How to get distinct values from an array of objects in JavaScript?

Question

Assuming I have the following:

var array = 
    [
        {\"name\":\"Joe\", \"age\":17}, 
        {\"name\":\"Bob\", \"age\":17}, 
        {\"name\":\"Carl\         


        

Answer 1

For those who want to return object with all properties unique by key

const array =
  [
    { "name": "Joe", "age": 17 },
    { "name": "Bob", "age": 17 },
    { "name": "Carl", "age": 35 }
  ]

const key = 'age';

const arrayUniqueByKey = [...new Map(array.map(item =>
  [item[key], item])).values()];

console.log(arrayUniqueByKey);

   /*OUTPUT
       [
        { "name": "Bob", "age": 17 },
        { "name": "Carl", "age": 35 }
       ]
   */

 // Note: this will pick the last duplicated item in the list.

Answer 2

Simple distinct filter using Maps :

let array = 
    [
        {"name":"Joe", "age":17}, 
        {"name":"Bob", "age":17}, 
        {"name":"Carl", "age": 35}
    ];

let data = new Map();

for (let obj of array) {
  data.set(obj.age, obj);
}

let out = [...data.values()];

console.log(out);

Answer 3

[...new Set([
    { "name": "Joe", "age": 17 },
    { "name": "Bob", "age": 17 },
    { "name": "Carl", "age": 35 }
  ].map(({ age }) => age))]

Answer 4

You may be interested in unique set of objects based on one of the keys:

let array = 
[
    {"name":"Joe", "age":17}, 
    {"name":"Bob", "age":17}, 
    {"name":"Carl", "age": 35}
]
let unq_objs = [...new Map(array.map(o =>[o["age"], o])).values()];
console.log(unq_objs)
//result
[{name: "Bob", age: 17},
{name: "Carl", age: 35}]

Answer 5

Simple one-liner with great performance. 6% faster than the ES6 solutions in my tests.

var ages = array.map(function(o){return o.age}).filter(function(v,i,a) {
    return a.indexOf(v)===i
});

Answer 6

If you are using ES6/ES2015 or later you can do it this way:

const data = [
  { group: 'A', name: 'SD' }, 
  { group: 'B', name: 'FI' }, 
  { group: 'A', name: 'MM' },
  { group: 'B', name: 'CO'}
];
const unique = [...new Set(data.map(item => item.group))]; // [ 'A', 'B']

Here is an example on how to do it.