Remove array in array javascript

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?

Ivar

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asked Nov 13, 2013 at 15:13

6

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 ] ];

answered Nov 13, 2013 at 15:16

SirkoSirko

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16

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].

JohnK

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answered May 26, 2017 at 14:35

Benny NeugebauerBenny Neugebauer

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2

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] ]]

answered Apr 29, 2019 at 9:45

5

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]];

Andriy

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answered Nov 13, 2013 at 15:17

Ashwin BalamohanAshwin Balamohan

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1

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. //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.includes[item]];
console.log[myArray]

MH2K9

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answered Aug 7, 2019 at 15:14

EggonEggon

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1

I just implemented as:

Array.prototype.exclude = function[list]{
        return this.filter[function[el]{return list.indexOf[el]