Get the Last Item in an Array

Get the Last Item in an Array

Here is my JavaScript code so far:

var linkElement = document.getElementById("BackButton");
var loc_array = document.location.href.split('/');
var newT = document.createTextNode(unescape(capWords(loc_array[loc_array.length-2]))); 
linkElement.appendChild(newT);

Currently it takes the second to last item in the array from the URL. However, I want to do a check for the last item in the array to be "index.html" and if so, grab the third to last item instead.

59 Answers

if (loc_array[loc_array.length - 1] === 'index.html') {
   // do something
} else {
   // something else
}

In the event that your server serves the same file for "index.html" and "inDEX.htML" you can also use: .toLowerCase().

Though, you might want to consider doing this server-side if possible: it will be cleaner and work for people without JS.


EDIT - ES-2022

Using ES-2022 Array.at(), the above may be written like this:

if (loc_array.at(-1) === 'index.html') {
   // do something
} else {
   // something else
}
9

Not sure if there's a drawback, but this seems quite concise:

arr.slice(-1)[0] 

or

arr.slice(-1).pop()

Both will return undefined if the array is empty.

6

Use Array.pop:

var lastItem = anArray.pop();

Important : This returns the last element and removes it from the array

3

A shorter version of what @chaiguy posted:

Array.prototype.last = function() {
    return this[this.length - 1];
}

Reading the -1 index returns undefined already.

EDIT:

These days the preference seems to be using modules and to avoid touching the prototype or using a global namespace.

export function last(array) {
    return array[array.length - 1];
}
8

Two options are:

var last = arr[arr.length - 1]

or

var last = arr.slice(-1)[0]

The former is faster, but the latter looks nicer

2

Performance

Today 2020.05.16 I perform tests of chosen solutions on Chrome v81.0, Safari v13.1 and Firefox v76.0 on MacOs High Sierra v10.13.6

Conclusions

  • arr[arr.length-1] (D) is recommended as fastest cross-browser solution
  • mutable solution arr.pop() (A) and immutable _.last(arr) (L) are fast
  • solutions I, J are slow for long strings
  • solutions H, K (jQuery) are slowest on all browsers

Details

I test two cases for solutions:

  • mutable: A, B, C,

  • immutable: D, E, F, G, H, I, J (my),

  • immutable from external libraries: K, L, M,

for two cases

  • short string - 10 characters - you can run test HERE
  • long string - 1M characters - you can run test HERE
function A(arr) {
  return arr.pop();
}

function B(arr) {  
  return arr.splice(-1,1);
}

function C(arr) {  
  return arr.reverse()[0]
}

function D(arr) {
  return arr[arr.length - 1];
}

function E(arr) {
  return arr.slice(-1)[0] ;
}

function F(arr) {
  let [last] = arr.slice(-1);
  return last;
}

function G(arr) {
  return arr.slice(-1).pop();
}

function H(arr) {
  return [...arr].pop();
}

function I(arr) {  
  return arr.reduceRight(a => a);
}

function J(arr) {  
  return arr.find((e,i,a)=> a.length==i+1);
}

function K(arr) {  
  return $(arr).get(-1);
}

function L(arr) {  
  return _.last(arr);
}

function M(arr) {  
  return _.nth(arr, -1);
}






// ----------
// TEST
// ----------

let loc_array=["domain","a","b","c","d","e","f","g","h","file"];

log = (f)=> console.log(`${f.name}: ${f([...loc_array])}`);

[A,B,C,D,E,F,G,H,I,J,K,L,M].forEach(f=> log(f));
<script src=""></script>
<script src="" integrity="sha256-VeNaFBVDhoX3H+gJ37DpT/nTuZTdjYro9yBruHjVmoQ=" crossorigin="anonymous"></script>

Example results for Chrome for short string

2

Here's how to get it with no effect on the original ARRAY

a = [1,2,5,6,1,874,98,"abc"];
a.length; //returns 8 elements

If you use pop(), it will modify your array

a.pop();  // will return "abc" AND REMOVES IT from the array 
a.length; // returns 7

But you can use this so it has no effect on the original array:

a.slice(-1).pop(); // will return "abc" won't do modify the array 
                   // because slice creates a new array object 
a.length;          // returns 8; no modification and you've got you last element 
2

The "cleanest" ES6 way (IMO) would be:

const foo = [1,2,3,4];
const bar = [...foo].pop();

This avoids mutating foo, as .pop() would had, if we didn't used the spread operator.
That said, I like aswell the foo.slice(-1)[0] solution.

4
const [y] = x.slice(-1)

Quick Explanation:

This syntax [y] = <array/object> is called destructuring assignment & according to Mozilla docs, the destructuring assingment makes possible to unpack values from an array or properties from an object into distinct variables

Read more about it: here

2

const [lastItem] = array.slice(-1);

Array.prototype.slice with -1 can be used to create a new Array containing only the last item of the original Array, you can then use Destructuring Assignment to create a variable using the first item of that new Array.

const lotteryNumbers = [12, 16, 4, 33, 41, 22];
const [lastNumber] = lotteryNumbers.slice(-1);

console.log(lotteryNumbers.slice(-1));
// => [22]
console.log(lastNumber);
// => 22
3

I'd rather use array.pop() than indexes.

while(loc_array.pop()!= "index.html"){
}
var newT = document.createTextNode(unescape(capWords(loc_array[loc_array.length])));

this way you always get the element previous to index.html (providing your array has isolated index.html as one item). Note: You'll lose the last elements from the array, though.

0

You can use relative indexing with Array#at:

const myArray = [1, 2, 3]

console.log(myArray.at(-1))
// => 3
6
const lastElement = myArray[myArray.length - 1];

This is the best options from performance point of view (~1000 times faster than arr.slice(-1)).

1

You can use this pattern...

let [last] = arr.slice(-1);

While it reads rather nicely, keep in mind it creates a new array so it's less efficient than other solutions but it'll almost never be the performance bottleneck of your application.

0

If one wants to get the last element in one go, he/she may use Array#splice():

lastElement = document.location.href.split('/').splice(-1,1);

Here, there is no need to store the split elements in an array, and then get to the last element. If getting last element is the only objective, this should be used.

Note: This changes the original array by removing its last element. Think of splice(-1,1) as a pop() function that pops the last element.

3

Multiple ways to find last value of an array in javascript

  • Without affecting original array
var arr = [1,2,3,4,5];

console.log(arr.slice(-1)[0])
console.log(arr[arr.length-1])
const [last] = [...arr].reverse();
console.log(last)

let copyArr = [...arr];
console.log(copyArr.reverse()[0]);
  • Modifies original array
var arr = [1,2,3,4,5];

console.log(arr.pop())
arr.push(5)
console.log(...arr.splice(-1));
  • By creating own helper method
let arr = [1, 2, 3, 4, 5];

Object.defineProperty(arr, 'last', 
{ get: function(){
  return this[this.length-1];
 }
})

console.log(arr.last);
1

Getting the last item of an array can be achieved by using the slice method with negative values.

You can read more about it here at the bottom.

var fileName = loc_array.slice(-1)[0];
if(fileName.toLowerCase() == "index.html")
{
  //your code...
}

Using pop() will change your array, which is not always a good idea.

2

This question has been around a long time, so I'm surprised that no one mentioned just putting the last element back on after a pop().

arr.pop() is exactly as efficient as arr[arr.length-1], and both are the same speed as arr.push().

Therefore, you can get away with:

---EDITED [check that thePop isn't undefined before pushing]---

let thePop = arr.pop()
thePop && arr.push(thePop)

---END EDIT---

Which can be reduced to this (same speed [EDIT: but unsafe!]):

arr.push(thePop = arr.pop())    //Unsafe if arr empty

This is twice as slow as arr[arr.length-1], but you don't have to stuff around with an index. That's worth gold on any day.

Of the solutions I've tried, and in multiples of the Execution Time Unit (ETU) of arr[arr.length-1]:

[Method]..............[ETUs 5 elems]...[ETU 1 million elems]

arr[arr.length - 1]      ------> 1              -----> 1

let myPop = arr.pop()
arr.push(myPop)          ------> 2              -----> 2

arr.slice(-1).pop()      ------> 36             -----> 924  

arr.slice(-1)[0]         ------> 36             -----> 924  

[...arr].pop()           ------> 120            -----> ~21,000,000 :)

The last three options, ESPECIALLY [...arr].pop(), get VERY much worse as the size of the array increases. On a machine without the memory limitations of my machine, [...arr].pop() probably maintains something like it's 120:1 ratio. Still, no one likes a resource hog.

1

Just putting another option here.

loc_array.splice(-1)[0] === 'index.html'

I found the above approach more clean and short onliner. Please, free feel to try this one.

Note: It will modify the original array, if you don't want to modify it you can use slice()

loc_array.slice(-1)[0] === 'index.html'

Thanks @VinayPai for pointing this out.

4

ES6 object destructuring is another way to go.

const {length, [length-1]: last}=[1,2,3,4,5]
console.log(last)

You extract length property from Array using object destructuring. You create another dynamic key using already extracted key by [length-1] and assign it to last, all in one line.

3

For those not afraid to overload the Array prototype (and with enumeration masking you shouldn't be):

Object.defineProperty( Array.prototype, "getLast", {
    enumerable: false,
    configurable: false,
    writable: false,
    value: function() {
        return this[ this.length - 1 ];
    }
} );
0

Here's more Javascript art if you came here looking for it

In the spirit of another answer that used reduceRight(), but shorter:

[3, 2, 1, 5].reduceRight(a => a);

It relies on the fact that, in case you don't provide an initial value, the very last element is selected as the initial one (check the docs here). Since the callback just keeps returning the initial value, the last element will be the one being returned in the end.

Beware that this should be considered Javascript art and is by no means the way I would recommend doing it, mostly because it runs in O(n) time, but also because it hurts readability.

Sarah Jenkins
Author

Sarah Jenkins

Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.