Array.prototype.lastIndexOf()

The lastIndexOf() method of Array instances 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.

Syntax

lastIndexOf(searchElement)
lastIndexOf(searchElement, fromIndex)

Parameters

searchElement
- : Element to locate in the array.
fromIndex
- : Zero-based index at which to start searching backwards, converted to an integer.
  - Negative index counts back from the end of the array — if fromIndex < 0, fromIndex + array.length is used.
  - If fromIndex < -array.length, the array is not searched and -1 is returned. You can think of it conceptually as starting at a nonexistent position before the beginning of the array and going backwards from there. There are no array elements on the way, so searchElement is never found.
  - If fromIndex >= array.length or fromIndex is omitted, array.length - 1 is used, causing the entire array to be searched. You can think of it conceptually as starting at a nonexistent position beyond the end of the array and going backwards from there. It eventually reaches the real end position of the array, at which point it starts searching backwards through the actual array elements.

Return value

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

Description

The lastIndexOf() method compares searchElement to elements of the array using strict equality (the same algorithm used by the === operator). NaN values are never compared as equal, so lastIndexOf() always returns -1 when searchElement is NaN.

The lastIndexOf() method skips empty slots in sparse arrays.

The lastIndexOf() method is generic. It only expects the this value to have a length property and integer-keyed properties.

Examples

Using lastIndexOf()

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

const 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

You cannot use lastIndexOf() to search for NaN.

const array = [NaN];
array.lastIndexOf(NaN); // -1

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.

const indices = [];
const array = ["a", "b", "a", "c", "a", "d"];
const element = "a";
let 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.

Using lastIndexOf() on sparse arrays

You cannot use lastIndexOf() to search for empty slots in sparse arrays.

console.log([1, , 3].lastIndexOf(undefined)); // -1

Calling lastIndexOf() on non-array objects

The lastIndexOf() method reads the length property of this and then accesses each property whose key is a nonnegative integer less than length.

const arrayLike = {
  length: 3,
  0: 2,
  1: 3,
  2: 2,
  3: 5, // ignored by lastIndexOf() since length is 3
};
console.log(Array.prototype.lastIndexOf.call(arrayLike, 2));
// 2
console.log(Array.prototype.lastIndexOf.call(arrayLike, 5));
// -1