Develop Reference

How to merge different properties into an object which has same id?

let data = [{id:1, value:10}, {id:2, value:20}, {id:1, name:’test’}]
Output i want is:
data = [{id:1, value:10, name:’test’}, {id:2, value:20}]
How to achieve this ? Please help. Thanks in advance

You can use reduce method to accumulate the array, then you can try findIndex if there is an item with this id exist in the accumulator if the index not equal -1 that's mean the item exist, So, if exist you can group the properties with rest operator, if not you can just push that item in the accumlator
let data = [{id:1, value:10}, {id:2, value:20}, {id:1, name:'test'}]
const result = data.reduce((acc, item) => {
const index = acc.findIndex(i => i.id === item.id);
index !== -1 ? (acc[index] = {...acc[index], ...item}) : acc.push(item)
return acc
}, [])
console.log(result)

The best possible way that I could come up with is using the Lodash's groupBy() method.
First you club all the objects basis on the id, and then keep merging all the objects with the same id.
import _ from "lodash";
let data = [
{
id: 1,
value: 10
},
{
id: 2,
value: 20
},
{
id: 1,
name: "test"
}
];
const groupedData = _.groupBy(data, "id");
const newData = Object.keys(groupedData).map((key) => {
let res = {};
groupedData[key].forEach((obj) => {
res = { ...res, ...obj };
});
return res;
});
console.log(newData);
You can access the working sandbox here.

The question has been answered already and I actually came up with almost exactly the same code as Mina, although it might be a bit hard to understand at first, here's a code sample that is probably a bit easier to understand :
let data = [{id:1, value: 10},{id:2, value: 20}, {id:1, name: 'test'}, {id:1, lastname: 'test2'}];
let transformedData = [];
for (const element of data) {
let matchedIndex = transformedData.findIndex(x => x.id === element.id);
if (matchedIndex < 0) {
matchedIndex = transformedData.length;
transformedData.push(element);
}
transformedData[matchedIndex] = {...transformedData[matchedIndex], ...element};
}
console.log("merged result" , transformedData);
It starts with an empty list and everytime an item is found that isn't matched yet it adds it to the list.
If it already exist in the list it just expands it with the additional data.
Keep in mind that in the end it does the same thing as the reduce approach highlighted by Mina.

let data = [{id:1, value:10}, {id:2, value:20}, {id:1, name:"test"}];
const cache={};
for (let ele of data) {
const {id}=ele;
cache[id] = Object.assign({},cache[id],ele);
}
console.log(Object.values(cache));

How can I do total count of values accordingly selected feild in array

I have a array of objects, for exemple:
let arr = [
{title:apple,quantity:2},
{title:banana,quantity:3},
{title:apple,quantity:5},
{title:banana,quantity:7}
];
array containe many same objects, and i want recived array with uniqe object :
let result = [
{title:apple,quantity:7},
{title:banana,quantity:10}
]
How can I do this?

You can iterate over your array and filter out all the object with same title. Then use reduce to add all the quantity and return a new object. Code is below,
let newArr = [];
arr.forEach((currentObj) => {
const alreadyExists = newArr.findIndex(item => currentObj.title === item.title) > -1;
if(!alreadyExists) {
const filtered = arr.filter(item => item.title === currentObj.title);
const newObject = filtered.reduce((acc, curr) => { return {...acc, quantity: acc.quantity += curr.quantity}}, {...currentObj, quantity: 0})
newArr.push(newObject);
}
})
console.log(newArr);

This is done on a phone so may have some typos but the gist is there:
const resultObj = arr.reduce((acc,curr) =>{
acc[curr.title] = acc[curr.title]== undefined? curr.quantity: acc[curr.title] + curr.quantity
return acc
},{})
const resultArr = Object.entries(resultObj).map([key,value]=>({title:key,quantity:value}))

You could do that in "one line" using arrow function expressions but it won't be very readable unless you know what's happening inside:
let arr = [
{title: "apple",quantity:2},
{title: "banana",quantity:3},
{title: "apple",quantity:5},
{title: "banana",quantity:7}
];
let newArr = [...arr.reduce((acc, {title, quantity}) =>
(acc.set(title, quantity + acc.get(title) || 0), acc), new Map())
].map(([title, quantity]) => ({title, quantity}));
console.log(newArr);
So basically the first part is the reduce method:
arr.reduce((acc, {title, quantity}) =>
(acc.set(title, quantity + acc.get(title) || 0), acc), new Map())
That will returns a Map object, where each title is a key (e.g. "apple") and the quantity is the value of the key.
At this point you have to convert the Map object into an array again, and you do it using the spread syntax.
After you got an array back, you will have it in the following form:
[["apple", 7], ["banana", 10]]
But that is not what you want yet, not in this form, so you have to convert it using the array's map method:
<array>.map(([title, quantity]) => ({title, quantity}))
To keep it concise it uses the destructuring assignment

Filter from Array of objects

I have a Vue.js web app that consumes data from an express.js API and shows the response array in a table. Each object of the array has unique id formed by Name_Version (for example P.90.001000-0004_2). The problem is that I need to show only the last version of each one. For example, if the API response is:
[{id: P.90.001000-0004_1}, {id: P.90.001000-0004_2}, {id: P.90.001000-0004_3}, {id: P.90.002222-0025_1}, {id: P.90.002222-0025_2}]
the result array to show in the table would be:
[{id: P.90.001000-0004_3}, {id: P.90.002222-0025_2}]
It's been impossible for me to achieve this result, can anyone please help me?

You can build an object that maps the portion of the id without the version to the entire object, and you can keep in this map only the latest version of each item:
const data = [{id: 'P.90.001000-0004_1'}, {id: 'P.90.001000-0004_2'}, {id: 'P.90.001000-0004_3'}, {id: 'P.90.002222-0025_1'}, {id: 'P.90.002222-0025_2'}];
const map = {};
data.forEach((item) => {
const [id, version] = item.id.split('_');
const latestItem = map[id];
if (!latestItem || latestItem.id.split('_')[1] < version) {
map[id] = item;
}
});
const filteredData = Object.values(map);
console.log(filteredData);

You can use reduce to get the latest version, and then use Object.entries with map and join to put them back into format:
const arr = [{id: 'P.90.001000-0004_1'}, {id: 'P.90.001000-0004_2'}, {id: 'P.90.001000-0004_3'}, {id: 'P.90.002222-0025_1'}, {id: 'P.90.002222-0025_2'}]
const latest = arr.reduce((a, {id}) => {
const [k,v] = id.split('_')
a[k] = +v > (a[k] || 0) ? v : a[k]
return a
}, {})
const output = Object.entries(latest).map(pair => ({ id: pair.join('_')}))
console.log(output)
If you need to keep the rest of the properties, you can use a map, and keep reference to the original object by index:
const arr = [{id: 'P.90.001000-0004_1'}, {id: 'P.90.001000-0004_2'}, {id: 'P.90.001000-0004_3'}, {id: 'P.90.002222-0025_1'}, {id: 'P.90.002222-0025_2'}]
const latest = arr.reduce((a, {id}, i) => {
const [k,v] = id.split('_')
if(!a.has(k) || +v > a.get(k)[0]) a.set(k, [+v, i])
return a
}, new Map())
const output = [...latest.values()].map(([,i]) => arr[i])
console.log(output)

How to create a new array position an element at first position based on a condition?

I have an array called data, I need to create a new array, where element matching variable toBeFirst is the first one.
Please note I cannot touch the original array data
const toBeFirst = 'b'
const data = [{name: 'a'},{name: 'b'},{name: 'e'}, {name: 'x'}]
// final result
const data = [{name: 'b'},{name: 'a'},{name: 'e'}, {name: 'x'}
any idea how to do it using latest js? can be done using reduce?

Just append the elements, and prepend if its that special element:
const result = data.reduce((arr, el) => el.name === toBeFirst ? [el].concat(arr) : arr.concat([el]), []);
But mutating the array is probably eaiser:
data.unshift(data.splice(data.findIndex(el => el.name === toBeFirst), 1)[0]);

Would've been nice if you supplied code of your trials. Anyway, you could use double filtering. To answer your second question: yes, you can use reduce, see the snippet.
const toBeFirst = 'b'
const data = [{name: 'a'},{name: 'b'},{name: 'e'}, {name: 'x'}];
const dataFirstB = data.filter(v => v.name === toBeFirst)
.concat(data.filter(v => v.name !== toBeFirst));
console.log(dataFirstB);
// using a reducer
const dataReduced = data.reduce( (a, v) =>
v.name !== toBeFirst ? [...a, v] : a,
data.filter(v => v.name === toBeFirst));
console.log(dataReduced);

Removing / Filtering Duplicates Array - Angular [duplicate]

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"}

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 it's 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.

One liners with filter ( Preserves order )
Find unique id's in an array.
arr.filter((v,i,a)=>a.findIndex(v2=>(v2.id===v.id))===i)
If the order is not important, map solutions will be faster: Solution with map
Unique by multiple properties ( place and name )
arr.filter((v,i,a)=>a.findIndex(v2=>['place','name'].every(k=>v2[k] ===v[k]))===i)
Unique by all properties (This will be slow for large arrays)
arr.filter((v,i,a)=>a.findIndex(v2=>(JSON.stringify(v2) === JSON.stringify(v)))===i)
Keep the last occurrence by replacing findIndex with findLastIndex.
arr.filter((v,i,a)=>a.findLastIndex(v2=>(v2.place === v.place))===i)

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()]

Simple and performant solution with a better runtime than the 70+ answers that already exist:
const ids = array.map(o => o.id)
const filtered = array.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(o => o.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.
This obviously also works for any other key that is not called id, multiple or even all keys.

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]);
}

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.

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"}]

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

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}),{})))

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);

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()

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');

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);

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;
}

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.
*/

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')

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 }]

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"
}
]

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'}];

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:-
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.
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.
Using .values() would give MapIterator with all values in a Map (obj in our case)
Finally, spread ... operator to give new Array with values from the above step.

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));
}

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

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.

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));
}

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.

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' }]

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);

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;
},[]);

Here I found a simple solution for removing duplicates from an array of objects using reduce method. I am filtering elements based on the position key of an object
const med = [
{name: 'name1', position: 'left'},
{name: 'name2', position: 'right'},
{name: 'name3', position: 'left'},
{name: 'name4', position: 'right'},
{name: 'name5', position: 'left'},
{name: 'name6', position: 'left1'}
]
const arr = [];
med.reduce((acc, curr) => {
if(acc.indexOf(curr.position) === -1) {
acc.push(curr.position);
arr.push(curr);
}
return acc;
}, [])
console.log(arr)

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()];