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"}
arrayWithNoDuplicates = Array.from(new Set(myArray))
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
))
)
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']);
Explanation
Start by understanding the two methods used: filter, findIndex
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)
t
on findIndex
stands for?
Using ES6+ in a single line you can get a unique list of objects by key:
const unique = [...new Map(arr.map((item, key) => [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()]
id
. The question needs the entire object to be unique across all fields such as place
and name
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.
{id: 1, name: 'one'}
and {namd: 'one', id: 1}
it would fail to detect the duplicate.
{ id }
is destructuring the object into only its id
-key. To illustrate let's look at this these two loops: 1. arr.forEach(object => console.log(object.id))
and 2. arr.forEach({id} => console.log(id))
. They are both doing exactly the same thing: printing the id
-key of all objects in arr
. However, one is using destructuring and the other one is using a more conventional key access via the dot notation.
const arr = [{id: 1, name: 'one'}, {id: 2, name: 'two'}, {id: 1, name: 'THREE'}]
And you don't wanna loose the names for id=1? Would it be possible to keep it in an array?
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.
uniqBy
instead of uniq
, e.g. _.uniqBy(data, 'name')
... documentation: lodash.com/docs#uniqBy
let data = [{'v': {'t':1, 'name':"foo"}}, {'v': {'t':1, 'name':"bar"}}];
do: let uniq = _.uniqBy(data, 'v.t');
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"}]
for(var i in array) { if(array[i][prop]){ //valid lookupObject[array[i][prop]] = array[i]; } else { console.log('falsy object'); } }
for (let i in originalArray) { if (lookupObject[originalArray[i]['id']] === undefined) { newArray.push(originalArray[i]); } lookupObject[originalArray[i]['id']] = originalArray[i]; }
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.
const distinct = (data, elements = []) => [...new Set(data.map(o => JSON.stringify(o, elements)))].map(o => JSON.parse(o));
Then when calling distinct
just pass in the property names for the elements array. For the original post that would be ['place', 'name']
. For @PirateApp's example that would be ['a', 'b']
.
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}),{})))
:cur
in cur.id]:cur
? I dont understand this piece of the code.
_.uniqBy(arr,'id')
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);
id
. The question needs the entire object to be unique across all fields such as place
and name
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;
}
function(_a,_b){return _a.a===_b.a && _a.b===_b.b;}
then the array won't be sorted.
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. */
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()
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'}];
id
. The question needs the entire object to be unique across all fields such as place
and name
place
and name
today. Anyone reading this thread is looking for an optimal way to dedup a list of objects, and this is a compact way of doing so.
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')
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"
}
]
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));
}
any
entirely defeats the purpose of TypeScript
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 }]
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.
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:..
add
method
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.
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' }]
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));
}
uniqueKey
should be keyof T
instead of string
to make it more precise.
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);
array
become one in the uniques
. Now should that object contain propThree
from array[0]
, or someOtherProperty
from array[1]
, or both, or something else? As long as we know exactly what to do in such case, what you asked for is doable for sure.
(({ propOne, propTwo }) => ({ propOne, propTwo }))(x)
?
(x)
is an arrow function which is unpacking the argument object into properties propOne
and propTwo
. Learn about object destructuring here. Now that I have read the code again, I think it should have been written a little more clearly. I have updated the code.
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;
},[]);
lodash.isequal
npm package as a lightweight object comparator to perform unique array filtering ...e.g. distinct array of objects. Just swapped in if (_.isEqual(a[i], b)) {
instead of looking @ a single property
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)
Continuing exploring ES6 ways of removing duplicates from array of objects: setting thisArg
argument of Array.prototype.filter
to new Set
provides a decent alternative:
const things = [ {place:"here",name:"stuff"}, {place:"there",name:"morestuff"}, {place:"there",name:"morestuff"} ]; const filtered = things.filter(function({place, name}) { const key =`${place}${name}`; return !this.has(key) && this.add(key); }, new Set); console.log(filtered);
However, it will not work with arrow functions () =>
, as this
is bound to their lexical scope.
Success story sharing
things.thing = things.thing.filter((thing, index, self) => self.findIndex(t => t.place === thing.place && t.name === thing.name) === index)
const uniqueArray = arrayOfObjects.filter((object,index) => index === arrayOfObjects.findIndex(obj => JSON.stringify(obj) === JSON.stringify(object)));
jsfiddle.net/x9ku0p7L/28