Remove all elements contained in another array

Question

I am looking for an efficient way to remove all elements from a javascript array if they are present in another array.

// If I have this array:
var myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];

// and this one:
var toRemove = ['b', 'c', 'g'];

I want to operate on myArray to leave it in this state: ['a', 'd', 'e', 'f']

With jQuery, I'm using grep() and inArray(), which works well:

myArray = $.grep(myArray, function(value) {
    return $.inArray(value, toRemove) < 0;
});

Is there a pure javascript way to do this without looping and splicing?

Answer 1

Use the Array.filter() method:

myArray = myArray.filter( function( el ) {
  return toRemove.indexOf( el ) < 0;
} );

---

Small improvement, as browser support for Array.includes() has increased:

myArray = myArray.filter( function( el ) {
  return !toRemove.includes( el );
} );

---

Next adaptation using arrow functions:

myArray = myArray.filter( ( el ) => !toRemove.includes( el ) );

Answer 2

The filter method should do the trick:

const myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
const toRemove = ['b', 'c', 'g'];

// ES5 syntax
const filteredArray = myArray.filter(function(x) { 
  return toRemove.indexOf(x) < 0;
});

If your toRemove array is large, this sort of lookup pattern can be inefficient. It would be more performant to create a map so that lookups are O(1) rather than O(n).

const toRemoveMap = toRemove.reduce(
  function(memo, item) {
    memo[item] = memo[item] || true;
    return memo;
  },
  {} // initialize an empty object
);

const filteredArray = myArray.filter(function (x) {
  return toRemoveMap[x];
});

// or, if you want to use ES6-style arrow syntax:
const toRemoveMap = toRemove.reduce((memo, item) => ({
  ...memo,
  [item]: true
}), {});

const filteredArray = myArray.filter(x => toRemoveMap[x]);

Answer 3

If you are using an array of objects. Then the below code should do the magic, where an object property will be the criteria to remove duplicate items.

In the below example, duplicates have been removed comparing name of each item.

Try this example. http://jsfiddle.net/deepak7641/zLj133rh/

var myArray = [
  {name: 'deepak', place: 'bangalore'}, 
  {name: 'chirag', place: 'bangalore'}, 
  {name: 'alok', place: 'berhampur'}, 
  {name: 'chandan', place: 'mumbai'}
];
var toRemove = [
  {name: 'deepak', place: 'bangalore'},
  {name: 'alok', place: 'berhampur'}
];

for( var i=myArray.length - 1; i>=0; i--){
  for( var j=0; j<toRemove.length; j++){
      if(myArray[i] && (myArray[i].name === toRemove[j].name)){
      myArray.splice(i, 1);
     }
    }
}

alert(JSON.stringify(myArray));

Answer 4

ECMAScript 6 sets can permit faster computing of the elements of one array that aren't in the other:

const myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
const toRemove = new Set(['b', 'c', 'g']);

const difference = myArray.filter( x => !toRemove.has(x) );

console.log(difference); // ["a", "d", "e", "f"]

Since the lookup complexity for the V8 engine browsers use these days is O(1), the time complexity of the whole algorithm is O(n).

Answer 5

var myArray = [
  {name: 'deepak', place: 'bangalore'}, 
  {name: 'chirag', place: 'bangalore'}, 
  {name: 'alok', place: 'berhampur'}, 
  {name: 'chandan', place: 'mumbai'}
];
var toRemove = [
  {name: 'deepak', place: 'bangalore'},
  {name: 'alok', place: 'berhampur'}
];



myArray = myArray.filter(ar => !toRemove.find(rm => (rm.name === ar.name && ar.place === rm.place) ))

Answer 6

Lodash has an utility function for this as well: https://lodash.com/docs#difference

Answer 7

How about the simplest possible:

var myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
var toRemove = ['b', 'c', 'g'];

var myArray = myArray.filter((item) => !toRemove.includes(item));
console.log(myArray)

Answer 8

I just implemented as:

Array.prototype.exclude = function(list){
        return this.filter(function(el){return list.indexOf(el)<0;})
}

Use as:

myArray.exclude(toRemove);

Answer 9

If you cannot use new ES5 stuff such filter I think you're stuck with two loops:

for( var i =myArray.length - 1; i>=0; i--){
  for( var j=0; j<toRemove.length; j++){
    if(myArray[i] === toRemove[j]){
      myArray.splice(i, 1);
    }
  }
}

Answer 10

Now in one-liner flavor:

console.log(['a', 'b', 'c', 'd', 'e', 'f', 'g'].filter(x => !~['b', 'c', 'g'].indexOf(x)))

Might not work on old browsers.

Answer 11

You can use _.differenceBy from lodash

const myArray = [
  {name: 'deepak', place: 'bangalore'}, 
  {name: 'chirag', place: 'bangalore'}, 
  {name: 'alok', place: 'berhampur'}, 
  {name: 'chandan', place: 'mumbai'}
];
const toRemove = [
  {name: 'deepak', place: 'bangalore'},
  {name: 'alok', place: 'berhampur'}
];
const sorted = _.differenceBy(myArray, toRemove, 'name');

Example code here: CodePen

Answer 12

Proper way to remove all elements contained in another array is to make source array same object by remove only elements:

Array.prototype.removeContained = function(array) {
  var i, results;
  i = this.length;
  results = [];
  while (i--) {
    if (array.indexOf(this[i]) !== -1) {
      results.push(this.splice(i, 1));
    }
  }
  return results;
};

Or CoffeeScript equivalent:

Array.prototype.removeContained = (array) ->
  i = @length
  @splice i, 1 while i-- when array.indexOf(@[i]) isnt -1

Testing inside chrome dev tools:

19:33:04.447 a=1
19:33:06.354 b=2
19:33:07.615 c=3
19:33:09.981 arr = [a,b,c]
19:33:16.460 arr1 = arr

19:33:20.317 arr1 === arr
19:33:20.331 true

19:33:43.592 arr.removeContained([a,c])
19:33:52.433 arr === arr1
19:33:52.438 true

Using Angular framework is the best way to keep pointer to source object when you update collections without large amount of watchers and reloads.

Answer 13

I build the logic without using any built-in methods, please let me know any optimization or modifications. I tested in JS editor it is working fine.

var myArray = [
            {name: 'deepak', place: 'bangalore'},
            {name: 'alok', place: 'berhampur'},
            {name: 'chirag', place: 'bangalore'},
            {name: 'chandan', place: 'mumbai'},

        ];
        var toRemove = [

            {name: 'chirag', place: 'bangalore'},
            {name: 'deepak', place: 'bangalore'},
            /*{name: 'chandan', place: 'mumbai'},*/
            /*{name: 'alok', place: 'berhampur'},*/


        ];
        var tempArr = [];
        for( var i=0 ; i < myArray.length; i++){
            for( var j=0; j<toRemove.length; j++){
                var toRemoveObj = toRemove[j];
                if(myArray[i] && (myArray[i].name === toRemove[j].name)) {
                    break;
                }else if(myArray[i] && (myArray[i].name !== toRemove[j].name)){
                        var fnd = isExists(tempArr,myArray[i]);
                        if(!fnd){
                            var idx = getIdex(toRemove,myArray[i])
                            if (idx === -1){
                                tempArr.push(myArray[i]);
                            }

                        }

                    }

                }
        }
        function isExists(source,item){
            var isFound = false;
            for( var i=0 ; i < source.length; i++){
                var obj = source[i];
                if(item && obj && obj.name === item.name){
                    isFound = true;
                    break;
                }
            }
            return isFound;
        }
        function getIdex(toRemove,item){
            var idex = -1;
            for( var i=0 ; i < toRemove.length; i++){
                var rObj =toRemove[i];
                if(rObj && item && rObj.name === item.name){
                    idex=i;
                    break;
                }
            }
            return idex;
        }

Answer 14

If you're using Typescript and want to match on a single property value, this should work based on Craciun Ciprian's answer above.

You could also make this more generic by allowing non-object matching and / or multi-property value matching.

/**
 *
 * @param arr1 The initial array
 * @param arr2 The array to remove
 * @param propertyName the key of the object to match on
 */
function differenceByPropVal<T>(arr1: T[], arr2: T[], propertyName: string): T[] {
  return arr1.filter(
    (a: T): boolean =>
      !arr2.find((b: T): boolean => b[propertyName] === a[propertyName])
  );
}