searchElement

Element to locate in the array.

fromIndexOptional

Zero-based index at which to start searching backwards,converted to an integer.

• Negative index counts back from the end of the array — if-array.length <= fromIndex < 0,fromIndex + array.length is used.
• IffromIndex < -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, sosearchElement is never found.
• IffromIndex >= array.length orfromIndex is omitted orundefined,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.

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

ThelastIndexOf() method comparessearchElement to elements of the array usingstrict equality (the same algorithm used by the=== operator).NaN values are never compared as equal, solastIndexOf() always returns-1 whensearchElement isNaN.

ThelastIndexOf() method skips empty slots insparse arrays.

ThelastIndexOf() method isgeneric. It only expects thethis value to have alength property and integer-keyed properties.

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

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

You cannot uselastIndexOf() to search forNaN.

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

The following example useslastIndexOf to find all the indices of anelement in a given array, usingpush() to add themto 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 caseidx === 0 separately here because theelement will always be found regardless of thefromIndex parameter if it isthe first element of the array. This is different from theindexOf() method.

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

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

ThelastIndexOf() method reads thelength property ofthis and then accesses each property whose key is a nonnegative integer less thanlength.

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));// 2console.log(Array.prototype.lastIndexOf.call(arrayLike, 5));// -1

• Polyfill ofArray.prototype.lastIndexOf incore-js
• es-shims polyfill ofArray.prototype.lastIndexOf
• Indexed collections guide
• Array
• Array.prototype.findIndex()
• Array.prototype.findLastIndex()
• Array.prototype.indexOf()
• TypedArray.prototype.lastIndexOf()
• String.prototype.lastIndexOf()