Selecting last element in JavaScript array
JavascriptArraysGoogle MapsGoogle Maps-MarkersJavascript Problem Overview
I'm making an application that updates a user's location and path in real time and displays this on a Google Map. I have functionality that allows multiple users to be tracked at the same time using an object, which is updated every second.
Right now, when a user pressed a button in the Android app, the coordinates are sent to a database and each time the location changes, a marker is updated on the map (and a polyline is formed).
Since I have multiple users, I send a unique and randomly generated alphanumeric string so that I can display an individual path for each user. When the JS pulls this data from the database, it checks if the user exists, if it does not, it creates a new key with the value being a list. It would look something like this:
loc = {f096012e-2497-485d-8adb-7ec0b9352c52: [new google.maps.LatLng(39, -86),
new google.maps.LatLng(38, -87),
new google.maps.LatLng(37, -88)],
44ed0662-1a9e-4c0e-9920-106258dcc3e7: [new google.maps.LatLng(40, -83),
new google.maps.LatLng(41, -82),
new google.maps.LatLng(42, -81)]}
What I'm doing is storing a list of coordinates as the value of the key, which is the user's ID. My program keeps updating this list each time the location is changed by adding to the list (this works properly).
What I need to do is update the marker's location each time the location changes. I would like to do this by selecting the last item in the array since that would be the last known location. Right now, each time the location is changed a new marker is added to the map (each one of the points in the example would show a marker at that location) so markers continue to be added.
I would use a ´for (x in loc)` statement each time the location updates to grab the last location from the list and use that to update the marker. How do I select this last element from the array within the hash?
Javascript Solutions
Solution 1 - Javascript
How to access last element of an array
It looks like that:
var my_array = /* some array here */;
var last_element = my_array[my_array.length - 1];
Which in your case looks like this:
var array1 = loc['f096012e-2497-485d-8adb-7ec0b9352c52'];
var last_element = array1[array1.length - 1];
or, in longer version, without creating new variables:
loc['f096012e-2497-485d-8adb-7ec0b9352c52'][loc['f096012e-2497-485d-8adb-7ec0b9352c52'].length - 1];
How to add a method for getting it simpler
If you are a fan for creating functions/shortcuts to fulfill such tasks, the following code:
if (!Array.prototype.last){
Array.prototype.last = function(){
return this[this.length - 1];
};
};
will allow you to get the last element of an array by invoking array's last()
method, in your case eg.:
loc['f096012e-2497-485d-8adb-7ec0b9352c52'].last();
You can check that it works here: http://jsfiddle.net/D4NRN/
Solution 2 - Javascript
Use the slice()
method:
my_array.slice(-1)[0]
Solution 3 - Javascript
You can also .pop
off the last element. Be careful, this will change the value of the array, but that might be OK for you.
var a = [1,2,3];
a.pop(); // 3
a // [1,2]
Solution 4 - Javascript
use es6 deconstruction array with the spread operator
var last = [...yourArray].pop();
note that yourArray doesn't change.
Solution 5 - Javascript
var arr = [1, 2, 3];
arr.slice(-1).pop(); // return 3 and arr = [1, 2, 3]
This will return undefined if the array is empty and this will not change the value of the array.
Solution 6 - Javascript
var last = array.slice(-1)[0];
I find slice at -1 useful for getting the last element (especially of an array of unknown length) and the performance is much better than calculating the length less 1.
Performance of the various methods for selecting last array element
Solution 7 - Javascript
Underscore and Lodash have the _.last(Array)
method, that returns the last element in an Array. They both work about the same
_.last([5, 4, 3, 2, 1]);
=> 1
Ramda also has a _.last
function
R.last(['fi', 'fo', 'fum']); //=> 'fum'
Solution 8 - Javascript
This worked:
array.reverse()[0]
Solution 9 - Javascript
You can define a getter on Array.prototype
:
if (!Array.prototype.hasOwnProperty("last")) {
Object.defineProperty(Array.prototype, "last", {
get() {
return this[this.length - 1];
}
});
}
console.log([9, 8, 7, 6].last); // => 6
As you can see, access doesn't look like a function call; the getter function is called internally.
Solution 10 - Javascript
In case you are using ES6 you can do:
const arr = [ 1, 2, 3 ];
[ ...arr ].pop(); // 3
arr; // [ 1, 2, 3 ] (wasn't changed)
Solution 11 - Javascript
So, a lot of people are answering with pop(), but most of them don't seem to realize that's a destructive method.
var a = [1,2,3]
a.pop()
//3
//a is now [1,2]
So, for a really silly, nondestructive method:
var a = [1,2,3]
a[a.push(a.pop())-1]
//3
a push pop, like in the 90s :)
push appends a value to the end of an array, and returns the length of the result. so
d=[]
d.push('life')
//=> 1
d
//=>['life']
pop returns the value of the last item of an array, prior to it removing that value at that index. so
c = [1,2,1]
c.pop()
//=> 1
c
//=> [1,2]
arrays are 0 indexed, so c.length => 3, c[c.length] => undefined (because you're looking for the 4th value if you do that(this level of depth is for any hapless newbs that end up here)).
Probably not the best, or even a good method for your application, what with traffic, churn, blah. but for traversing down an array, streaming it onto another, just being silly with inefficient methods, this. Totally this.
Solution 12 - Javascript
Use JavaScript objects if this is critical to your application. You shouldn't be using raw primitives to manage critical parts of your application. As this seems to be the core of your application, you should use objects instead. I've written some code below to help get you started. The method lastLocation
would return the last location.
function User(id) {
this.id = id;
this.locations = [];
this.getId = function() {
return this.id;
};
this.addLocation = function(latitude, longitude) {
this.locations[this.locations.length] = new google.maps.LatLng(latitude, longitude);
};
this.lastLocation = function() {
return this.locations[this.locations.length - 1];
};
this.removeLastLocation = function() {
return this.locations.pop();
};
}
function Users() {
this.users = {};
this.generateId = function() {
return Math.random();
};
this.createUser = function() {
var id = this.generateId();
this.users[id] = new User(id);
return this.users[id];
};
this.getUser = function(id) {
return this.users[id];
};
this.removeUser = function(id) {
var user = this.getUser(id);
delete this.users[id];
return user;
};
}
var users = new Users();
var user = users.createUser();
user.addLocation(0, 0);
user.addLocation(0, 1);
Solution 13 - Javascript
var last = function( obj, key ) {
var a = obj[key];
return a[a.length - 1];
};
last(loc, 'f096012e-2497-485d-8adb-7ec0b9352c52');