Removing duplicates in an Array of Objects in JS with Sets

The other day at work I was faced with what I think is a rather common problem when dealing with data coming from an API.

I was getting from my async http call an array of objects (addresses in this case), but for unimportant reasons this array was returning objects that could be duplicate (so two identical ids).

My first instinct was to make a new array, and loop through the addresses' array and do a .findIndex on each for my "copy". If the "copy" didn't have the address, then I would push it. In the end the code was readable, but I wanted a simpler more straightforward way.

There are many ways to solve this particular problem!

Now that we got that out of the way 😋, I wanted to show this solution that uses the power of JavaScript Set.

const addresses = [...]; // Some array I got from async call

const uniqueAddresses = Array.from(new Set(addresses.map(a => a.id)))
 .map(id => {
   return addresses.find(a => a.id === id)
 })

Let's dissect this mess:

1. Array.from(new Set()) I'm going to make a new set, and I want to turn it back into an array with Array.from so that I can later re-map it.
2. new Set(addresses.map(a => a.id)) Set will only allow unique values in it, so i'm going to pass it the ids of each object. If the loop tries to add the same value again, it'll get ignored for free.
3. .map(id => [...]) With the array of ids I got on step 1, I run a map function on it and return the actual address from the original address array

That's it! Simple, clear, and I had fun learning about/using Set 😎

---

Huge shout out to my bae Natalia Tepluhina who endured and helped me come up with crazy ideas for this, and her awesome solution to do it with Array.reduce showcased below ❤️👩‍💻

const arr = [
  { id: 1, name: "test1" },
  { id: 2, name: "test2" },
  { id: 2, name: "test3" },
  { id: 3, name: "test4" },
  { id: 4, name: "test5" },
  { id: 5, name: "test6" },
  { id: 5, name: "test7" },
  { id: 6, name: "test8" }
];

const filteredArr = arr.reduce((acc, current) => {
  const x = acc.find(item => item.id === current.id);
  if (!x) {
    return acc.concat([current]);
  } else {
    return acc;
  }
}, []);

Comments

[Matt]

Without getting into cryptic one-liners, there's a pretty straight forward linear time solution as well.

const seen = new Set();
const arr = [
  { id: 1, name: "test1" },
  { id: 2, name: "test2" },
  { id: 2, name: "test3" },
  { id: 3, name: "test4" },
  { id: 4, name: "test5" },
  { id: 5, name: "test6" },
  { id: 5, name: "test7" },
  { id: 6, name: "test8" }
];

const filteredArr = arr.filter(el => {
  const duplicate = seen.has(el.id);
  seen.add(el.id);
  return !duplicate;
});

[Marina Mosti]

Hey Matt, nice solution! Yeah, all ways leads to Rome :)

[ajv]

Well... yes and no.

I feel it's important to distinguish that the OP-s solution loops over the whole array twice. I know there are times to make a trade-off between performance and readability, but I don't feel this needs to be one of those times :)

[spiritupbro]

wow crazy matt
thanks for this

[John Fewell]

Great answer, thank you Matt!

[Tony Childs]

Here's another possibility using the Map class constructor and values method:

const arr = [
  { id: 1, name: "test1" },
  { id: 2, name: "test2" },
  { id: 2, name: "test3" },
  { id: 3, name: "test4" },
  { id: 4, name: "test5" },
  { id: 5, name: "test6" },
  { id: 5, name: "test7" },
  { id: 6, name: "test8" }
];

const uniqueObjects = [...new Map(arr.map(item => [item.id, item])).values()]

[Corfitz]

There is a difference between Tony's and Matt's approach in how the final array will look like.

Matt's approach is adding the id for each entry it loops through to a Set and checks whether or not it has been ´seen´ before or not, and return the object if 'no' is the case. So if we look at the returned object with ID: 2, Matt will return the object with name: "test2" as it will consider the next object a duplicate and skip it.

Tony's approach is by creating a new map using ID as a key - which has to be unique - and then extracts the values. E.g. [1: { id: 1, name: "test1" }, 2: { id: 2, name: "test2" }....] etc. What this means though, is that even though id: 2 has been added to the map, it is simply overwritten by the third item in the array, thus Tony will return name: "test3" for ID: 2.

Just keep this in mind whether you want the first object or the last object by a duplicated identifier to be the truth.

[alanwong9179]

I have to create an account and say thank you! This answer save my day.

[Mayank Soni]

Hi Tony. Your answer helped me, thanks!
BTW, can you please help me achieve this array?
[
{ id: 1, name: "test1" },
{ id: 2, name: "test2" },
{ id: 2, name: "test3" },
{ id: 3, name: "test2" }
]

I mean even if 'id' or 'name' already exists, it should not be omitted because either of the value is different (like in the case of 'name: "test2"') in the whole array.

[Corfitz]

Hey @mayanxoni ,

Might be late with a reply here, but in your case I would probably map through your array of objects as stringified content, as you are not relying on a single identifier but an entire object.

NB: Though this might not be performant when you are dealing with bigger objects and large arrays.

const arr = [
{ id: 1, name: "test1" },
{ id: 2, name: "test2" },
{ id: 2, name: "test2" },
{ id: 2, name: "test2" },
{ id: 2, name: "test3" },
{ id: 3, name: "test2" }
];

const newArray = Array.from(new Set(arr.map(el => JSON.stringify(el)))).map(el => JSON.parse(el));

[Tony Childs]

Hi Mayank,

I'm not sure I follow. If no duplicates are removed then that is just the original array is it not? Or do you mean you want an array with just ids 1, 2, and 3? If so, you can use Array.prototype.filter and only return true for the ids you want to keep.

[brian]

Great

[Sam Clark]

wow, i love that

[marlon zayro arias v]

Thanks !!

[Jonas Winzen]

Interesting challenge. So I've three very similar solutions. They are all based on the same principle of reducing the array into a key-value structure and re-creating the array from the values only.

Approach 1: Classical Reducer

(reducer maintains immutability)

/**
 * classic reducer
 **/
const uniqByProp = prop => arr =>
  Object.values(
    arr.reduce(
      (acc, item) =>
        item && item[prop]
          ? { ...acc, [item[prop]]: item } // just include items with the prop
          : acc,
      {}
    )
  );

// usage:

const uniqueById = uniqByProp("id");

const unifiedArray = uniqueById(arrayWithDuplicates);

Depending on your array size, this approach might easily become a bottleneck in your app. More performant is to mutate your accumulator object directly in the reducer.

Approach 2: Reducer with object-mutation

/**
 * using object mutation
 **/
const uniqByProp = prop => arr =>
  Object.values(
    arr.reduce(
      (acc, item) => (
        item && item[prop] && (acc[item[prop]] = item), acc
      ), // using object mutation (faster)
      {}
    )
  );

// usage (same as above):

const uniqueById = uniqByProp("id");

const unifiedArray = uniqueById(arrayWithDuplicates);

The larger your input array, the more performance gain you'll have from the second approach. In my benchmark (for an input array of length 500 - with a duplicate element probability of 0.5), the second approach is ~440 x as fast as the first approach.

Approach 3: Using ES6 Map

My favorite approach uses a map, instead of an object to accumulate the elements. This has the advantage of preserving the ordering of the original array:

/**
 * using ES6 Map
 **/
const uniqByProp_map = prop => arr =>
  Array.from(
    arr
      .reduce(
        (acc, item) => (
          item && item[prop] && acc.set(item[prop], item),
          acc
        ), // using map (preserves ordering)
        new Map()
      )
      .values()
  );

// usage (still the same):

const uniqueById = uniqByProp("id");

const unifiedArray = uniqueById(arrayWithDuplicates);

Using the the same benchmark conditions as above, this approach is ~2 x as fast as the second approach and ~900 x as fast as the first approach.

Conclusion

Even if all three approaches are looking quite similar, they have surprisingly different performance footprints.

You'll find the benchmarks I used here: jsperf.com/uniq-by-prop

[Drozerah]

Hi ! One other one line path to Rome, from France :

arr = arr.filter((power, toThe, yellowVests) => yellowVests.map(updateDemocracy => updateDemocracy['id']).indexOf(power['id']) === toThe)

console.log(arr)

`

[Marina Mosti]

Merci! :)

[ArthurCat]

More JS but too slow.

[Itachi Uchiha]

I have also this solution O.o

const dupAddress = [
    {
        id: 1,
        name: 'Istanbul'
    },
    {
        id: 2,
        name: 'Kocaeli'
    },
    {
        id: 3,
        name: 'Ankara'
    },
    {
        id: 1,
        name: 'Istanbul'
    }
]

let addresses = [...new Set([...dupAddress.map(address => dupAddress[address.id])])]

console.log(addresses)

But this only works with address.id, so this doesn't work with address.name

Really, why this doesn't work like that?

let addresses = [...new Set([...dupAddress.map(address => dupAddress[address.name])])]

[Marina Mosti]

Well, you're passing [address.id] as an index to the dupAddress array, that's just not going to work because the id !== index. Try changing it to address.id or address.name without accessing the array

[Itachi Uchiha]

Okay, I tried it didn't work actually I was wonder why that didn't work. Thanks.

[waqas]

let addresses = [...new Set([...dupAddress.map(address => dupAddress[address.name])])]

[hasan]

let array = [];
let singleEle = [];
const arr = [
{ id: 1, {questionId : { _id: "5e2016a1560d8c2aa842e65d"} }},
{ id: 1, {questionId : { _id: "5e1c211cc201f33834e7baf1"} }},
{ id: 1, {questionId : { _id: "5e201733560d8c2aa842e65e"} }}
];
arr.forEach(item => {
if (array[item.questionId._id] ) {

}
else{
array[item.questionId._id] = true;
singleEle.push(item)
}
});

[Johnny Reina]

I found myself with this issue recently and though I've always used the same code to find distinct primitives (before we had the Set object), this code required me to adhere to the C# API where you pass in a comparison function T -> T -> boolean. This solution felt relatively clean though obviously not in linear time.
github.com/jreina/ShittyLINQ.js/bl...

[Aissa0347]

Thanks for chairing, I found this also helpful

const arr = [
  { id: 1, name: "test1" },
  { id: 2, name: "test2" },
  { id: 2, name: "test3" },
  { id: 3, name: "test4" },
  { id: 4, name: "test5" },
  { id: 5, name: "test6" },
  { id: 5, name: "test7" },
  { id: 6, name: "test8" }
];
filteredArr= arr.filter((currentUser, index) => {
      return (
        arr.findIndex((user) => user.id === currentUser.id) === index
      );
    });

[JuanDouek]

Other solutions...

Leave last appearance:

const arr = [
    { id: 1, name: "test1" },
    { id: 2, name: "test2" },
    { id: 2, name: "test3" },
    { id: 2, name: "test4" },
    { id: 3, name: "test5" },
    { id: 4, name: "test5" }
];

const by_id = {};

for (item of arr) by_id[item.id] = item;

const uniques = Object.values(by_id);

Leave first appearance:

const arr = [
    { id: 1, name: "test1" },
    { id: 2, name: "test2" },
    { id: 2, name: "test3" },
    { id: 2, name: "test4" },
    { id: 3, name: "test5" },
    { id: 4, name: "test5" }
];

const by_id = {};

for (item of arr)
    if(!by_id[item.id]) by_id[item.id] = item;

const uniques = Object.values(by_id);

Remove duplicated names also:

const arr = [
    { id: 1, name: "test1" },
    { id: 2, name: "test2" },
    { id: 2, name: "test3" },
    { id: 2, name: "test4" },
    { id: 3, name: "test5" },
    { id: 4, name: "test5" }
];

const by_id = {};
const by_name = {};

for (item of arr)
    if(!by_id[item.id] && !by_name[item.name])
    {
        by_id[item.id] = item;
        by_name[item.name] = 1;
    }

const uniques = Object.values(by_id);

[DuaneQ]

Thanks for the code snippet, Marina...I'm getting to errors when I attempt to use it. The first is "Set is only referred to a type but is being used as a value here"

when I use the "REDUCE" example I get the following in the console:

Maximum call stack size exceeded
at Array.reduce