Array.prototype.lastIndexOf()

The lastIndexOf() method returns the last index at which a given element can be found in the array, or -1 if it is not present. The array is searched backwards, starting at fromIndex.

The source for this interactive example is stored in a GitHub repository. If you'd like to contribute to the interactive examples project, please clone https://github.com/mdn/interactive-examples and send us a pull request.

Syntax

arr.lastIndexOf(searchElement[, fromIndex])

Parameters

searchElement

Element to locate in the array.

fromIndex Optional

The index at which to start searching backwards. Defaults to the array's length minus one (arr.length - 1), i.e. the whole array will be searched. If the index is greater than or equal to the length of the array, the whole array will be searched. If negative, it is taken as the offset from the end of the array. Note that even when the index is negative, the array is still searched from back to front. If the calculated index is less than 0, -1 is returned, i.e. the array will not be searched.

Return value

The last index of the element in the array; -1 if not found.

Description

lastIndexOf compares searchElement to elements of the Array using strict equality (the same method used by the ===, or triple-equals, operator).

Polyfill

lastIndexOf was added to the ECMA-262 standard in the 5th edition; as such it may not be present in other implementations of the standard. You can work around this by inserting the following code at the beginning of your scripts, allowing use of lastIndexOf in implementations which do not natively support it. This algorithm is exactly the one specified in ECMA-262, 5th edition, assuming Object, TypeError, Number, Math.floor, Math.abs, and Math.min have their original values.

// Production steps of ECMA-262, Edition 5, 15.4.4.15
// Reference: http://es5.github.io/#x15.4.4.15
if (!Array.prototype.lastIndexOf) {
  Array.prototype.lastIndexOf = function(searchElement /*, fromIndex*/) {
    'use strict';

    if (this === void 0 || this === null) {
      throw new TypeError();
    }

    var n, k,
      t = Object(this),
      len = t.length >>> 0;
    if (len === 0) {
      return -1;
    }

    n = len - 1;
    if (arguments.length > 1) {
      n = Number(arguments[1]);
      if (n != n) {
        n = 0;
      }
      else if (n != 0 && n != (1 / 0) && n != -(1 / 0)) {
        n = (n > 0 || -1) * Math.floor(Math.abs(n));
      }
    }

    for (k = n >= 0 ? Math.min(n, len - 1) : len - Math.abs(n); k >= 0; k--) {
      if (k in t && t[k] === searchElement) {
        return k;
      }
    }
    return -1;
  };
}

Again, note that this implementation aims for absolute compatibility with lastIndexOf in Firefox and the SpiderMonkey JavaScript engine, including in several cases which are arguably edge cases. If you intend to use this in real-world applications, you may be able to calculate from with less complicated code if you ignore those cases.

Examples

Using lastIndexOf

The following example uses lastIndexOf to locate values in an array.

var numbers = [2, 5, 9, 2];
numbers.lastIndexOf(2);     // 3
numbers.lastIndexOf(7);     // -1
numbers.lastIndexOf(2, 3);  // 3
numbers.lastIndexOf(2, 2);  // 0
numbers.lastIndexOf(2, -2); // 0
numbers.lastIndexOf(2, -1); // 3

Finding all the occurrences of an element

The following example uses lastIndexOf to find all the indices of an element in a given array, using push to add them to another array as they are found.

var indices = [];
var array = ['a', 'b', 'a', 'c', 'a', 'd'];
var element = 'a';
var idx = array.lastIndexOf(element);
while (idx != -1) {
  indices.push(idx);
  idx = (idx > 0 ? array.lastIndexOf(element, idx - 1) : -1);
}

console.log(indices);
// [4, 2, 0]

Note that we have to handle the case idx == 0 separately here because the element will always be found regardless of the fromIndex parameter if it is the first element of the array. This is different from the indexOf method.

Specifications

Browser compatibility

See also

• Array.prototype.indexOf()
• TypedArray.prototype.lastIndexOf()