 37
How do I get a new list without duplicates? [1, 2, 3, 1] → [1, 2, 3] How do I get a new list where items that are duplicated are entirely removed? [1, 2, 3, 1] → [2, 3] 5 57
Use the retainAll() method of Set: Set<String> s1; Set<String> s2; s1.retainAll(s2); // s1 now contains only elements in both sets If you want to preserve the sets, create a new set to hold the intersection: Set<String> intersection = new HashSet<String>(s1); // use the copy constructor intersection.retainAll(s2); The javadoc of retainAll() says it’s exactly what you want: Retains only the elements in this set that … Read more