Как найти все индексы элемента в массиве

asked Dec 27, 2013 at 9:52

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); // []

answered Oct 25, 2018 at 8:20

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)

answered Feb 12, 2019 at 11:21

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.

answered May 16, 2018 at 17:53

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]

answered Feb 10, 2019 at 3:05

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'));

answered May 15, 2019 at 9:08

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]

answered May 17, 2020 at 8:19

When both parameter passed as array

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

answered Mar 30, 2021 at 10:07

Also, findIndex() will be useful:

var cars = ['Nano', 'Volvo', 'BMW', 'Nano', 'VW', 'Nano'];

const indexes = [];
const searchedItem = 'NaNo';

cars.findIndex((value, index) => {
  if (value.toLowerCase() === searchedItem.toLowerCase()) {
    indexes.push(index);
  }
});

console.log(indexes); //[ 0, 3, 5 ]

Bonus:

This custom solution using Object.entries() and forEach()

var cars = ['Nano', 'Volvo', 'BMW', 'Nano', 'VW', 'Nano'];

const indexes = [];
const searchableItem = 'Nano';

Object.entries(cars).forEach((item, index) => {
  if (item[1].toLowerCase() === searchableItem.toLowerCase())
    indexes.push(index);
});

console.log(indexes);

Note: I did not run run all tests

answered Dec 19, 2022 at 19:52

findIndex retrieves only the first index which matches callback output. You can implement your own findIndexes by extending Array , then casting your arrays to the new structure .

class EnhancedArray extends Array {
  findIndexes(where) {
    return this.reduce((a, e, i) => (where(e, i) ? a.concat(i) : a), []);
  }
}
   /*----Working with simple data structure (array of numbers) ---*/

//existing array
let myArray = [1, 3, 5, 5, 4, 5];

//cast it :
myArray = new EnhancedArray(...myArray);

//run
console.log(
   myArray.findIndexes((e) => e===5)
)
/*----Working with Array of complex items structure-*/

let arr = [{name: 'Ahmed'}, {name: 'Rami'}, {name: 'Abdennour'}];

arr= new EnhancedArray(...arr);


console.log(
  arr.findIndexes((o) => o.name.startsWith('A'))
)

answered Aug 12, 2017 at 2:32