Remove Object from Array using JavaScript

Question

How can I remove an object from an array? I wish to remove the object that includes name Kristian from someArray. For example:

someArray = [{name:"Kristian", lines:"2,5,10"},
             {name:"John", lines:"1,19,26,96"}];

I want to achieve:

someArray = [{name:"John", lines:"1,19,26,96"}];

Answer 1

You can use several methods to remove item(s) from an Array:

//1
someArray.shift(); // first element removed
//2
someArray = someArray.slice(1); // first element removed
//3
someArray.splice(0, 1); // first element removed
//4
someArray.pop(); // last element removed
//5
someArray = someArray.slice(0, someArray.length - 1); // last element removed
//6
someArray.length = someArray.length - 1; // last element removed

If you want to remove element at position x, use:

someArray.splice(x, 1);

Or

someArray = someArray.slice(0, x).concat(someArray.slice(-x));

Reply to the comment of @chill182: you can remove one or more elements from an array using Array.filter, or Array.splice combined with Array.findIndex (see MDN).

See this Stackblitz project or the snippet below:

// non destructive filter > noJohn = John removed, but someArray will not change
let someArray = getArray();
let noJohn = someArray.filter( el => el.name !== "John" ); 
log(`let noJohn = someArray.filter( el => el.name !== "John")`,
  `non destructive filter [noJohn] =`, format(noJohn));
log(`**someArray.length ${someArray.length}`);

// destructive filter/reassign John removed > someArray2 =
let someArray2 = getArray();
someArray2 = someArray2.filter( el => el.name !== "John" );
log("", 
  `someArray2 = someArray2.filter( el => el.name !== "John" )`,
  `destructive filter/reassign John removed [someArray2] =`, 
  format(someArray2));
log(`**someArray2.length after filter ${someArray2.length}`);

// destructive splice /w findIndex Brian remains > someArray3 =
let someArray3 = getArray();
someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1);
someArray3.splice(someArray3.findIndex(v => v.name === "John"), 1);
log("",
  `someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1),`,
  `destructive splice /w findIndex Brian remains [someArray3] =`, 
  format(someArray3));
log(`**someArray3.length after splice ${someArray3.length}`);

// if you're not sure about the contents of your array, 
// you should check the results of findIndex first
let someArray4 = getArray();
const indx = someArray4.findIndex(v => v.name === "Michael");
someArray4.splice(indx, indx >= 0 ? 1 : 0);
log("", `someArray4.splice(indx, indx >= 0 ? 1 : 0)`,
  `check findIndex result first [someArray4] = (nothing is removed)`,
  format(someArray4));
log(`**someArray4.length (should still be 3) ${someArray4.length}`);

// -- helpers -- 
function format(obj) {
  return JSON.stringify(obj, null, " ");
}

function log(...txt) {
  document.querySelector("pre").textContent += `${txt.join("\n")}\n`
}

function getArray() {
  return [ {name: "Kristian", lines: "2,5,10"},
           {name: "John", lines: "1,19,26,96"},
           {name: "Brian", lines: "3,9,62,36"} ];
}

<pre>
**Results**

</pre>

Answer 2

The clean solution would be to use Array.filter:

var filtered = someArray.filter(function(el) { return el.Name != "Kristian"; }); 

The problem with this is that it does not work on IE < 9. However, you can include code from a Javascript library (e.g. underscore.js) that implements this for any browser.

Answer 3

I recommend using lodash.js or sugar.js for common tasks like this:

// lodash.js
someArray = _.reject(someArray, function(el) { return el.Name === "Kristian"; });

// sugar.js
someArray.remove(function(el) { return el.Name === "Kristian"; });

in most projects, having a set of helper methods that is provided by libraries like these is quite useful.

Answer 4

ES2015

let someArray = [
               {name:"Kristian", lines:"2,5,10"},
               {name:"John", lines:"1,19,26,96"},
               {name:"Kristian", lines:"2,58,160"},
               {name:"Felix", lines:"1,19,26,96"}
            ];

someArray = someArray.filter(person => person.name != 'John');

It will remove John!

Answer 5

How about this?

$.each(someArray, function(i){
    if(someArray[i].name === 'Kristian') {
        someArray.splice(i,1);
        return false;
    }
});

Answer 6

Your "array" as shown is invalid JavaScript syntax. Curly brackets {} are for objects with property name/value pairs, but square brackets [] are for arrays - like so:

someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];

In that case, you can use the .splice() method to remove an item. To remove the first item (index 0), say:

someArray.splice(0,1);

// someArray = [{name:"John", lines:"1,19,26,96"}];

If you don't know the index but want to search through the array to find the item with name "Kristian" to remove you could to this:

for (var i =0; i < someArray.length; i++)
   if (someArray[i].name === "Kristian") {
      someArray.splice(i,1);
      break;
   }

EDIT: I just noticed your question is tagged with "jQuery", so you could try the $.grep() method:

someArray = $.grep(someArray,
                   function(o,i) { return o.name === "Kristian"; },
                   true);

Answer 7

You could use array.filter().

e.g.

        someArray = [{name:"Kristian", lines:"2,5,10"},
                     {name:"John", lines:"1,19,26,96"}];

        someArray = someArray.filter(function(returnableObjects){
               return returnableObjects.name !== 'Kristian';
        });

        //someArray will now be = [{name:"John", lines:"1,19,26,96"}];

Arrow functions:

someArray = someArray.filter(x => x.name !== 'Kristian')

Answer 8

const someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];

We get the index of the object which have name property value as "Kristian"

const index = someArray.findIndex(key => key.name === "Kristian");
console.log(index); // 0

By using splice function we are removing the object which have the name property value as "Kristian"

someArray.splice(index,1);
console.log(someArray); // [{name:"John", lines:"1,19,26,96"}]

Answer 9

I have made a dynamic function takes the objects Array, Key and value and returns the same array after removing the desired object:

function removeFunction (myObjects,prop,valu)
        {
             return myObjects.filter(function (val) {
              return val[prop] !== valu;
          });

        }

Full Example: DEMO

var obj = {
            "results": [
              {
                  "id": "460",
                  "name": "Widget 1",
                  "loc": "Shed"
              }, {
                  "id": "461",
                  "name": "Widget 2",
                  "loc": "Kitchen"
              }, {
                  "id": "462",
                  "name": "Widget 3",
                  "loc": "bath"
              }
            ]
            };


        function removeFunction (myObjects,prop,valu)
        {
             return myObjects.filter(function (val) {
              return val[prop] !== valu;
          });

        }


console.log(removeFunction(obj.results,"id","460"));

Answer 10

This is a function that works for me:

function removeFromArray(array, value) {
    var idx = array.indexOf(value);
    if (idx !== -1) {
        array.splice(idx, 1);
    }
    return array;
}

Answer 11

You could also try doing something like this:

var myArray = [{'name': 'test'}, {'name':'test2'}];
var myObject = {'name': 'test'};
myArray.splice(myArray.indexOf(myObject),1);

Answer 12

someArray = jQuery.grep(someArray , function (value) {
        return value.name != 'Kristian';
});

Answer 13

Use splice function on arrays. Specify the position of the start element and the length of the subsequence you want to remove.

someArray.splice(pos, 1);

Answer 14

Here is an example with map and splice

const arrayObject = [
  { name: "name1", value: "value1" },
  { name: "name2", value: "value2" },
  { name: "name3", value: "value3" },
];

let index = arrayObject.map((item) => item.name).indexOf("name1");
if (index > -1) {
  arrayObject.splice(index, 1);
  console.log("Result", arrayObject);
}

Output

Result [
  {
    "name": "name2",
    "value": "value2"
  },
  {
    "name": "name3",
    "value": "value3"
  }
]

Answer 15

Vote for the UndercoreJS for simple work with arrays.

_.without() function helps to remove an element:

 _.without([1, 2, 1, 0, 3, 1, 4], 0, 1);
    => [2, 3, 4]

Answer 16

Performance

Today 2021.01.27 I perform tests on MacOs HighSierra 10.13.6 on Chrome v88, Safari v13.1.2 and Firefox v84 for chosen solutions.

Results

For all browsers:

• fast/fastest solutions when element not exists: A and B
• fast/fastest solutions for big arrays: C
• fast/fastest solutions for big arrays when element exists: H
• quite slow solutions for small arrays: F and G
• quite slow solutions for big arrays: D, E and F

Details

I perform 4 tests cases:

• small array (10 elements) and element exists - you can run it HERE
• small array (10 elements) and element NOT exists - you can run it HERE
• big array (milion elements) and element exists - you can run it HERE
• big array (milion elements) and element NOT exists - you can run it HERE

Below snippet presents differences between solutions A B C D E F G H I

function A(arr, name) {
  let idx = arr.findIndex(o => o.name==name);
  if(idx>=0) arr.splice(idx, 1);
  return arr;
}


function B(arr, name) {
  let idx = arr.findIndex(o => o.name==name);
  return idx<0 ? arr : arr.slice(0,idx).concat(arr.slice(idx+1,arr.length));
}


function C(arr, name) {
  let idx = arr.findIndex(o => o.name==name);
  delete arr[idx];
  return arr;
}


function D(arr, name) {
  return arr.filter(el => el.name != name);
}


function E(arr, name) {
  let result = [];
  arr.forEach(o => o.name==name || result.push(o));
  return result;
}


function F(arr, name) {
  return _.reject(arr, el => el.name == name);
}


function G(arr, name) {
  let o = arr.find(o => o.name==name);
  return _.without(arr,o);
}


function H(arr, name) {
  $.each(arr, function(i){
      if(arr[i].name === 'Kristian') {
          arr.splice(i,1);
          return false;
      }
  });
  return arr;
}


function I(arr, name) {
  return $.grep(arr,o => o.name!=name);
}








// Test
let test1 = [   
    {name:"Kristian", lines:"2,5,10"},
    {name:"John", lines:"1,19,26,96"},  
];

let test2 = [   
    {name:"John3", lines:"1,19,26,96"},
    {name:"Kristian", lines:"2,5,10"},
    {name:"John", lines:"1,19,26,96"},
  {name:"Joh2", lines:"1,19,26,96"},
];

let test3 = [   
    {name:"John3", lines:"1,19,26,96"},
    {name:"John", lines:"1,19,26,96"},
  {name:"Joh2", lines:"1,19,26,96"},
];

console.log(`
Test1: original array from question
Test2: array with more data
Test3: array without element which we want to delete
`);

[A,B,C,D,E,F,G,H,I].forEach(f=> console.log(`
Test1 ${f.name}: ${JSON.stringify(f([...test1],"Kristian"))}
Test2 ${f.name}: ${JSON.stringify(f([...test2],"Kristian"))}
Test3 ${f.name}: ${JSON.stringify(f([...test3],"Kristian"))}
`));

<script src="https://code.jquery.com/jquery-3.5.1.min.js" integrity="sha256-9/aliU8dGd2tb6OSsuzixeV4y/faTqgFtohetphbbj0=" crossorigin="anonymous"></script>
  
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.20/lodash.min.js" integrity="sha512-90vH1Z83AJY9DmlWa8WkjkV79yfS2n2Oxhsi2dZbIv0nC4E6m5AbH8Nh156kkM7JePmqD6tcZsfad1ueoaovww==" crossorigin="anonymous"> </script>
  
This shippet only presents functions used in performance tests - it not perform tests itself!

And here are example results for chrome

Answer 17

With ES 6 arrow function

let someArray = [
                 {name:"Kristian", lines:"2,5,10"},
                 {name:"John", lines:"1,19,26,96"}
                ];
let arrayToRemove={name:"Kristian", lines:"2,5,10"};
someArray=someArray.filter((e)=>e.name !=arrayToRemove.name && e.lines!= arrayToRemove.lines)

Answer 18

Although this is probably not that appropriate for this situation I found out the other day that you can also use the delete keyword to remove an item from an array if you don't need to alter the size of the array e.g.

var myArray = [1,2,3];

delete myArray[1];

console.log(myArray[1]); //undefined

console.log(myArray.length); //3 - doesn't actually shrink the array down

Answer 19

Simplest solution would be to create a map that stores the indexes for each object by name, like this:

//adding to array
var newPerson = {name:"Kristian", lines:"2,5,10"}
someMap[ newPerson.name ] = someArray.length;
someArray.push( newPerson );

//deleting from the array
var index = someMap[ 'Kristian' ];
someArray.splice( index, 1 );

Answer 20

You can use map function also.

someArray = [{name:"Kristian", lines:"2,5,10"},{name:"John",lines:"1,19,26,96"}];
newArray=[];
someArray.map(function(obj, index){
    if(obj.name !== "Kristian"){
       newArray.push(obj);
    }
});
someArray = newArray;
console.log(someArray);

Answer 21

If you want to remove all occurrences of a given object (based on some condition) then use the javascript splice method inside a for the loop.

Since removing an object would affect the array length, make sure to decrement the counter one step, so that length check remains intact.

var objArr=[{Name:"Alex", Age:62},
  {Name:"Robert", Age:18},
  {Name:"Prince", Age:28},
  {Name:"Cesar", Age:38},
  {Name:"Sam", Age:42},
  {Name:"David", Age:52}
];

for(var i = 0;i < objArr.length; i ++)
{
  if(objArr[i].Age > 20)
  {
    objArr.splice(i, 1);
    i--;  //re-adjust the counter.
  }
}

The above code snippet removes all objects with age greater than 20.

Answer 22

This answer

for (var i =0; i < someArray.length; i++)
   if (someArray[i].name === "Kristian") {
      someArray.splice(i,1);
   }

is not working for multiple records fulfilling the condition. If you have two such consecutive records, only the first one is removed, and the other one skipped. You have to use:

for (var i = someArray.length - 1; i>= 0; i--)
   ...

instead .

Answer 23

I guess the answers are very branched and knotted.

You can use the following path to remove an array object that matches the object given in the modern JavaScript jargon.

coordinates = [
    { lat: 36.779098444109145, lng: 34.57202827508546 },
    { lat: 36.778754712956506, lng: 34.56898128564454 },
    { lat: 36.777414146732426, lng: 34.57179224069215 }
];

coordinate = { lat: 36.779098444109145, lng: 34.57202827508546 };

removeCoordinate(coordinate: Coordinate): Coordinate {
    const found = this.coordinates.find((coordinate) => coordinate == coordinate);
    if (found) {
      this.coordinates.splice(found, 1);
    }
    return coordinate;
  }

Answer 24

There seems to be an error in your array syntax so assuming you mean an array as opposed to an object, Array.splice is your friend here:

someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];
someArray.splice(1,1)

Answer 25

Use javascript's splice() function.

This may help: http://www.w3schools.com/jsref/jsref_splice.asp

Answer 26

You could also use some:

someArray = [{name:"Kristian", lines:"2,5,10"},
             {name:"John", lines:"1,19,26,96"}];

someArray.some(item => { 
    if(item.name === "Kristian") // Case sensitive, will only remove first instance
        someArray.splice(someArray.indexOf(item),1) 
})

Answer 27

This is what I use.

Array.prototype.delete = function(pos){
    this[pos] = undefined;
    var len = this.length - 1;
    for(var a = pos;a < this.length - 1;a++){
      this[a] = this[a+1];
    }
    this.pop();
  }

Then it is as simple as saying

var myArray = [1,2,3,4,5,6,7,8,9];
myArray.delete(3);

Replace any number in place of three. After the expected output should be:

console.log(myArray); //Expected output 1,2,3,5,6,7,8,9

Answer 28

splice(i, 1) where i is the incremental index of the array will remove the object. But remember splice will also reset the array length so watch out for 'undefined'. Using your example, if you remove 'Kristian', then in the next execution within the loop, i will be 2 but someArray will be a length of 1, therefore if you try to remove "John" you will get an "undefined" error. One solution to this albeit not elegant is to have separate counter to keep track of index of the element to be removed.

Answer 29

Returns only objects from the array whose property name is not "Kristian"

var noKristianArray = $.grep(someArray, function (el) { return el.name!= "Kristian"; });

---

Demo:

 var someArray = [
                {name:"Kristian", lines:"2,5,10"},
                {name:"John", lines:"1,19,26,96"},
                {name:"Kristian", lines:"2,58,160"},
                {name:"Felix", lines:"1,19,26,96"}
                ];
    
var noKristianArray = $.grep(someArray, function (el) { return el.name!= "Kristian"; });

console.log(noKristianArray);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 30

you can filter like this:

const someArray = [{
    name: "Kristian",
    lines: "2,5,10"
  },
  {
    name: "John",
    lines: "1,19,26,96"
  }
];
var filtered = someArray.filter((el) => el.name != "Kristian");
console.log(filtered)

Answer 31

This Concepts using Kendo Grid

var grid = $("#addNewAllergies").data("kendoGrid");

var selectedItem = SelectedCheckBoxList;

for (var i = 0; i < selectedItem.length; i++) {
    if(selectedItem[i].boolKendoValue==true)
    {
        selectedItem.length= 0;
    }
}

Answer 32

If you don't have any properties on the objects in the array that you know (or perhaps that are unique), but you have a reference to the object that you want to remove, you can do what's in the unregisterObject method below:

let registeredObjects = [];

function registerObject(someObject) { registeredObjects.push(someObject); }
function unregisterObject(someObject) { registeredObjects = registeredObjects.filter(obj => obj !== someObject); }

let myObject1 = {hidden: "someValue1"}; // Let's pretend we don't know the hidden attribute
let myObject2 = {hidden: "someValue2"};

registerObject(myObject1);
registerObject(myObject2);
console.log(`There are ${registeredObjects.length} objects registered. They are: ${JSON.stringify(registeredObjects)}`);

unregisterObject(myObject1);
console.log(`There are ${registeredObjects.length} objects registered. They are: ${JSON.stringify(registeredObjects)}`);