How to find index of all occurrences of element in array?

Question

I am trying to find the index of all the instances of an element, say, "Nano", in a JavaScript array.

var Cars = ["Nano", "Volvo", "BMW", "Nano", "VW", "Nano"];

I tried jQuery.inArray, or similarly, .indexOf(), but it only gave the index of the last instance of the element, i.e. 5 in this case.

How do I get it for all instances?

Answer 1

The .indexOf() method has an optional second parameter that specifies the index to start searching from, so you can call it in a loop to find all instances of a particular value:

function getAllIndexes(arr, val) {
    var indexes = [], i = -1;
    while ((i = arr.indexOf(val, i+1)) != -1){
        indexes.push(i);
    }
    return indexes;
}

var indexes = getAllIndexes(Cars, "Nano");

You don't really make it clear how you want to use the indexes, so my function returns them as an array (or returns an empty array if the value isn't found), but you could do something else with the individual index values inside the loop.

UPDATE: As per VisioN's comment, a simple for loop would get the same job done more efficiently, and it is easier to understand and therefore easier to maintain:

function getAllIndexes(arr, val) {
    var indexes = [], i;
    for(i = 0; i < arr.length; i++)
        if (arr[i] === val)
            indexes.push(i);
    return indexes;
}

Answer 2

Another alternative solution is to use Array.prototype.reduce():

["Nano","Volvo","BMW","Nano","VW","Nano"].reduce(function(a, e, i) {
    if (e === 'Nano')
        a.push(i);
    return a;
}, []);   // [0, 3, 5]

N.B.: Check the browser compatibility for reduce method and use polyfill if required.

Answer 3

Another approach using Array.prototype.map() and Array.prototype.filter():

var indices = array.map((e, i) => e === value ? i : '').filter(String)

Answer 4

You can write a simple readable solution to this by using both map and filter:

const nanoIndexes = Cars
  .map((car, i) => car === 'Nano' ? i : -1)
  .filter(index => index !== -1);

---

EDIT: If you don't need to support IE/Edge (or are transpiling your code), ES2019 gave us flatMap, which lets you do this in a simple one-liner:

const nanoIndexes = Cars.flatMap((car, i) => car === 'Nano' ? i : []);

Answer 5

More simple way with es6 style.

const indexOfAll = (arr, val) => arr.reduce((acc, el, i) => (el === val ? [...acc, i] : acc), []);


//Examples:
var cars = ["Nano", "Volvo", "BMW", "Nano", "VW", "Nano"];
indexOfAll(cars, "Nano"); //[0, 3, 5]
indexOfAll([1, 2, 3, 1, 2, 3], 1); // [0,3]
indexOfAll([1, 2, 3], 4); // []

Answer 6

Note: MDN gives a method using a while loop:

var indices = [];
var array = ['a', 'b', 'a', 'c', 'a', 'd'];
var element = 'a';
var idx = array.indexOf(element);
while (idx != -1) {
  indices.push(idx);
  idx = array.indexOf(element, idx + 1);
}

I wouldn't say it's any better than other answers. Just interesting.

Answer 7

const indexes = cars
    .map((car, i) => car === "Nano" ? i : null)
    .filter(i => i !== null)

Answer 8

I just want to update with another easy method.

You can also use forEach method.

var Cars = ["Nano", "Volvo", "BMW", "Nano", "VW", "Nano"];

var result = [];

Cars.forEach((car, index) => car === 'Nano' ? result.push(index) : null)

Answer 9

This worked for me:

let array1 = [5, 12, 8, 130, 44, 12, 45, 12, 56];
let numToFind = 12
let indexesOf12 = [] // the number whose occurrence in the array we want to find

array1.forEach(function(elem, index, array) {
    if (elem === numToFind) {indexesOf12.push(index)}
    return indexesOf12
})

console.log(indexesOf12) // outputs [1, 5, 7]

Answer 10

Just to share another method, you can use Function Generators to achieve the result as well:

function findAllIndexOf(target, needle) {
  return [].concat(...(function*(){
    for (var i = 0; i < target.length; i++) if (target[i] === needle) yield [i];
  })());
}

var target = "hellooooo";
var target2 = ['w','o',1,3,'l','o'];

console.log(findAllIndexOf(target, 'o'));
console.log(findAllIndexOf(target2, 'o'));

Answer 11

["a", "b", "a", "b"]
   .map((val, index) => ({ val, index }))
   .filter(({val, index}) => val === "a")
   .map(({val, index}) => index)

=> [0, 2]

Answer 12

You can use Polyfill

if (!Array.prototype.filterIndex) 
{
    Array.prototype.filterIndex = function (func, thisArg) {

        'use strict';
        if (!((typeof func === 'Function' || typeof func === 'function') && this))
            throw new TypeError();

        let len = this.length >>> 0,
            res = new Array(len), // preallocate array
            t = this, c = 0, i = -1;

        let kValue;
        if (thisArg === undefined) {
            while (++i !== len) {
                // checks to see if the key was set
                if (i in this) {
                    kValue = t[i]; // in case t is changed in callback
                    if (func(t[i], i, t)) {
                        res[c++] = i;
                    }
                }
            }
        }
        else {
            while (++i !== len) {
                // checks to see if the key was set
                if (i in this) {
                    kValue = t[i];
                    if (func.call(thisArg, t[i], i, t)) {
                        res[c++] = i;
                    }
                }
            }
        }

        res.length = c; // shrink down array to proper size
        return res;
    };
}

Use it like this:

[2,23,1,2,3,4,52,2].filterIndex(element => element === 2)

result: [0, 3, 7]