Subtract arrays - javascript [duplicate]

Question

I have arrays like;

var john = { name: "John Smith", age: 23 };
var mary = { name: "Mary Key", age: 18 };
var bob = { name: "Bob-small", age: 6 };
var people = [john, mary, bob];

var john2 = { name: "John Smith", age: 23 };
var people2 = [john2];

What I would like to do is subtract people2 from people and get result;

[mary, bob];

How can I achieve this? TIA

Answer 1

The difference of two sets, A and B, is defined as the set of all those elements of A which are not in B. If we implement it naively, computing the difference of two sets of sizes m and n respectively would take O(m * n) time. Not very efficient:

const john1 = { name: "John Smith", age: 23 };
const john2 = { name: "John Smith", age: 23 };
const mary = { name: "Mary Key", age: 18 };
const bob = { name: "Bob-small", age: 6 };

const people1 = [john1, mary, bob];
const people2 = [john2];

const eqPerson = (p, q) => p.name === q.name && p.age === q.age;

const result = people1.filter(p => people2.every(q => !eqPerson(p, q)));

console.log(result); // [mary, bob]

---

Fortunately, there's a faster way to compute the set difference for large sets using hashing.

const john1 = { name: "John Smith", age: 23 };
const john2 = { name: "John Smith", age: 23 };
const mary = { name: "Mary Key", age: 18 };
const bob = { name: "Bob-small", age: 6 };

const people1 = [john1, mary, bob];
const people2 = [john2];

const hashPerson = ({ name, age }) => `${name} ${age}`;

const hashSet = new Set(people2.map(hashPerson));

const result = people1.filter(p => !hashSet.has(hashPerson(p)));

console.log(result); // [mary, bob]

The advantage is that creating a hash set takes O(n) time and calculating the difference takes O(m) time. Hence in total it only takes O(m + n) time instead of O(m * n) time. In addition, you can reuse the hash set in the future.

Answer 2

Here is an easy solution:

var diff = people.filter(function(item) {
  return !people2.some(function(test){
    return test.name === item.name && test.age === item.age;
  });
});

Make sure the function passed to people2.some correctly checks that the two objects are equal, since == would fail, as you have references to different objects with identical properties.