How to get a number of random elements from an array?

Question

I am working on 'how to access elements randomly from an array in javascript'. I found many links regarding this. Like: Get random item from JavaScript array

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

But in this, we can choose only one item from the array. If we want more than one elements then how can we achieve this? How can we get more than one element from an array?

Answer 1

Just two lines :

// Shuffle array
const shuffled = array.sort(() => 0.5 - Math.random());

// Get sub-array of first n elements after shuffled
let selected = shuffled.slice(0, n);

---

DEMO:

Answer 2

Try this non-destructive (and fast) function:

function getRandom(arr, n) {
    var result = new Array(n),
        len = arr.length,
        taken = new Array(len);
    if (n > len)
        throw new RangeError("getRandom: more elements taken than available");
    while (n--) {
        var x = Math.floor(Math.random() * len);
        result[n] = arr[x in taken ? taken[x] : x];
        taken[x] = --len in taken ? taken[len] : len;
    }
    return result;
}

Answer 3

There is a one-liner unique solution here

 array.sort(() => Math.random() - Math.random()).slice(0, n)

Answer 4

lodash _.sample and _.sampleSize.

Gets one or n random elements at unique keys from collection up to the size of collection.

_.sample([1, 2, 3, 4]);
// => 2

_.sampleSize([1, 2, 3], 2);
// => [3, 1]
 
_.sampleSize([1, 2, 3], 3);
// => [2, 3, 1]

Answer 5

create a funcion which does that:

var getMeRandomElements = function(sourceArray, neededElements) {
    var result = [];
    for (var i = 0; i < neededElements; i++) {
        result.push(sourceArray[Math.floor(Math.random()*sourceArray.length)]);
    }
    return result;
}

you should also check if the sourceArray has enough elements to be returned. and if you want unique elements returned, you should remove selected element from the sourceArray.

Answer 6

Porting .sample from the Python standard library:

function sample(population, k){
    /*
        Chooses k unique random elements from a population sequence or set.

        Returns a new list containing elements from the population while
        leaving the original population unchanged.  The resulting list is
        in selection order so that all sub-slices will also be valid random
        samples.  This allows raffle winners (the sample) to be partitioned
        into grand prize and second place winners (the subslices).

        Members of the population need not be hashable or unique.  If the
        population contains repeats, then each occurrence is a possible
        selection in the sample.

        To choose a sample in a range of integers, use range as an argument.
        This is especially fast and space efficient for sampling from a
        large population:   sample(range(10000000), 60)

        Sampling without replacement entails tracking either potential
        selections (the pool) in a list or previous selections in a set.

        When the number of selections is small compared to the
        population, then tracking selections is efficient, requiring
        only a small set and an occasional reselection.  For
        a larger number of selections, the pool tracking method is
        preferred since the list takes less space than the
        set and it doesn't suffer from frequent reselections.
    */

    if(!Array.isArray(population))
        throw new TypeError("Population must be an array.");
    var n = population.length;
    if(k < 0 || k > n)
        throw new RangeError("Sample larger than population or is negative");

    var result = new Array(k);
    var setsize = 21;   // size of a small set minus size of an empty list

    if(k > 5)
        setsize += Math.pow(4, Math.ceil(Math.log(k * 3) / Math.log(4)))

    if(n <= setsize){
        // An n-length list is smaller than a k-length set
        var pool = population.slice();
        for(var i = 0; i < k; i++){          // invariant:  non-selected at [0,n-i)
            var j = Math.random() * (n - i) | 0;
            result[i] = pool[j];
            pool[j] = pool[n - i - 1];       // move non-selected item into vacancy
        }
    }else{
        var selected = new Set();
        for(var i = 0; i < k; i++){
            var j = Math.random() * n | 0;
            while(selected.has(j)){
                j = Math.random() * n | 0;
            }
            selected.add(j);
            result[i] = population[j];
        }
    }

    return result;
}

Implementation ported from Lib/random.py.

Notes:

• setsize is set based on characteristics in Python for efficiency. Although it has not been adjusted for JavaScript, the algorithm will still function as expected.
• Some other answers described in this page are not safe according to the ECMAScript specification due to the misuse of Array.prototype.sort. This algorithm however is guaranteed to terminate in finite time.
• For older browsers that do not have Set implemented, the set can be replaced with an Array and .has(j) replaced with .indexOf(j) > -1.

Performance against the accepted answer:

• https://jsperf.com/pick-random-elements-from-an-array
• The performance difference is the greatest on Safari.

Answer 7

Getting 5 random items without changing the original array:

const n = 5;
const sample = items
  .map(x => ({ x, r: Math.random() }))
  .sort((a, b) => a.r - b.r)
  .map(a => a.x)
  .slice(0, n);

(Don't use this for big lists)

Answer 8

If you want to randomly get items from the array in a loop without repetitions you can remove the selected item from the array with splice:

var items = [1, 2, 3, 4, 5];
var newItems = [];

for (var i = 0; i < 3; i++) {
  var idx = Math.floor(Math.random() * items.length);
  newItems.push(items[idx]);
  items.splice(idx, 1);
}

console.log(newItems);

Answer 9

ES6 syntax

const pickRandom = (arr,count) => {
  let _arr = [...arr];
  return[...Array(count)].map( ()=> _arr.splice(Math.floor(Math.random() * _arr.length), 1)[0] ); 
}

Answer 10

I can't believe that no one didn't mention this method, pretty clean and straightforward.

const getRnd = (a, n) => new Array(n).fill(null).map(() => a[Math.floor(Math.random() * a.length)]);

Answer 11

Array.prototype.getnkill = function() {
    var a = Math.floor(Math.random()*this.length);
    var dead = this[a];
    this.splice(a,1);
    return dead;
}

//.getnkill() removes element in the array 
//so if you like you can keep a copy of the array first:

//var original= items.slice(0); 


var item = items.getnkill();

var anotheritem = items.getnkill();

Answer 12

Here's a nicely typed version. It doesn't fail. Returns a shuffled array if sample size is larger than original array's length.

function sampleArr<T>(arr: T[], size: number): T[] {
  const setOfIndexes = new Set<number>();
  while (setOfIndexes.size < size && setOfIndexes.size < arr.length) {
    setOfIndexes.add(randomIntFromInterval(0, arr.length - 1));
  }
  return Array.from(setOfIndexes.values()).map(i => arr[i]);
}

const randomIntFromInterval = (min: number, max: number): number =>
  Math.floor(Math.random() * (max - min + 1) + min);

Answer 13

In this answer, I want to share with you the test that I have to know the best method that gives equal chances for all elements to have random subarray.

Method 01

array.sort(() => Math.random() - Math.random()).slice(0, n)

using this method, some elements have higher chances comparing with others.

calculateProbability = function(number=0 ,iterations=10000,arraySize=100) { 
let occ = 0 
for (let index = 0; index < iterations; index++) {
   const myArray= Array.from(Array(arraySize).keys()) //=> [0, 1, 2, 3, 4, ... arraySize]
   
  /** Wrong Method */
    const arr = myArray.sort(function() {
     return val= .5 - Math.random();
      });
     
  if(arr[0]===number) {
    occ ++
    }

    
}

console.log("Probability of ",number, " = ",occ*100 /iterations,"%")

}

calculateProbability(0)
calculateProbability(0)
calculateProbability(0)
calculateProbability(50)
calculateProbability(50)
calculateProbability(50)
calculateProbability(25)
calculateProbability(25)
calculateProbability(25)

Method 2

Using this method, the elements have the same probability:

 const arr = myArray
      .map((a) => ({sort: Math.random(), value: a}))
      .sort((a, b) => a.sort - b.sort)
      .map((a) => a.value)

calculateProbability = function(number=0 ,iterations=10000,arraySize=100) { 
let occ = 0 
for (let index = 0; index < iterations; index++) {
   const myArray= Array.from(Array(arraySize).keys()) //=> [0, 1, 2, 3, 4, ... arraySize]
   

  /** Correct Method */
  const arr = myArray
  .map((a) => ({sort: Math.random(), value: a}))
  .sort((a, b) => a.sort - b.sort)
  .map((a) => a.value)
    
  if(arr[0]===number) {
    occ ++
    }

    
}

console.log("Probability of ",number, " = ",occ*100 /iterations,"%")

}

calculateProbability(0)
calculateProbability(0)
calculateProbability(0)
calculateProbability(50)
calculateProbability(50)
calculateProbability(50)
calculateProbability(25)
calculateProbability(25)
calculateProbability(25)

The correct answer is posted in in the following link: https://stackoverflow.com/a/46545530/3811640

Answer 14

2020
non destructive functional programing style, working in a immutable context.

const _randomslice = (ar, size) => {
  let new_ar = [...ar];
  new_ar.splice(Math.floor(Math.random()*ar.length),1);
  return ar.length <= (size+1) ? new_ar : _randomslice(new_ar, size);
}


console.log(_randomslice([1,2,3,4,5],2));

Answer 15

EDIT: This solution is slower than others presented here (which splice the source array) if you want to get only a few elements. The speed of this solution depends only on the number of elements in the original array, while the speed of the splicing solution depends on the number of elements required in the output array.

If you want non-repeating random elements, you can shuffle your array then get only as many as you want:

function shuffle(array) {
    var counter = array.length, temp, index;

    // While there are elements in the array
    while (counter--) {
        // Pick a random index
        index = (Math.random() * counter) | 0;

        // And swap the last element with it
        temp = array[counter];
        array[counter] = array[index];
        array[index] = temp;
    }

    return array;
}

var arr = [0,1,2,3,4,5,7,8,9];

var randoms = shuffle(arr.slice(0)); // array is cloned so it won't be destroyed
randoms.length = 4; // get 4 random elements

DEMO: http://jsbin.com/UHUHuqi/1/edit

Shuffle function taken from here: https://stackoverflow.com/a/6274398/1669279

Answer 16

I needed a function to solve this kind of issue so I'm sharing it here.

    const getRandomItem = function(arr) {
        return arr[Math.floor(Math.random() * arr.length)];
    }

    // original array
    let arr = [4, 3, 1, 6, 9, 8, 5];

    // number of random elements to get from arr
    let n = 4;

    let count = 0;
    // new array to push random item in
    let randomItems = []
    do {
        let item = getRandomItem(arr);
        randomItems.push(item);
        // update the original array and remove the recently pushed item
        arr.splice(arr.indexOf(item), 1);
        count++;
    } while(count < n);

    console.log(randomItems);
    console.log(arr);

Note: if n = arr.length then basically you're shuffling the array arr and randomItems returns that shuffled array.

Demo

Answer 17

Here's an optimized version of the code ported from Python by @Derek, with the added destructive (in-place) option that makes it the fastest algorithm possible if you can go with it. Otherwise it either makes a full copy or, for a small number of items requested from a large array, switches to a selection-based algorithm.

// Chooses k unique random elements from pool.
function sample(pool, k, destructive) {
    var n = pool.length;
    
    if (k < 0 || k > n)
        throw new RangeError("Sample larger than population or is negative");
    
    if (destructive || n <= (k <= 5 ? 21 : 21 + Math.pow(4, Math.ceil(Math.log(k*3) / Math.log(4))))) {
        if (!destructive)
            pool = Array.prototype.slice.call(pool);
        for (var i = 0; i < k; i++) { // invariant: non-selected at [i,n)
            var j = i + Math.random() * (n - i) | 0;
            var x = pool[i];
            pool[i] = pool[j];
            pool[j] = x;
        }
        pool.length = k; // truncate
        return pool;
    } else {
        var selected = new Set();
        while (selected.add(Math.random() * n | 0).size < k) {}
        return Array.prototype.map.call(selected, i => pool[i]);
    }
}

In comparison to Derek's implementation, the first algorithm is much faster in Firefox while being a bit slower in Chrome, although now it has the destructive option - the most performant one. The second algorithm is simply 5-15% faster. I try not to give any concrete numbers since they vary depending on k and n and probably won't mean anything in the future with the new browser versions.

The heuristic that makes the choice between algorithms originates from Python code. I've left it as is, although it sometimes selects the slower one. It should be optimized for JS, but it's a complex task since the performance of corner cases is browser- and their version-dependent. For example, when you try to select 20 out of 1000 or 1050, it will switch to the first or the second algorithm accordingly. In this case the first one runs 2x faster than the second one in Chrome 80 but 3x slower in Firefox 74.

Answer 18

Sampling with possible duplicates:

const sample_with_duplicates = Array(sample_size).fill().map(() => items[~~(Math.random() * items.length)])

Sampling without duplicates:

const sample_without_duplicates = [...Array(items.length).keys()].sort(() => 0.5 - Math.random()).slice(0, sample_size).map(index => items[index]);

Since without duplicates requires sorting the whole index array first, it is considerably slow than with possible duplicates for big items input arrays.

Obviously, the max size of without duplicates is <= items.length

Check this fiddle: https://jsfiddle.net/doleron/5zw2vequ/30/

Answer 19

It extracts random elements from srcArray one by one while it get's enough or there is no more elements in srcArray left for extracting. Fast and reliable.

function getNRandomValuesFromArray(srcArr, n) {
    // making copy to do not affect original srcArray
    srcArr = srcArr.slice();
    resultArr = [];
    // while srcArray isn't empty AND we didn't enough random elements
    while (srcArr.length && resultArr.length < n) {
        // remove one element from random position and add this element to the result array
        resultArr = resultArr.concat( // merge arrays
            srcArr.splice( // extract one random element
                Math.floor(Math.random() * srcArr.length),
                1
            )
        );
    }

    return resultArr;
}

Answer 20

Here's a function I use that allows you to easily sample an array with or without replacement:

  // Returns a random sample (either with or without replacement) from an array
  const randomSample = (arr, k, withReplacement = false) => {
    let sample;
    if (withReplacement === true) {  // sample with replacement
      sample = Array.from({length: k}, () => arr[Math.floor(Math.random() *  arr.length)]);
    } else { // sample without replacement
      if (k > arr.length) {
        throw new RangeError('Sample size must be less than or equal to array length         when sampling without replacement.')
      }
      sample = arr.map(a => [a, Math.random()]).sort((a, b) => {
        return a[1] < b[1] ? -1 : 1;}).slice(0, k).map(a => a[0]); 
      };
    return sample;
  };

Using it is simple:

Without Replacement (default behavior)

randomSample([1, 2, 3], 2) may return [2, 1]

With Replacement

randomSample([1, 2, 3, 4, 5, 6], 4) may return [2, 3, 3, 2]

Answer 21

var getRandomElements = function(sourceArray, requiredLength) {
    var result = [];
    while(result.length<requiredLength){
        random = Math.floor(Math.random()*sourceArray.length);
        if(result.indexOf(sourceArray[random])==-1){
            result.push(sourceArray[random]);
        }
    }
    return result;
}

Answer 22

Here is the most correct answer and it will give you Random + Unique elements.

function randomize(array, n)
{
    var final = [];
    array = array.filter(function(elem, index, self) {
        return index == self.indexOf(elem);
    }).sort(function() { return 0.5 - Math.random() });

    var len = array.length,
    n = n > len ? len : n;

    for(var i = 0; i < n; i ++)
    {
        final[i] = array[i];
    }

    return final;
}

// randomize([1,2,3,4,5,3,2], 4);
// Result: [1, 2, 3, 5] // Something like this

Answer 23

2019

This is same as Laurynas Mališauskas answer, just that the elements are unique (no duplicates).

var getMeRandomElements = function(sourceArray, neededElements) {
    var result = [];
    for (var i = 0; i < neededElements; i++) {
    var index = Math.floor(Math.random() * sourceArray.length);
        result.push(sourceArray[index]);
        sourceArray.splice(index, 1);
    }
    return result;
}

Now to answer original question "How to get multiple random elements by jQuery", here you go:

var getMeRandomElements = function(sourceArray, neededElements) {
    var result = [];
    for (var i = 0; i < neededElements; i++) {
    var index = Math.floor(Math.random() * sourceArray.length);
        result.push(sourceArray[index]);
        sourceArray.splice(index, 1);
    }
    return result;
}

var $set = $('.someClass');// <<<<< change this please

var allIndexes = [];
for(var i = 0; i < $set.length; ++i) {
    allIndexes.push(i);
}

var totalRandom = 4;// <<<<< change this please
var randomIndexes = getMeRandomElements(allIndexes, totalRandom);

var $randomElements = null;
for(var i = 0; i < randomIndexes.length; ++i) {
    var randomIndex = randomIndexes[i];
    if($randomElements === null) {
        $randomElements = $set.eq(randomIndex);
    } else {
        $randomElements.add($set.eq(randomIndex));
    }
}

// $randomElements is ready
$randomElements.css('backgroundColor', 'red');

Answer 24

items.sort(() => (Math.random() > 0.5 ? 1 : -1)).slice(0, count);