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", "age": 35}
    ]

What is the best way to be able to get an array of all of the distinct ages such that I get an result array of:

[17, 35]

Is there some way I could alternatively structure the data or better method such that I would not have to iterate through each array checking the value of "age" and check against another array for its existence, and add it if not?

If there was some way I could just pull out the distinct ages without iterating...

Current inefficent way I would like to improve... If it means that instead of "array" being an array of objects, but a "map" of objects with some unique key (i.e. "1,2,3") that would be okay too. Im just looking for the most performance efficient way.

The following is how I currently do it, but for me, iteration appears to just be crummy for efficiency even though it does work...

var distinct = []
for (var i = 0; i < array.length; i++)
   if (array[i].age not in distinct)
      distinct.push(array[i].age)

Answer 1

If this were PHP I'd build an array with the keys and take array_keys at the end, but JS has no such luxury. Instead, try this:

var flags = [], output = [], l = array.length, i;
for( i=0; i<l; i++) {
    if( flags[array[i].age]) continue;
    flags[array[i].age] = true;
    output.push(array[i].age);
}

Answer 2

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.

Answer 3

using ES6

let array = [
  { "name": "Joe", "age": 17 },
  { "name": "Bob", "age": 17 },
  { "name": "Carl", "age": 35 }
];
array.map(item => item.age)
  .filter((value, index, self) => self.indexOf(value) === index)

> [17, 35]

Answer 4

You could use a dictionary approach like this one. Basically you assign the value you want to be distinct as a key in the "dictionary" (here we use an array as an object to avoid dictionary-mode). If the key did not exist then you add that value as distinct.

Here is a working demo:

var array = [{"name":"Joe", "age":17}, {"name":"Bob", "age":17}, {"name":"Carl", "age": 35}];
var unique = [];
var distinct = [];
for( let i = 0; i < array.length; i++ ){
  if( !unique[array[i].age]){
    distinct.push(array[i].age);
    unique[array[i].age] = 1;
  }
}
var d = document.getElementById("d");
d.innerHTML = "" + distinct;

<div id="d"></div>

This will be O(n) where n is the number of objects in array and m is the number of unique values. There is no faster way than O(n) because you must inspect each value at least once.

The previous version of this used an object, and for in. These were minor in nature, and have since been minorly updated above. However, the reason for a seeming advance in performance between the two versions in the original jsperf was due to the data sample size being so small. Thus, the main comparison in the previous version was looking at the difference between the internal map and filter use versus the dictionary mode lookups.

I have updated the code above, as noted, however, I have also updated the jsperf to look through 1000 objects instead of 3. 3 overlooked many of the performance pitfalls involved (obsolete jsperf).

Performance

https://jsperf.com/filter-vs-dictionary-more-data When I ran this dictionary was 96% faster.

Answer 5

Using ES6 features, you could do something like:

const uniqueAges = [...new Set( array.map(obj => obj.age)) ];

Answer 6

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 7

This is how you would solve this using new Set via ES6 for Typescript as of August 25th, 2017

Array.from(new Set(yourArray.map((item: any) => item.id)))

Answer 8

I'd just map and remove dups:

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

console.log(ages); //=> [17, 35]

Edit: Aight! Not the most efficient way in terms of performance, but the simplest most readable IMO. If you really care about micro-optimization or you have huge amounts of data then a regular for loop is going to be more "efficient".

Answer 9

var unique = array
    .map(p => p.age)
    .filter((age, index, arr) => arr.indexOf(age) == index)
    .sort(); // sorting is optional

// or in ES6

var unique = [...new Set(array.map(p => p.age))];

// or with lodash

var unique = _.uniq(_.map(array, 'age'));

ES6 example

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

const arr = data.map(p => p.age); // [17, 17, 35]
const s = new Set(arr); // {17, 35} a set removes duplications, but it's still a set
const unique = [...s]; // [17, 35] Use the spread operator to transform a set into an Array
// or use Array.from to transform a set into an array
const unique2 = Array.from(s); // [17, 35]

Answer 10

There are many valid answers already, but I wanted to add one that uses only the reduce() method because it is clean and simple.

function uniqueBy(arr, prop){
  return arr.reduce((a, d) => {
    if (!a.includes(d[prop])) { a.push(d[prop]); }
    return a;
  }, []);
}

Use it like this:

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

var ages = uniqueBy(array, "age");
console.log(ages); // [17, 35]

Answer 11

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

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

const distinctItems = [...new Map(array.map(item => [item["age"], item])).values()];

Result:

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

Answer 12

The forEach version of @travis-j's answer (helpful on modern browsers and Node JS world):

var unique = {};
var distinct = [];
array.forEach(function (x) {
  if (!unique[x.age]) {
    distinct.push(x.age);
    unique[x.age] = true;
  }
});

34% faster on Chrome v29.0.1547: http://jsperf.com/filter-versus-dictionary/3

And a generic solution that takes a mapper function (tad slower than direct map, but that's expected):

function uniqueBy(arr, fn) {
  var unique = {};
  var distinct = [];
  arr.forEach(function (x) {
    var key = fn(x);
    if (!unique[key]) {
      distinct.push(key);
      unique[key] = true;
    }
  });
  return distinct;
}

// usage
uniqueBy(array, function(x){return x.age;}); // outputs [17, 35]

Answer 13

const x = [
  {"id":"93","name":"CVAM_NGP_KW"},
  {"id":"94","name":"CVAM_NGP_PB"},
  {"id":"93","name":"CVAM_NGP_KW"},
  {"id":"94","name":"CVAM_NGP_PB"}
].reduce(
  (accumulator, current) => {
    if(!accumulator.some(x => x.id === current.id)) {
      accumulator.push(current)
    }
    return accumulator;
  }, []
)

console.log(x)

/* output 
[ 
  { id: '93', name: 'CVAM_NGP_KW' },
  { id: '94', name: 'CVAM_NGP_PB' } 
]
*/

Answer 14

I've started sticking Underscore in all new projects by default just so I never have to think about these little data-munging problems.

var array = [{"name":"Joe", "age":17}, {"name":"Bob", "age":17}, {"name":"Carl", "age": 35}];
console.log(_.chain(array).map(function(item) { return item.age }).uniq().value());

Produces [17, 35].

Answer 15

Here's another way to solve this:

var result = {};
for(var i in array) {
    result[array[i].age] = null;
}
result = Object.keys(result);

I have no idea how fast this solution is compared to the others, but I like the cleaner look. ;-)

---

EDIT: Okay, the above seems to be the slowest solution of all here.

I've created a performance test case here: http://jsperf.com/distinct-values-from-array

Instead of testing for the ages (Integers), I chose to compare the names (Strings).

Method 1 (TS's solution) is very fast. Interestingly enough, Method 7 outperforms all other solutions, here I just got rid of .indexOf() and used a "manual" implementation of it, avoiding looped function calling:

var result = [];
loop1: for (var i = 0; i < array.length; i++) {
    var name = array[i].name;
    for (var i2 = 0; i2 < result.length; i2++) {
        if (result[i2] == name) {
            continue loop1;
        }
    }
    result.push(name);
}

The difference in performance using Safari & Firefox is amazing, and it seems like Chrome does the best job on optimization.

I'm not exactly sure why the above snippets is so fast compared to the others, maybe someone wiser than me has an answer. ;-)

Answer 16

I have a small solution

let data = [{id: 1}, {id: 2}, {id: 3}, {id: 2}, {id: 3}];

let result = data.filter((value, index, self) => self.findIndex((m) => m.id === value.id) === index);

Answer 17

using lodash

var array = [
    { "name": "Joe", "age": 17 },
    { "name": "Bob", "age": 17 },
    { "name": "Carl", "age": 35 }
];
_.chain(array).pluck('age').unique().value();
> [17, 35]

Answer 18

Using Lodash

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

_.chain(array).map('age').unique().value();

Returns [17,35]

Answer 19

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 20

function get_unique_values_from_array_object(array,property){
    var unique = {};
    var distinct = [];
    for( var i in array ){
       if( typeof(unique[array[i][property]]) == "undefined"){
          distinct.push(array[i]);
       }
       unique[array[i][property]] = 0;
    }
    return distinct;
}

Answer 21

underscore.js _.uniq(_.pluck(array,"age"))

Answer 22

Here's a versatile solution that uses reduce, allows for mapping, and maintains insertion order.

items: An array

mapper: A unary function that maps the item to the criteria, or empty to map the item itself.

function distinct(items, mapper) {
    if (!mapper) mapper = (item)=>item;
    return items.map(mapper).reduce((acc, item) => {
        if (acc.indexOf(item) === -1) acc.push(item);
        return acc;
    }, []);
}

Usage

const distinctLastNames = distinct(items, (item)=>item.lastName);
const distinctItems = distinct(items);

You can add this to your Array prototype and leave out the items parameter if that's your style...

const distinctLastNames = items.distinct( (item)=>item.lastName) ) ;
const distinctItems = items.distinct() ;

You can also use a Set instead of an Array to speed up the matching.

function distinct(items, mapper) {
    if (!mapper) mapper = (item)=>item;
    return items.map(mapper).reduce((acc, item) => {
        acc.add(item);
        return acc;
    }, new Set());
}

Answer 23

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

      const ages = [...new Set(array.reduce((a, c) => [...a, c.age], []))];
    
      console.log(ages);

Answer 24

Just found this and I thought it's useful

_.map(_.indexBy(records, '_id'), function(obj){return obj})

Again using underscore, so if you have an object like this

var records = [{_id:1,name:'one', _id:2,name:'two', _id:1,name:'one'}]

it will give you the unique objects only.

What happens here is that indexBy returns a map like this

{ 1:{_id:1,name:'one'}, 2:{_id:2,name:'two'} }

and just because it's a map, all keys are unique.

Then I'm just mapping this list back to array.

In case you need only the distinct values

_.map(_.indexBy(records, '_id'), function(obj,key){return key})

Keep in mind that the key is returned as a string so, if you need integers instead, you should do

_.map(_.indexBy(records, '_id'), function(obj,key){return parseInt(key)})

Answer 25

i think you are looking for groupBy function (using Lodash)

_personsList = [{"name":"Joe", "age":17}, 
                {"name":"Bob", "age":17}, 
                {"name":"Carl", "age": 35}];
_uniqAgeList = _.groupBy(_personsList,"age");
_uniqAges = Object.keys(_uniqAgeList);

produces result:

17,35

jsFiddle demo:http://jsfiddle.net/4J2SX/201/

Answer 26

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

Answer 27

If you have Array.prototype.includes or are willing to polyfill it, this works:

var ages = []; array.forEach(function(x) { if (!ages.includes(x.age)) ages.push(x.age); });

Answer 28

I know my code is little length and little time complexity but it's understandable so I tried this way.

I'm trying to develop prototype based function here and code also change.

Here,Distinct is my own prototype function.

<script>
  var array = [{
      "name": "Joe",
      "age": 17
    },
    {
      "name": "Bob",
      "age": 17
    },
    {
      "name": "Carl",
      "age": 35
    }
  ]

  Array.prototype.Distinct = () => {
    var output = [];
    for (let i = 0; i < array.length; i++) {
      let flag = true;
      for (let j = 0; j < output.length; j++) {
        if (array[i].age == output[j]) {
          flag = false;
          break;
        }
      }
      if (flag)
        output.push(array[i].age);
    }
    return output;
  }
  //Distinct is my own function
  console.log(array.Distinct());
</script>

Answer 29

If like me you prefer a more "functional" without compromising speed, this example uses fast dictionary lookup wrapped inside reduce closure.

var array = 
[
    {"name":"Joe", "age":17}, 
    {"name":"Bob", "age":17}, 
    {"name":"Carl", "age": 35}
]
var uniqueAges = array.reduce((p,c,i,a) => {
    if(!p[0][c.age]) {
        p[1].push(p[0][c.age] = c.age);
    }
    if(i<a.length-1) {
        return p
    } else {
        return p[1]
    }
}, [{},[]])

According to this test my solution is twice as fast as the proposed answer

Answer 30

There are a lot of great answers here, but none of them have addressed the following line:

Is there some way I could alternatively structure the data

I would create an object whose keys are the ages, each pointing to an array of names.

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

var map = array.reduce(function(result, item) {
  result[item.age] = result[item.age] || [];
  result[item.age].push(item.name);
  return result;
}, {});

console.log(Object.keys(map));
console.log(map);

This way you've converted the data structure into one that is very easy to retrieve the distinct ages from.

Here is a more compact version that also stores the entire object instead of just the name (in case you are dealing with objects with more than 2 properties so they cant be stored as key and value).

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

var map = array.reduce((r, i) => ((r[i.age] = r[i.age] || []).push(i), r), {});

console.log(Object.keys(map));
console.log(map);