Get a random item from a JavaScript array [duplicate]

This question already has answers here: Getting a random value from a JavaScript开发者_开发问答 array (28 answers) Closed 7 years ago.

var items = Array(523, 3452, 334, 31, ..., 5346);

How do I get random item from items?

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var item = items[Math.floor(Math.random()*items.length)];

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1. solution: define Array prototype

Array.prototype.random = function () {
  return this[Math.floor((Math.random()*this.length))];
}

that will work on inline arrays

[2,3,5].random()

and of course predefined arrays

var list = [2,3,5]
list.random()

2. solution: define custom function that accepts list and returns element

function get_random (list) {
  return list[Math.floor((Math.random()*list.length))];
}


get_random([2,3,5])

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Use underscore (or loDash :)):

var randomArray = [
   '#cc0000','#00cc00', '#0000cc'
];

// use _.sample
var randomElement = _.sample(randomArray);

// manually use _.random
var randomElement = randomArray[_.random(randomArray.length-1)];

Or to shuffle an entire array:

// use underscore's shuffle function
var firstRandomElement = _.shuffle(randomArray)[0];

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If you really must use jQuery to solve this problem (NB: you shouldn't):

(function($) {
    $.rand = function(arg) {
        if ($.isArray(arg)) {
            return arg[$.rand(arg.length)];
        } else if (typeof arg === "number") {
            return Math.floor(Math.random() * arg);
        } else {
            return 4;  // chosen by fair dice roll
        }
    };
})(jQuery);

var items = [523, 3452, 334, 31, ..., 5346];
var item = jQuery.rand(items);

This plugin will return a random element if given an array, or a value from [0 .. n) given a number, or given anything else, a guaranteed random value!

For extra fun, the array return is generated by calling the function recursively based on the array's length :)

Working demo at http://jsfiddle.net/2eyQX/

---

Here's yet another way:

function rand(items) {
    // "~~" for a closest "int"
    return items[~~(items.length * Math.random())];
}

Or as recommended below by @1248177:

function rand(items) {
    // "|" for a kinda "int div"
    return items[items.length * Math.random() | 0];
}

---

var random = items[Math.floor(Math.random()*items.length)]

---

jQuery is JavaScript! It's just a JavaScript framework. So to find a random item, just use plain old JavaScript, for example,

var randomItem = items[Math.floor(Math.random()*items.length)]

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// 1. Random shuffle items
items.sort(function() {return 0.5 - Math.random()})

// 2. Get first item
var item = items[0]

Shorter:

var item = items.sort(function() {return 0.5 - Math.random()})[0];

Even shoter (by José dB.):

let item = items.sort(() => 0.5 - Math.random())[0];

---

var rndval=items[Math.floor(Math.random()*items.length)];

---

var items = Array(523,3452,334,31,...5346);

function rand(min, max) {
  var offset = min;
  var range = (max - min) + 1;

  var randomNumber = Math.floor( Math.random() * range) + offset;
  return randomNumber;
}


randomNumber = rand(0, items.length - 1);

randomItem = items[randomNumber];

credit:

Javascript Function: Random Number Generator

---

If you are using node.js, you can use unique-random-array. It simply picks something random from an array.

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An alternate way would be to add a method to the Array prototype:

 Array.prototype.random = function (length) {
       return this[Math.floor((Math.random()*length))];
 }

 var teams = ['patriots', 'colts', 'jets', 'texans', 'ravens', 'broncos']
 var chosen_team = teams.random(teams.length)
 alert(chosen_team)

---

const ArrayRandomModule = {
  // get random item from array
  random: function (array) {
    return array[Math.random() * array.length | 0];
  },

  // [mutate]: extract from given array a random item
  pick: function (array, i) {
    return array.splice(i >= 0 ? i : Math.random() * array.length | 0, 1)[0];
  },

  // [mutate]: shuffle the given array
  shuffle: function (array) {
    for (var i = array.length; i > 0; --i)
      array.push(array.splice(Math.random() * i | 0, 1)[0]);
    return array;
  }
}