I have an array like this:
var arr1 = ["a", "b", "c", "d"];
How can I randomize / shuffle it?
arr1.sort(() => (Math.random() > .5) ? 1 : -1);
- anyone a.sort(() => Math.random() - 0.5)
- anyone The de-facto unbiased shuffle algorithm is the Fisher-Yates (aka Knuth) Shuffle.
You can see a great visualization here (and the original post linked to this)
function shuffle(array) {
let currentIndex = array.length, randomIndex;
// While there remain elements to shuffle.
while (currentIndex > 0) {
// Pick a remaining element.
randomIndex = Math.floor(Math.random() * currentIndex);
currentIndex--;
// And swap it with the current element.
[array[currentIndex], array[randomIndex]] = [
array[randomIndex], array[currentIndex]];
}
return array;
}
// Used like so
var arr = [2, 11, 37, 42];
shuffle(arr);
console.log(arr);
Some more info about the algorithm used.
Answered 2023-09-20 21:00:48
i--
not --i
. Also, the test if (i==0)...
is superfluous since if i == 0
the while loop will never be entered. The call to Math.floor
can be done faster using ...| 0
. Either tempi or tempj can be removed and the value be directly assigned to myArray[i] or j as appropriate. - anyone for
loop, incorrectly using !=
with !==
, the infinite loop if passed an empty array, and the modification and return of a parameter. - anyone Here's a JavaScript implementation of the Durstenfeld shuffle, an optimized version of Fisher-Yates:
/* Randomize array in-place using Durstenfeld shuffle algorithm */
function shuffleArray(array) {
for (var i = array.length - 1; i > 0; i--) {
var j = Math.floor(Math.random() * (i + 1));
var temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
It picks a random element for each original array element, and excludes it from the next draw, like picking randomly from a deck of cards.
This clever exclusion swaps the picked element with the current one, then picks the next random element from the remainder, looping backwards for optimal efficiency, ensuring the random pick is simplified (it can always start at 0), and thereby skipping the final element.
Algorithm runtime is O(n)
. Note that the shuffle is done in-place so if you don't want to modify the original array, first make a copy of it with .slice(0)
.
The new ES6 allows us to assign two variables at once. This is especially handy when we want to swap the values of two variables, as we can do it in one line of code. Here is a shorter form of the same function, using this feature.
function shuffleArray(array) {
for (let i = array.length - 1; i > 0; i--) {
const j = Math.floor(Math.random() * (i + 1));
[array[i], array[j]] = [array[j], array[i]];
}
}
Answered 2023-09-20 21:00:48
Math.random() should not be multiplied with the loop counter + 1, but with
array.lengt()`. See Generating random whole numbers in JavaScript in a specific range? for a very comprehensive explanation. - anyone return array;
? - anyone return array
will only allow for easier chaining with other . - anyone You can do it easily with map and sort:
let unshuffled = ['hello', 'a', 't', 'q', 1, 2, 3, {cats: true}]
let shuffled = unshuffled
.map(value => ({ value, sort: Math.random() }))
.sort((a, b) => a.sort - b.sort)
.map(({ value }) => value)
console.log(shuffled)
You can shuffle polymorphic arrays, and the sort is as random as Math.random, which is good enough for most purposes.
Since the elements are sorted against consistent keys that are not regenerated each iteration, and each comparison pulls from the same distribution, any non-randomness in the distribution of Math.random is canceled out.
Speed
Time complexity is O(N log N), same as quick sort. Space complexity is O(N). This is not as efficient as a Fischer Yates shuffle but, in my opinion, the code is significantly shorter and more functional. If you have a large array you should certainly use Fischer Yates. If you have a small array with a few hundred items, you might do this.
Answered 2023-09-20 21:00:48
.sort
algorithm - anyone Warning!
The use of this algorithm is not recommended, because it is inefficient and strongly biased; see comments. It is being left here for future reference, because the idea is not that rare.
[1,2,3,4,5,6].sort( () => .5 - Math.random() );
This https://javascript.info/array-methods#shuffle-an-array tutorial explains the differences straightforwardly.
Answered 2023-09-20 21:00:48
Use the underscore.js library. The method _.shuffle()
is nice for this case.
Here is an example with the method:
var _ = require("underscore");
var arr = [1,2,3,4,5,6];
// Testing _.shuffle
var testShuffle = function () {
var indexOne = 0;
var stObj = {
'0': 0,
'1': 1,
'2': 2,
'3': 3,
'4': 4,
'5': 5
};
for (var i = 0; i < 1000; i++) {
arr = _.shuffle(arr);
indexOne = _.indexOf(arr, 1);
stObj[indexOne] ++;
}
console.log(stObj);
};
testShuffle();
Answered 2023-09-20 21:00:48
One could (but should NOT) use it as a protoype from Array:
From ChristopheD:
Array.prototype.shuffle = function() {
var i = this.length, j, temp;
if ( i == 0 ) return this;
while ( --i ) {
j = Math.floor( Math.random() * ( i + 1 ) );
temp = this[i];
this[i] = this[j];
this[j] = temp;
}
return this;
}
Answered 2023-09-20 21:00:48
NEW!
Shorter & probably *faster Fisher-Yates shuffle algorithm
function fy(a,b,c,d){//array,placeholder,placeholder,placeholder
c=a.length;while(c)b=Math.random()*(--c+1)|0,d=a[c],a[c]=a[b],a[b]=d
}
script size (with fy as function name): 90bytes
DEMO http://jsfiddle.net/vvpoma8w/
*faster probably on all browsers except chrome.
If you have any questions just ask.
EDIT
yes it is faster
PERFORMANCE: http://jsperf.com/fyshuffle
using the top voted functions.
EDIT There was a calculation in excess (don't need --c+1) and noone noticed
shorter(4bytes)&faster(test it!).
function fy(a,b,c,d){//array,placeholder,placeholder,placeholder
c=a.length;while(c)b=Math.random()*c--|0,d=a[c],a[c]=a[b],a[b]=d
}
Caching somewhere else var rnd=Math.random
and then use rnd()
would also increase slightly the performance on big arrays.
http://jsfiddle.net/vvpoma8w/2/
Readable version (use the original version. this is slower, vars are useless, like the closures & ";", the code itself is also shorter ... maybe read this How to 'minify' Javascript code , btw you are not able to compress the following code in a javascript minifiers like the above one.)
function fisherYates( array ){
var count = array.length,
randomnumber,
temp;
while( count ){
randomnumber = Math.random() * count-- | 0;
temp = array[count];
array[count] = array[randomnumber];
array[randomnumber] = temp
}
}
Answered 2023-09-20 21:00:48
fy
and shuffle prototype
, I get fy
consistently at the bottom in Chrome 37 on OS X 10.9.5 (81% slower ~20k ops compared to ~100k) and Safari 7.1 it's up to ~8% slower. YMMV, but it's not always faster. jsperf.com/fyshuffle/3 - anyone Shuffle Array In place
function shuffleArr (array){
for (var i = array.length - 1; i > 0; i--) {
var rand = Math.floor(Math.random() * (i + 1));
[array[i], array[rand]] = [array[rand], array[i]]
}
}
ES6 Pure, Iterative
const getShuffledArr = arr => {
const newArr = arr.slice()
for (let i = newArr.length - 1; i > 0; i--) {
const rand = Math.floor(Math.random() * (i + 1));
[newArr[i], newArr[rand]] = [newArr[rand], newArr[i]];
}
return newArr
};
Reliability and Performance Test
Some solutions on this page aren't reliable (they only partially randomise the array). Other solutions are significantly less efficient. With testShuffleArrayFun
(see below) we can test array shuffling functions for reliability and performance.
function testShuffleArrayFun(getShuffledArrayFun){
const arr = [0,1,2,3,4,5,6,7,8,9]
var countArr = arr.map(el=>{
return arr.map(
el=> 0
)
}) // For each possible position in the shuffledArr and for
// each possible value, we'll create a counter.
const t0 = performance.now()
const n = 1000000
for (var i=0 ; i<n ; i++){
// We'll call getShuffledArrayFun n times.
// And for each iteration, we'll increment the counter.
var shuffledArr = getShuffledArrayFun(arr)
shuffledArr.forEach(
(value,key)=>{countArr[key][value]++}
)
}
const t1 = performance.now()
console.log(`Count Values in position`)
console.table(countArr)
const frequencyArr = countArr.map( positionArr => (
positionArr.map(
count => count/n
)
))
console.log("Frequency of value in position")
console.table(frequencyArr)
console.log(`total time: ${t1-t0}`)
}
Other solutions just for fun.
ES6 Pure, Recursive
const getShuffledArr = arr => {
if (arr.length === 1) {return arr};
const rand = Math.floor(Math.random() * arr.length);
return [arr[rand], ...getShuffledArr(arr.filter((_, i) => i != rand))];
};
ES6 Pure using array.map
function getShuffledArr (arr){
return [...arr].map( (_, i, arrCopy) => {
var rand = i + ( Math.floor( Math.random() * (arrCopy.length - i) ) );
[arrCopy[rand], arrCopy[i]] = [arrCopy[i], arrCopy[rand]]
return arrCopy[i]
})
}
ES6 Pure using array.reduce
function getShuffledArr (arr){
return arr.reduce(
(newArr, _, i) => {
var rand = i + ( Math.floor( Math.random() * (newArr.length - i) ) );
[newArr[rand], newArr[i]] = [newArr[i], newArr[rand]]
return newArr
}, [...arr]
)
}
Answered 2023-09-20 21:00:48
[array[i], array[rand]]=[array[rand], array[i]]
? Maybe you can outline how that works. Why do you choose to iterate downwards? - anyone Edit: This answer is incorrect
See comments and https://stackoverflow.com/a/18650169/28234. It is being left here for reference because the idea isn't rare.
A very simple way for small arrays is simply this:
const someArray = [1, 2, 3, 4, 5];
someArray.sort(() => Math.random() - 0.5);
It's probably not very efficient, but for small arrays this works just fine. Here's an example so you can see how random (or not) it is, and whether it fits your usecase or not.
const resultsEl = document.querySelector('#results');
const buttonEl = document.querySelector('#trigger');
const generateArrayAndRandomize = () => {
const someArray = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
someArray.sort(() => Math.random() - 0.5);
return someArray;
};
const renderResultsToDom = (results, el) => {
el.innerHTML = results.join(' ');
};
buttonEl.addEventListener('click', () => renderResultsToDom(generateArrayAndRandomize(), resultsEl));
<h1>Randomize!</h1>
<button id="trigger">Generate</button>
<p id="results">0 1 2 3 4 5 6 7 8 9</p>
Answered 2023-09-20 21:00:48
Adding to @Laurens Holsts answer. This is 50% compressed.
function shuffleArray(d) {
for (var c = d.length - 1; c > 0; c--) {
var b = Math.floor(Math.random() * (c + 1));
var a = d[c];
d[c] = d[b];
d[b] = a;
}
return d
};
Answered 2023-09-20 21:00:48
var b =
in a loop instead of declaring b outside loop and assigning it with b =
in a loop? - anyone Warning!
Using this answer for randomizing large arrays, cryptography, or any other application requiring true randomness is not recommended, due to its bias and inefficiency. Elements position is only semi-randomized, and they will tend to stay closer to their original position. See https://stackoverflow.com/a/18650169/28234.
You can arbitrarily decide whether to return 1 : -1
by using Math.random
:
[1, 2, 3, 4].sort(() => (Math.random() > 0.5) ? 1 : -1)
Try running the following example:
const array = [1, 2, 3, 4];
// Based on the value returned by Math.Random,
// the decision is arbitrarily made whether to return 1 : -1
const shuffeled = array.sort(() => {
const randomTrueOrFalse = Math.random() > 0.5;
return randomTrueOrFalse ? 1 : -1
});
console.log(shuffeled);
Answered 2023-09-20 21:00:48
I found this variant hanging out in the "deleted by author" answers on a duplicate of this question. Unlike some of the other answers that have many upvotes already, this is:
shuffled
name rather than shuffle
)Here's a jsfiddle showing it in use.
Array.prototype.shuffled = function() {
return this.map(function(n){ return [Math.random(), n] })
.sort().map(function(n){ return n[1] });
}
Answered 2023-09-20 21:00:48
[1,2,3,4,5,6].sort(function() { return .5 - Math.random(); });
- it doesn't give a random sort, and if you use it you can end up embarrassed: robweir.com/blog/2010/02/microsoft-random-browser-ballot.html - anyone .sort(function(a,b){ return a[0] - b[0]; })
if you want the sort to compare values numerically. The default .sort()
comparator is lexicographic, meaning it will consider 10
to be less than 2
since 1
is less than 2
. - anyone Math.random()
produces. (that is, lexicographic order is the same as numeric order when dealing with numbers from 0 (inclusive) to 1 (exclusive)) - anyone With ES2015 you can use this one:
Array.prototype.shuffle = function() {
let m = this.length, i;
while (m) {
i = (Math.random() * m--) >>> 0;
[this[m], this[i]] = [this[i], this[m]]
}
return this;
}
Usage:
[1, 2, 3, 4, 5, 6, 7].shuffle();
Answered 2023-09-20 21:00:48
n >>> 0
instead of ~~n
. Array indices can be higher than 2³¹-1. - anyone benchmarks
Let's first see the results then we'll look at each implementation of shuffle
below -
splice is slow
Any solution using splice
or shift
in a loop is going to be very slow. Which is especially noticeable when we increase the size of the array. In a naive algorithm we -
rand
position, i
, in the input array, t
t[i]
to the outputsplice
position i
from array t
To exaggerate the slow effect, we'll demonstrate this on an array of one million elements. The following script almost 30 seconds -
const shuffle = t =>
Array.from(sample(t, t.length))
function* sample(t, n)
{ let r = Array.from(t)
while (n > 0 && r.length)
{ const i = rand(r.length) // 1
yield r[i] // 2
r.splice(i, 1) // 3
n = n - 1
}
}
const rand = n =>
0 | Math.random() * n
function swap (t, i, j)
{ let q = t[i]
t[i] = t[j]
t[j] = q
return t
}
const size = 1e6
const bigarray = Array.from(Array(size), (_,i) => i)
console.time("shuffle via splice")
const result = shuffle(bigarray)
console.timeEnd("shuffle via splice")
document.body.textContent = JSON.stringify(result, null, 2)
body::before {
content: "1 million elements via splice";
font-weight: bold;
display: block;
}
pop is fast
The trick is not to splice
and instead use the super efficient pop
. To do this, in place of the typical splice
call, you -
i
t[i]
with the last element, t[t.length - 1]
t.pop()
to the resultNow we can shuffle
one million elements in less than 100 milliseconds -
const shuffle = t =>
Array.from(sample(t, t.length))
function* sample(t, n)
{ let r = Array.from(t)
while (n > 0 && r.length)
{ const i = rand(r.length) // 1
swap(r, i, r.length - 1) // 2
yield r.pop() // 3
n = n - 1
}
}
const rand = n =>
0 | Math.random() * n
function swap (t, i, j)
{ let q = t[i]
t[i] = t[j]
t[j] = q
return t
}
const size = 1e6
const bigarray = Array.from(Array(size), (_,i) => i)
console.time("shuffle via pop")
const result = shuffle(bigarray)
console.timeEnd("shuffle via pop")
document.body.textContent = JSON.stringify(result, null, 2)
body::before {
content: "1 million elements via pop";
font-weight: bold;
display: block;
}
even faster
The two implementations of shuffle
above produce a new output array. The input array is not modified. This is my preferred way of working however you can increase the speed even more by shuffling in place.
Below shuffle
one million elements in less than 10 milliseconds -
function shuffle (t)
{ let last = t.length
let n
while (last > 0)
{ n = rand(last)
swap(t, n, --last)
}
}
const rand = n =>
0 | Math.random() * n
function swap (t, i, j)
{ let q = t[i]
t[i] = t[j]
t[j] = q
return t
}
const size = 1e6
const bigarray = Array.from(Array(size), (_,i) => i)
console.time("shuffle in place")
shuffle(bigarray)
console.timeEnd("shuffle in place")
document.body.textContent = JSON.stringify(bigarray, null, 2)
body::before {
content: "1 million elements in place";
font-weight: bold;
display: block;
}
Answered 2023-09-20 21:00:48
//one line solution
shuffle = (array) => array.sort(() => Math.random() - 0.5);
//Demo
let arr = [1, 2, 3];
shuffle(arr);
alert(arr);
https://javascript.info/task/shuffle
Math.random() - 0.5
is a random number that may be positive or negative, so the sorting function reorders elements randomly.
Answered 2023-09-20 21:00:48
var shuffle = function(array) {
temp = [];
originalLength = array.length;
for (var i = 0; i < originalLength; i++) {
temp.push(array.splice(Math.floor(Math.random()*array.length),1));
}
return temp;
};
Answered 2023-09-20 21:00:48
A recursive solution:
function shuffle(a,b){
return a.length==0?b:function(c){
return shuffle(a,(b||[]).concat(c));
}(a.splice(Math.floor(Math.random()*a.length),1));
};
Answered 2023-09-20 21:00:48
Modern short inline solution using ES6 features:
['a','b','c','d'].map(x => [Math.random(), x]).sort(([a], [b]) => a - b).map(([_, x]) => x);
(for educational purposes)
Answered 2023-09-20 21:00:48
Math.random()
function, you will get an uniform distribution (each item has the same chance to be at any position). - anyone Fisher-Yates shuffle in javascript. I'm posting this here because the use of two utility functions (swap and randInt) clarifies the algorithm compared to the other answers here.
function swap(arr, i, j) {
// swaps two elements of an array in place
var temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
function randInt(max) {
// returns random integer between 0 and max-1 inclusive.
return Math.floor(Math.random()*max);
}
function shuffle(arr) {
// For each slot in the array (starting at the end),
// pick an element randomly from the unplaced elements and
// place it in the slot, exchanging places with the
// element in the slot.
for(var slot = arr.length - 1; slot > 0; slot--){
var element = randInt(slot+1);
swap(arr, element, slot);
}
}
Answered 2023-09-20 21:00:48
Using Fisher-Yates shuffle algorithm and ES6:
// Original array
let array = ['a', 'b', 'c', 'd'];
// Create a copy of the original array to be randomized
let shuffle = [...array];
// Defining function returning random value from i to N
const getRandomValue = (i, N) => Math.floor(Math.random() * (N - i) + i);
// Shuffle a pair of two elements at random position j
shuffle.forEach( (elem, i, arr, j = getRandomValue(i, arr.length)) => [arr[i], arr[j]] = [arr[j], arr[i]] );
console.log(shuffle);
// ['d', 'a', 'b', 'c']
Answered 2023-09-20 21:00:48
First of all, have a look here for a great visual comparison of different sorting methods in javascript.
Secondly, if you have a quick look at the link above you'll find that the random order
sort seems to perform relatively well compared to the other methods, while being extremely easy and fast to implement as shown below:
function shuffle(array) {
var random = array.map(Math.random);
array.sort(function(a, b) {
return random[array.indexOf(a)] - random[array.indexOf(b)];
});
}
Edit: as pointed out by @gregers, the compare function is called with values rather than indices, which is why you need to use indexOf
. Note that this change makes the code less suitable for larger arrays as indexOf
runs in O(n) time.
Answered 2023-09-20 21:00:48
Array.prototype.sort
passes in two values as a
and b
, not the index. So this code doesn't work. - anyone All the other answers are based on Math.random() which is fast but not suitable for cryptgraphic level randomization.
The below code is using the well known Fisher-Yates
algorithm while utilizing Web Cryptography API
for cryptographic level of randomization.
var d = [1,2,3,4,5,6,7,8,9,10];
function shuffle(a) {
var x, t, r = new Uint32Array(1);
for (var i = 0, c = a.length - 1, m = a.length; i < c; i++, m--) {
crypto.getRandomValues(r);
x = Math.floor(r / 65536 / 65536 * m) + i;
t = a [i], a [i] = a [x], a [x] = t;
}
return a;
}
console.log(shuffle(d));
Answered 2023-09-20 21:00:48
Disclaimer
Please note that this solution is not suitable for large arrays! If you are shuffling large datasets, you should use the Durstenfeld algorithm suggested above.
Solution
function shuffle(array) {
const result = [], itemsLeft = array.concat([]);
while (itemsLeft.length) {
const randomIndex = Math.floor(Math.random() * itemsLeft.length);
const [randomItem] = itemsLeft.splice(randomIndex, 1); // take out a random item from itemsLeft
result.push(randomItem); // ...and add it to the result
}
return result;
}
How it works
copies the initial array
into itemsLeft
picks up a random index from itemsLeft
, adds the corresponding element to the result
array and deletes it from the itemsLeft
array
repeats step (2) until itemsLeft
array gets empty
returns result
Answered 2023-09-20 21:00:48
splice
being a horribly inefficient way to do what they called "striking out". If you don't want to mutate the original array, then just copy it, and then shuffle that copy in place using the much more efficient Durstenfeld variant. - anyone splice
method to create a copy like so: source = array.slice();
. - anyone For those of us who are not very gifted but have access to the wonders of lodash, there is such a thing as lodash.shuffle.
Answered 2023-09-20 21:00:48
Here is the EASIEST one,
function shuffle(array) {
return array.sort(() => Math.random() - 0.5);
}
for further example, you can check it here
Answered 2023-09-20 21:00:48
We're still shuffling arrays in 2019, so here goes my approach, which seems to be neat and fast to me:
const src = [...'abcdefg'];
const shuffle = arr =>
[...arr].reduceRight((res,_,__,s) =>
(res.push(s.splice(0|Math.random()*s.length,1)[0]), res),[]);
console.log(shuffle(src));
.as-console-wrapper {min-height: 100%}
Answered 2023-09-20 21:00:48
You can do it easily with:
// array
var fruits = ["Banana", "Orange", "Apple", "Mango"];
// random
fruits.sort(function(a, b){return 0.5 - Math.random()});
// out
console.log(fruits);
Please reference at JavaScript Sorting Arrays
Answered 2023-09-20 21:00:48
A simple modification of CoolAJ86's answer that does not modify the original array:
/**
* Returns a new array whose contents are a shuffled copy of the original array.
* @param {Array} The items to shuffle.
* https://stackoverflow.com/a/2450976/1673761
* https://stackoverflow.com/a/44071316/1673761
*/
const shuffle = (array) => {
let currentIndex = array.length;
let temporaryValue;
let randomIndex;
const newArray = array.slice();
// While there remains elements to shuffle...
while (currentIndex) {
randomIndex = Math.floor(Math.random() * currentIndex);
currentIndex -= 1;
// Swap it with the current element.
temporaryValue = newArray[currentIndex];
newArray[currentIndex] = newArray[randomIndex];
newArray[randomIndex] = temporaryValue;
}
return newArray;
};
Answered 2023-09-20 21:00:48
I found this useful:
const shuffle = (array: any[]) => {
return array.slice().sort(() => Math.random() - 0.5);
}
console.log(shuffle([1,2,3,4,5,6,7,8,9,10]));
// Output: [4, 3, 8, 10, 1, 7, 9, 2, 6, 5]
Answered 2023-09-20 21:00:48
yet another implementation of Fisher-Yates, using strict mode:
function shuffleArray(a) {
"use strict";
var i, t, j;
for (i = a.length - 1; i > 0; i -= 1) {
t = a[i];
j = Math.floor(Math.random() * (i + 1));
a[i] = a[j];
a[j] = t;
}
return a;
}
Answered 2023-09-20 21:00:48