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 … Read more
If order is not important and you don’t need to worry about duplicates then you can use set intersection: >>> a = [1,2,3,4,5] >>> b = [1,3,5,6] >>> list(set(a) & set(b)) [1, 3, 5]
You don’t need to define intersection. It’s already a first-class part of set. >>> b1 = [1,2,3,4,5,9,11,15] >>> b2 = [4,5,6,7,8] >>> set(b1).intersection(b2) set([4, 5])
Use a combination of Array.prototype.filter and Array.prototype.includes: const filteredArray = array1.filter(value => array2.includes(value)); For older browsers, with Array.prototype.indexOf and without an arrow function: var filteredArray = array1.filter(function(n) { return array2.indexOf(n) !== -1; }); NB! Both .includes and .indexOf internally compares elements in the array by using ===, so if the array contains objects it will … Read more
The common approach to get a unique collection of items is to use a set. Sets are unordered collections of distinct objects. To create a set from any iterable, you can simply pass it to the built-in set() function. If you later need a real list again, you can similarly pass the set to the … Read more