How to remove all duplicates from an array of objects?

Question

I have an object that contains an array of objects.

obj = {};

obj.arr = new Array();

obj.arr.push({place:"here",name:"stuff"});
obj.arr.push({place:"there",name:"morestuff"});
obj.arr.push({place:"there",name:"morestuff"});

I'm wondering what is the best method to remove duplicate objects from an array. So for example, obj.arr would become...

{place:"here",name:"stuff"},
{place:"there",name:"morestuff"}

Answer 1

How about with some ES6 magic?

obj.arr = obj.arr.filter((value, index, self) =>
  index === self.findIndex((t) => (
    t.place === value.place && t.name === value.name
  ))
)

Reference URL

A more generic solution would be:

const uniqueArray = obj.arr.filter((value, index) => {
  const _value = JSON.stringify(value);
  return index === obj.arr.findIndex(obj => {
    return JSON.stringify(obj) === _value;
  });
});

Using the above property strategy instead of JSON.stringify:

const isPropValuesEqual = (subject, target, propNames) =>
  propNames.every(propName => subject[propName] === target[propName]);

const getUniqueItemsByProperties = (items, propNames) => 
  items.filter((item, index, array) =>
    index === array.findIndex(foundItem => isPropValuesEqual(foundItem, item, propNames))
  );

You can add a wrapper if you want the propNames property to be either an array or a value:

const getUniqueItemsByProperties = (items, propNames) => {
  const propNamesArray = Array.from(propNames);

  return items.filter((item, index, array) =>
    index === array.findIndex(foundItem => isPropValuesEqual(foundItem, item, propNamesArray))
  );
};

allowing both getUniqueItemsByProperties('a') and getUniqueItemsByProperties(['a']);

Stackblitz Example

Explanation

• Start by understanding the two methods used:
  • filter, findIndex
• Next take your idea of what makes your two objects equal and keep that in mind.
• We can detect something as a duplicate, if it satisfies the criterion that we have just thought of, but its position is not at the first instance of an object with the criterion.
• Therefore we can use the above criterion to determine if something is a duplicate.

Answer 2

One liners with filter() (Preserves order)

If you have some identifier in the objects which signifies uniqueness (e.g. id), then we can use filter() with findIndex() to work through the list and verify that the index of each object with that id value matches only itself. This means that there's only one such object in the list, i.e. no duplicates.

myArr.filter((obj1, i, arr) => 
  arr.findIndex(obj2 => (obj2.id === obj1.id)) === i
)

(Note that this solution keeps the first instance of detected duplicates in the result. You can instead take the last instance by replacing findIndex with findLastIndex in the above.)

If the order is not important, then map solutions will be faster: Solution with map

---

The above format can be applied to other cases by altering how we check for duplicates (i.e. replacing obj2.id === obj1.id with something else).

Unique by multiple properties (e.g. place and name, as in the question)

myArr.filter((obj1, i, arr) => 
  arr.findIndex(obj2 => 
    ['place', 'name'].every(key => obj2[key] === obj1[key])
  ) === i
)

Unique by all properties

myArr.filter((obj1, i, arr) => 
  arr.findIndex(obj2 => 
    JSON.stringify(obj2) === JSON.stringify(obj1)
  ) === i
)

Caveats:

• This may get slow, depending on object & array sizes
• JSON.stringify() key order is generally consistent, but is only guaranteed in ES2015 and later
  • This means that your mileage may vary, and you may want to prefer something more robust (like comparing specific keys)

Answer 3

Using ES6+ in a single line you can get a unique list of objects by key:

const key = 'place';
const unique = [...new Map(arr.map(item => [item[key], item])).values()]

It can be put into a function:

function getUniqueListBy(arr, key) {
    return [...new Map(arr.map(item => [item[key], item])).values()]
}

Here is a working example:

const arr = [
    {place: "here",  name: "x", other: "other stuff1" },
    {place: "there", name: "x", other: "other stuff2" },
    {place: "here",  name: "y", other: "other stuff4" },
    {place: "here",  name: "z", other: "other stuff5" }
]

function getUniqueListBy(arr, key) {
    return [...new Map(arr.map(item => [item[key], item])).values()]
}

const arr1 = getUniqueListBy(arr, 'place')

console.log("Unique by place")
console.log(JSON.stringify(arr1))

console.log("\nUnique by name")
const arr2 = getUniqueListBy(arr, 'name')

console.log(JSON.stringify(arr2))

How does it work

First the array is remapped in a way that it can be used as an input for a Map.

arr.map(item => [item[key], item]);

which means each item of the array will be transformed in another array with 2 elements; the selected key as first element and the entire initial item as second element, this is called an entry (ex. array entries, map entries). And here is the official doc with an example showing how to add array entries in Map constructor.

Example when key is place:

[["here", {place: "here",  name: "x", other: "other stuff1" }], ...]

Secondly, we pass this modified array to the Map constructor and here is the magic happening. Map will eliminate the duplicate keys values, keeping only last inserted value of the same key. Note: Map keeps the order of insertion. (check difference between Map and object)

new Map(entry array just mapped above)

Third we use the map values to retrieve the original items, but this time without duplicates.

new Map(mappedArr).values()

And last one is to add those values into a fresh new array so that it can look as the initial structure and return that:

return [...new Map(mappedArr).values()]

Answer 4

Simple and performant solution with a better runtime than the 70+ answers that already exist:

const ids = arr.map(({ id }) => id);
const filtered = arr.filter(({ id }, index) => !ids.includes(id, index + 1));

---

Example:

const arr = [{
  id: 1,
  name: 'one'
}, {
  id: 2,
  name: 'two'
}, {
  id: 1,
  name: 'one'
}];

const ids = arr.map(({ id }) => id);
const filtered = arr.filter(({ id }, index) => !ids.includes(id, index + 1));

console.log(filtered);

How it works:

Array.filter() removes all duplicate objects by checking if the previously mapped id-array includes the current id ({id} destructs the object into only its id). To only filter out actual duplicates, it is using Array.includes()'s second parameter fromIndex with index + 1 which will ignore the current object and all previous.

Since every iteration of the filter callback method will only search the array beginning at the current index + 1, this also dramatically reduces the runtime because only objects not previously filtered get checked.

What if you don't have a single unique identifier like id?

Just create a temporary one:

const objToId = ({ name, city, birthyear }) => `${name}-${city}-${birthyear}`;


const ids = arr.map(objToId);
const filtered = arr.filter((item, index) => !ids.includes(objToId(item), index + 1));

Answer 5

A primitive method would be:

const obj = {};

for (let i = 0, len = things.thing.length; i < len; i++) {
  obj[things.thing[i]['place']] = things.thing[i];
}

things.thing = new Array();

 for (const key in obj) { 
   things.thing.push(obj[key]);
}

Answer 6

If you can use Javascript libraries such as underscore or lodash, I recommend having a look at _.uniq function in their libraries. From lodash:

_.uniq(array, [isSorted=false], [callback=_.identity], [thisArg])

Basically, you pass in the array that in here is an object literal and you pass in the attribute that you want to remove duplicates with in the original data array, like this:

var data = [{'name': 'Amir', 'surname': 'Rahnama'}, {'name': 'Amir', 'surname': 'Stevens'}];
var non_duplidated_data = _.uniq(data, 'name'); 

UPDATE: Lodash now has introduced a .uniqBy as well.

Answer 7

I had this exact same requirement, to remove duplicate objects in a array, based on duplicates on a single field. I found the code here: Javascript: Remove Duplicates from Array of Objects

So in my example, I'm removing any object from the array that has a duplicate licenseNum string value.

var arrayWithDuplicates = [
    {"type":"LICENSE", "licenseNum": "12345", state:"NV"},
    {"type":"LICENSE", "licenseNum": "A7846", state:"CA"},
    {"type":"LICENSE", "licenseNum": "12345", state:"OR"},
    {"type":"LICENSE", "licenseNum": "10849", state:"CA"},
    {"type":"LICENSE", "licenseNum": "B7037", state:"WA"},
    {"type":"LICENSE", "licenseNum": "12345", state:"NM"}
];

function removeDuplicates(originalArray, prop) {
     var newArray = [];
     var lookupObject  = {};

     for(var i in originalArray) {
        lookupObject[originalArray[i][prop]] = originalArray[i];
     }

     for(i in lookupObject) {
         newArray.push(lookupObject[i]);
     }
      return newArray;
 }

var uniqueArray = removeDuplicates(arrayWithDuplicates, "licenseNum");
console.log("uniqueArray is: " + JSON.stringify(uniqueArray));

The results:

uniqueArray is:

[{"type":"LICENSE","licenseNum":"10849","state":"CA"},
{"type":"LICENSE","licenseNum":"12345","state":"NM"},
{"type":"LICENSE","licenseNum":"A7846","state":"CA"},
{"type":"LICENSE","licenseNum":"B7037","state":"WA"}]

Answer 8

One liner using Set

var things = new Object();

things.thing = new Array();

things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});

// assign things.thing to myData for brevity
var myData = things.thing;

things.thing = Array.from(new Set(myData.map(JSON.stringify))).map(JSON.parse);

console.log(things.thing)

Explanation:

1. new Set(myData.map(JSON.stringify)) creates a Set object using the stringified myData elements.
2. Set object will ensure that every element is unique.
3. Then I create an array based on the elements of the created set using Array.from.
4. Finally, I use JSON.parse to convert stringified element back to an object.

Answer 9

ES6 one liner is here

let arr = [
  {id:1,name:"sravan ganji"},
  {id:2,name:"pinky"},
  {id:4,name:"mammu"},
  {id:3,name:"avy"},
  {id:3,name:"rashni"},
];

console.log(Object.values(arr.reduce((acc,cur)=>Object.assign(acc,{[cur.id]:cur}),{})))

Answer 10

To remove all duplicates from an array of objects, the simplest way is use filter:

var uniq = {};
var arr  = [{"id":"1"},{"id":"1"},{"id":"2"}];
var arrFiltered = arr.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true));
console.log('arrFiltered', arrFiltered);

Answer 11

One liners with Map ( High performance, Does not preserve order )

Find unique id's in array arr.

const arrUniq = [...new Map(arr.map(v => [v.id, v])).values()]

If the order is important check out the solution with filter: Solution with filter

---

Unique by multiple properties ( place and name ) in array arr

const arrUniq = [...new Map(arr.map(v => [JSON.stringify([v.place,v.name]), v])).values()]

Unique by all properties in array arr

const arrUniq = [...new Map(arr.map(v => [JSON.stringify(v), v])).values()]

Keep the first occurrence in array arr

const arrUniq = [...new Map(arr.slice().reverse().map(v => [v.id, v])).values()].reverse()

Answer 12

Here's another option to do it using Array iterating methods if you need comparison only by one field of an object:

    function uniq(a, param){
        return a.filter(function(item, pos, array){
            return array.map(function(mapItem){ return mapItem[param]; }).indexOf(item[param]) === pos;
        })
    }

    uniq(things.thing, 'place');

Answer 13

This is a generic way of doing this: you pass in a function that tests whether two elements of an array are considered equal. In this case, it compares the values of the name and place properties of the two objects being compared.

ES5 answer

function removeDuplicates(arr, equals) {
    var originalArr = arr.slice(0);
    var i, len, val;
    arr.length = 0;

    for (i = 0, len = originalArr.length; i < len; ++i) {
        val = originalArr[i];
        if (!arr.some(function(item) { return equals(item, val); })) {
            arr.push(val);
        }
    }
}

function thingsEqual(thing1, thing2) {
    return thing1.place === thing2.place
        && thing1.name === thing2.name;
}

var things = [
  {place:"here",name:"stuff"},
  {place:"there",name:"morestuff"},
  {place:"there",name:"morestuff"}
];

removeDuplicates(things, thingsEqual);
console.log(things);

Original ES3 answer

function arrayContains(arr, val, equals) {
    var i = arr.length;
    while (i--) {
        if ( equals(arr[i], val) ) {
            return true;
        }
    }
    return false;
}

function removeDuplicates(arr, equals) {
    var originalArr = arr.slice(0);
    var i, len, j, val;
    arr.length = 0;

    for (i = 0, len = originalArr.length; i < len; ++i) {
        val = originalArr[i];
        if (!arrayContains(arr, val, equals)) {
            arr.push(val);
        }
    }
}

function thingsEqual(thing1, thing2) {
    return thing1.place === thing2.place
        && thing1.name === thing2.name;
}

removeDuplicates(things.thing, thingsEqual);

Answer 14

If you can wait to eliminate the duplicates until after all the additions, the typical approach is to first sort the array and then eliminate duplicates. The sorting avoids the N * N approach of scanning the array for each element as you walk through them.

The "eliminate duplicates" function is usually called unique or uniq. Some existing implementations may combine the two steps, e.g., prototype's uniq

This post has few ideas to try (and some to avoid :-) ) if your library doesn't already have one! Personally I find this one the most straight forward:

    function unique(a){
        a.sort();
        for(var i = 1; i < a.length; ){
            if(a[i-1] == a[i]){
                a.splice(i, 1);
            } else {
                i++;
            }
        }
        return a;
    }  

    // Provide your own comparison
    function unique(a, compareFunc){
        a.sort( compareFunc );
        for(var i = 1; i < a.length; ){
            if( compareFunc(a[i-1], a[i]) === 0){
                a.splice(i, 1);
            } else {
                i++;
            }
        }
        return a;
    }

Answer 15

I think the best approach is using reduce and Map object. This is a single line solution.

const data = [
  {id: 1, name: 'David'},
  {id: 2, name: 'Mark'},
  {id: 2, name: 'Lora'},
  {id: 4, name: 'Tyler'},
  {id: 4, name: 'Donald'},
  {id: 5, name: 'Adrian'},
  {id: 6, name: 'Michael'}
]

const uniqueData = [...data.reduce((map, obj) => map.set(obj.id, obj), new Map()).values()];

console.log(uniqueData)

/*
  in `map.set(obj.id, obj)`
  
  'obj.id' is key. (don't worry. we'll get only values using the .values() method)
  'obj' is whole object.
*/

Answer 16

Considering lodash.uniqWith

const objects = [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }, { 'x': 1, 'y': 2 }];
 
_.uniqWith(objects, _.isEqual);
// => [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }]

Answer 17

To add one more to the list. Using ES6 and Array.reduce with Array.find.
In this example filtering objects based on a guid property.

let filtered = array.reduce((accumulator, current) => {
  if (! accumulator.find(({guid}) => guid === current.guid)) {
    accumulator.push(current);
  }
  return accumulator;
}, []);

Extending this one to allow selection of a property and compress it into a one liner:

const uniqify = (array, key) => array.reduce((prev, curr) => prev.find(a => a[key] === curr[key]) ? prev : prev.push(curr) && prev, []);

To use it pass an array of objects and the name of the key you wish to de-dupe on as a string value:

const result = uniqify(myArrayOfObjects, 'guid')

Answer 18

You could also use a Map:

const dedupThings = Array.from(things.thing.reduce((m, t) => m.set(t.place, t), new Map()).values());

Full sample:

const things = new Object();

things.thing = new Array();

things.thing.push({place:"here",name:"stuff"});
things.thing.push({place:"there",name:"morestuff"});
things.thing.push({place:"there",name:"morestuff"});

const dedupThings = Array.from(things.thing.reduce((m, t) => m.set(t.place, t), new Map()).values());

console.log(JSON.stringify(dedupThings, null, 4));

Result:

[
    {
        "place": "here",
        "name": "stuff"
    },
    {
        "place": "there",
        "name": "morestuff"
    }
]

Answer 19

Dang, kids, let's crush this thing down, why don't we?

let uniqIds = {}, source = [{id:'a'},{id:'b'},{id:'c'},{id:'b'},{id:'a'},{id:'d'}];
let filtered = source.filter(obj => !uniqIds[obj.id] && (uniqIds[obj.id] = true));
console.log(filtered);
// EXPECTED: [{id:'a'},{id:'b'},{id:'c'},{id:'d'}];

Answer 20

let myData = [{place:"here",name:"stuff"}, 
 {place:"there",name:"morestuff"},
 {place:"there",name:"morestuff"}];


let q = [...new Map(myData.map(obj => [JSON.stringify(obj), obj])).values()];

console.log(q)

One-liner using ES6 and new Map().

// assign things.thing to myData
let myData = things.thing;

[...new Map(myData.map(obj => [JSON.stringify(obj), obj])).values()];

Details:-

1. Doing .map() on the data list and converting each individual object into a [key, value] pair array(length =2), the first element(key) would be the stringified version of the object and second(value) would be an object itself.
2. Adding above created array list to new Map() would have the key as stringified object and any same key addition would result in overriding the already existing key.
3. Using .values() would give MapIterator with all values in a Map (obj in our case)
4. Finally, spread ... operator to give new Array with values from the above step.

Answer 21

A TypeScript solution

This will remove duplicate objects and also preserve the types of the objects.

function removeDuplicateObjects(array: any[]) {
  return [...new Set(array.map(s => JSON.stringify(s)))]
    .map(s => JSON.parse(s));
}

Answer 22

 const things = [
  {place:"here",name:"stuff"},
  {place:"there",name:"morestuff"},
  {place:"there",name:"morestuff"}
];
const filteredArr = things.reduce((thing, current) => {
  const x = thing.find(item => item.place === current.place);
  if (!x) {
    return thing.concat([current]);
  } else {
    return thing;
  }
}, []);
console.log(filteredArr)

Solution Via Set Object | According to the data type

const seen = new Set();
 const things = [
  {place:"here",name:"stuff"},
  {place:"there",name:"morestuff"},
  {place:"there",name:"morestuff"}
];

const filteredArr = things.filter(el => {
  const duplicate = seen.has(el.place);
  seen.add(el.place);
  return !duplicate;
});
console.log(filteredArr)

Set Object Feature

Each value in the Set Object has to be unique, the value equality will be checked

The Purpose of Set object storing unique values according to the Data type , whether primitive values or object references.it has very useful four Instance methods add, clear , has & delete.

Unique & data Type feature:..

addmethod

it's push unique data into collection by default also preserve data type .. that means it prevent to push duplicate item into collection also it will check data type by default...

has method

sometime needs to check data item exist into the collection and . it's handy method for the collection to cheek unique id or item and data type..

delete method

it will remove specific item from the collection by identifying data type..

clear method

it will remove all collection items from one specific variable and set as empty object

Set object has also Iteration methods & more feature..

Better Read from Here : Set - JavaScript | MDN

Answer 23

If array contains objects, then you can use this to remove duplicate

const persons= [
      { id: 1, name: 'John',phone:'23' },
      { id: 2, name: 'Jane',phone:'23'},
      { id: 1, name: 'Johnny',phone:'56' },
      { id: 4, name: 'Alice',phone:'67' },
    ];
const unique = [...new Map(persons.map((m) => [m.id, m])).values()];

if remove duplicates on the basis of phone, just replace m.id with m.phone

const unique = [...new Map(persons.map((m) => [m.phone, m])).values()];

Answer 24

removeDuplicates() takes in an array of objects and returns a new array without any duplicate objects (based on the id property).

const allTests = [
  {name: 'Test1', id: '1'}, 
  {name: 'Test3', id: '3'},
  {name: 'Test2', id: '2'},
  {name: 'Test2', id: '2'},
  {name: 'Test3', id: '3'}
];

function removeDuplicates(array) {
  let uniq = {};
  return array.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true))
}

removeDuplicates(allTests);

Expected outcome:

[
  {name: 'Test1', id: '1'}, 
  {name: 'Test3', id: '3'},
  {name: 'Test2', id: '2'}
];

First, we set the value of variable uniq to an empty object.

Next, we filter through the array of objects. Filter creates a new array with all elements that pass the test implemented by the provided function.

return array.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true));

Above, we use the short-circuiting functionality of &&. If the left side of the && evaluates to true, then it returns the value on the right of the &&. If the left side is false, it returns what is on the left side of the &&.

For each object(obj) we check uniq for a property named the value of obj.id (In this case, on the first iteration it would check for the property '1'.) We want the opposite of what it returns (either true or false) which is why we use the ! in !uniq[obj.id]. If uniq has the id property already, it returns true which evaluates to false (!) telling the filter function NOT to add that obj. However, if it does not find the obj.id property, it returns false which then evaluates to true (!) and returns everything to the right of the &&, or (uniq[obj.id] = true). This is a truthy value, telling the filter method to add that obj to the returned array, and it also adds the property {1: true} to uniq. This ensures that any other obj instance with that same id will not be added again.

Answer 25

Fast (less runtime) and type-safe answer for lazy Typescript developers:

export const uniqueBy = <T>( uniqueKey: keyof T, objects: T[]): T[] => {
  const ids = objects.map(object => object[uniqueKey]);
  return objects.filter((object, index) => !ids.includes(object[uniqueKey], index + 1));
} 

Answer 26

This way works well for me:

function arrayUnique(arr, uniqueKey) {
  const flagList = new Set()
  return arr.filter(function(item) {
    if (!flagList.has(item[uniqueKey])) {
      flagList.add(item[uniqueKey])
      return true
    }
  })
}
const data = [
  {
    name: 'Kyle',
    occupation: 'Fashion Designer'
  },
  {
    name: 'Kyle',
    occupation: 'Fashion Designer'
  },
  {
    name: 'Emily',
    occupation: 'Web Designer'
  },
  {
    name: 'Melissa',
    occupation: 'Fashion Designer'
  },
  {
    name: 'Tom',
    occupation: 'Web Developer'
  },
  {
    name: 'Tom',
    occupation: 'Web Developer'
  }
]
console.table(arrayUnique(data, 'name'))// work well

printout

┌─────────┬───────────┬────────────────────┐
│ (index) │   name    │     occupation     │
├─────────┼───────────┼────────────────────┤
│    0    │  'Kyle'   │ 'Fashion Designer' │
│    1    │  'Emily'  │   'Web Designer'   │
│    2    │ 'Melissa' │ 'Fashion Designer' │
│    3    │   'Tom'   │  'Web Developer'   │
└─────────┴───────────┴────────────────────┘

ES5:

function arrayUnique(arr, uniqueKey) {
  const flagList = []
  return arr.filter(function(item) {
    if (flagList.indexOf(item[uniqueKey]) === -1) {
      flagList.push(item[uniqueKey])
      return true
    }
  })
}

These two ways are simpler and more understandable.

Answer 27

Here is a solution for ES6 where you only want to keep the last item. This solution is functional and Airbnb style compliant.

const things = {
  thing: [
    { place: 'here', name: 'stuff' },
    { place: 'there', name: 'morestuff1' },
    { place: 'there', name: 'morestuff2' }, 
  ],
};

const removeDuplicates = (array, key) => {
  return array.reduce((arr, item) => {
    const removed = arr.filter(i => i[key] !== item[key]);
    return [...removed, item];
  }, []);
};

console.log(removeDuplicates(things.thing, 'place'));
// > [{ place: 'here', name: 'stuff' }, { place: 'there', name: 'morestuff2' }]

Answer 28

I know there is a ton of answers in this question already, but bear with me...

Some of the objects in your array may have additional properties that you are not interested in, or you simply want to find the unique objects considering only a subset of the properties.

Consider the array below. Say you want to find the unique objects in this array considering only propOne and propTwo, and ignore any other properties that may be there.

The expected result should include only the first and last objects. So here goes the code:

const array = [{
    propOne: 'a',
    propTwo: 'b',
    propThree: 'I have no part in this...'
},
{
    propOne: 'a',
    propTwo: 'b',
    someOtherProperty: 'no one cares about this...'
},
{
    propOne: 'x',
    propTwo: 'y',
    yetAnotherJunk: 'I am valueless really',
    noOneHasThis: 'I have something no one has'
}];

const uniques = [...new Set(
    array.map(x => JSON.stringify(((o) => ({
        propOne: o.propOne,
        propTwo: o.propTwo
    }))(x))))
].map(JSON.parse);

console.log(uniques);

Answer 29

Another option would be to create a custom indexOf function, which compares the values of your chosen property for each object and wrap this in a reduce function.

var uniq = redundant_array.reduce(function(a,b){
      function indexOfProperty (a, b){
          for (var i=0;i<a.length;i++){
              if(a[i].property == b.property){
                   return i;
               }
          }
         return -1;
      }

      if (indexOfProperty(a,b) < 0 ) a.push(b);
        return a;
    },[]);

Answer 30

This solution worked best for me , by utilising Array.from Method, And also its shorter and readable.

let person = [
{name: "john"}, 
{name: "jane"}, 
{name: "imelda"}, 
{name: "john"},
{name: "jane"}
];

const data = Array.from(new Set(person.map(JSON.stringify))).map(JSON.parse);
console.log(data);