CODEWITHSUNDEEP
javascriptarrays
Gaurav joshi
Gaurav joshi

Reputation: 1799

How to remove more than one elements from an array from different indexes at same time in JavaScript?

I have an array of integers, which I'm using the .push() method to add to.

I know that splice() method adds/removes items to/from an array, and returns the removed item(s).

But is there is a way to remove multiple elements from the array at different indexes at the same time?

I know we can also use filter() method but The filter() method creates a new array with all elements that pass the test implemented by the provided function which I don't want.

Ex:- 

var array = [1,2,3,4,5];

I know to remove 3 and 5 I can follow following steps:-

  1. array.splice(2, 1);
  2. array.splice(3, 1);

But Can I achieve this in a single step without using splice() method twice?

Upvotes: 4

Views: 3758

Answers (4)

guest271314
guest271314

Reputation: 1

You can use trailing comma at destructuring assignment to select specific elements from array, assign resulting values within array to original array reference.

var array = [1,2,3,4,5];
{let [a,b,,c,,] = array; array = [a,b,c]};

console.log(array);

You can use object destructuring on an array to get only specific indexes from array 

var array = [1,2,3,4,5];
{let {0:a,1:b,3:c} = array; array = [a,b,c]};

console.log(array);

You can also use .forEach() to iterate an array of elements to match, .indexOf(), .splice() to remove elements which match value of element within array

var array = [1,2,3,4,5];
[3, 5].forEach(p => array.splice(array.indexOf(p), 1));

console.log(array);

Using for..of loop, Array.prototype.findIndex(), Array.prototype.splice() 

var array = [1,2,3,4,5];
var not = [3, 5];
for (let n of not) array.splice(array.findIndex(v => v === n), 1);

console.log(array);

Using for..of loop with Array.prototype.entries(), Array.prototype.some(), Array.prototype.splice()

var array = [1,2,3,4,5];
var not = [3,5];
for (let [k, p] of array.entries()) not.some(n => !(n-p)) && array.splice(k, 1);

console.log(array);

You can also use for loop, Object.assign() to set indexes to keep at beginning of array, call .splice() with parameter set to - .length of the number of elements to remove from array

var array = [1,2,3,4,5];
var not = [3,5];
for (var o = {}, k = -1, i = 0; i < array.length; i++) {
  if (!not.some(n => n === array[i])) o[++k] = array[i];
}
Object.assign(array, o).splice(-not.length);

console.log(array);

Using Array.prototype.reduce() with Object.assign()

var array = [1,2,3,4,5];
var not = [3,5];
Object.assign(array, array.reduce(([o, not, k], p) => 
  [Object.assign(o, !not.some(n => n === p) ? {[++k]:p} : void 0), not, k]
  , [{}, not, -1]).shift()
).splice(-not.length);

console.log(array);

Another option, if the numbers are unique, is to use Set, .delete(), convert Set object to Array using rest element at destructuring assignment

var array = new Set;
var not = [3, 5];
for (let n = 1; n <= 5; n++) array.add(n);

for (let n of not) array.delete(n);
[...array] = array;

console.log(array); // `array = new Set(array)`

Upvotes: 3

Redu
Redu

Reputation: 26161

May be you can do like this in place;

var          arr = [1,2,3,4,5,17,23,37,61,15],
itemsToGetRidOff = [3,17,15,1,23];
arr.reduceRight((_,e,i,a) => itemsToGetRidOff.includes(e) && a.splice(i,1),[]);
console.log(arr);

Upvotes: 1

Lakshmi Swetha G
Lakshmi Swetha G

Reputation: 2839

Hard code when you know the specific values you need to remove and when the array is small:

var array = [1,2,3,4,5,6,7];
array.splice(2,3,4);
console.log(array);

Upvotes: 1

Deividas
Deividas

Reputation: 6507

If you know the exact values you want removed you can use array.filter().

This removes more than one element from an array at different indexes at same time.

Example in your case:

var filtered = [1,2,3,4,5].filter(value => { return value !== 3 && value !== 5 });

console.log(filtered)

Upvotes: 3

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