我有一个像这样的数组：
var arr1 = ["a", "b", "c", "d"];
如何将其随机/随机播放？
如何随机化（随机播放）JavaScript数组？
Funny enough there was no non mutating recursive answer:
var shuffle = arr => {
const recur = (arr,currentIndex)=>{
console.log("What?",JSON.stringify(arr))
if(currentIndex===0){
return arr;
}
const randomIndex = Math.floor(Math.random() * currentIndex);
const swap = arr[currentIndex];
arr[currentIndex] = arr[randomIndex];
arr[randomIndex] = swap;
return recur(
arr,
currentIndex - 1
);
}
return recur(arr.map(x=>x),arr.length-1);
};
var arr = [1,2,3,4,5,[6]];
console.log(shuffle(arr));
console.log(arr);From a theoretical point of view, the most elegant way of doing it, in my humble opinion, is to get a single random number between 0 and n!-1 and to compute a one to one mapping from {0, 1, …, n!-1} to all permutations of (0, 1, 2, …, n-1). As long as you can use a (pseudo-)random generator reliable enough for getting such a number without any significant bias, you have enough information in it for achieving what you want without needing several other random numbers.
When computing with IEEE754 double precision floating numbers, you can expect your random generator to provide about 15 decimals. Since you have 15!=1,307,674,368,000 (with 13 digits), you can use the following functions with arrays containing up to 15 elements and assume there will be no significant bias with arrays containing up to 14 elements. If you work on a fixed-size problem requiring to compute many times this shuffle operation, you may want to try the following code which may be faster than other codes since it uses Math.random only once (it involves several copy operations however).
The following function will not be used, but I give it anyway; it returns the index of a given permutation of (0, 1, 2, …, n-1) according to the one to one mapping used in this message (the most natural one when enumerating permuations); it is intended to work with up to 16 elements:
function permIndex(p) {
var fact = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000];
var tail = [];
var i;
if (p.length == 0) return 0;
for(i=1;i<(p.length);i++) {
if (p[i] > p[0]) tail.push(p[i]-1);
else tail.push(p[i]);
}
return p[0] * fact[p.length-1] + permIndex(tail);
}
The reciprocal of the previous function (required for your own question) is below; it is intended to work with up to 16 elements; it returns the permutation of order n of (0, 1, 2, …, s-1):
function permNth(n, s) {
var fact = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000];
var i, j;
var p = [];
var q = [];
for(i=0;i<s;i++) p.push(i);
for(i=s-1; i>=0; i--) {
j = Math.floor(n / fact[i]);
n -= j*fact[i];
q.push(p[j]);
for(;j<i;j++) p[j]=p[j+1];
}
return q;
}
Now, what you want merely is:
function shuffle(p) {
var fact = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000];
return permNth(Math.floor(Math.random()*fact[p.length]), p.length).map(
function(i) { return p[i]; });
}
It should work for up to 16 elements with a little theoretical bias (though unnoticeable from a practical point of view); it can be seen as fully usable for 15 elements; with arrays containing less than 14 elements, you can safely consider there will be absolutely no bias.
the shortest arrayShuffle function
function arrayShuffle(o) {
for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
return o;
}
Just to have a finger in the pie. Here i present a recursive implementation of Fisher Yates shuffle (i think). It gives uniform randomness.
Note: The ~~ (double tilde operator) is in fact behaves like Math.floor() for positive real numbers. Just a short cut it is.
var shuffle = a => a.length ? a.splice(~~(Math.random()*a.length),1).concat(shuffle(a))
: a;
console.log(JSON.stringify(shuffle([0,1,2,3,4,5,6,7,8,9])));Edit: The above code is O(n^2) due to the employment of .splice() but we can eliminate splice and shuffle in O(n) by the swap trick.
var shuffle = (a, l = a.length, r = ~~(Math.random()*l)) => l ? ([a[r],a[l-1]] = [a[l-1],a[r]], shuffle(a, l-1))
: a;
var arr = Array.from({length:3000}, (_,i) => i);
console.time("shuffle");
shuffle(arr);
console.timeEnd("shuffle");The problem is, JS can not coop on with big recursions. In this particular case you array size is limited with like 3000~7000 depending on your browser engine and some unknown facts.
arr1.sort(() => Math.random() - 0.5);
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;
}
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));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));
};
您可以轻松地做到这一点：
// array
var fruits = ["Banana", "Orange", "Apple", "Mango"];
// random
fruits.sort(function(a, b){return 0.5 - Math.random()});
// out
console.log(fruits);我在此问题的副本中的“作者删除”答案中发现了这种变体。与已经有许多投票的其他一些答案不同，这是：
- 其实随机
- 不到位（因此，
shuffled名称而不是shuffle） - 这里尚不存在多种变体
Array.prototype.shuffled = function() {
return this.map(function(n){ return [Math.random(), n] })
.sort().map(function(n){ return n[1] });
}
借助ES2015，您可以使用以下工具：
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;
}
用法：
[1, 2, 3, 4, 5, 6, 7].shuffle();
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;
};
编辑：此答案不正确
参见https://stackoverflow.com/a/18650169/28234。它被留在这里作为参考，因为这个想法并不罕见。
//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是一个可以为正数或为负数的随机数，因此排序功能会随机对元素进行重新排序。
添加到@Laurens Holsts答案。这是50％压缩的。
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
};
一个人可以（或应该）将其用作Array的原型：
来自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;
}
这是Durstenfeld shuffle的JavaScript实现，它是Fisher-Yates的优化版本：
/* 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;
}
}
它为每个原始数组元素选择一个随机元素，并将其从下一次绘制中排除，就像从一副纸牌中随机选择一样。
这种巧妙的排除方式将选择的元素与当前元素交换，然后从其余元素中选择下一个随机元素，向后循环以实现最佳效率，确保简化了随机选择（它始终可以从0开始），从而跳过了最后一个元素。
算法运行时为O(n)。请注意，随机播放是就地完成的，因此，如果您不想修改原始数组，请先使用复制它.slice(0)。
编辑：更新到ES6 / ECMAScript 2015
新的ES6允许我们一次分配两个变量。当我们要交换两个变量的值时，这特别方便，因为我们可以在一行代码中完成它。这是使用此功能的相同功能的缩写。
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]];
}
}
[社区编辑：此答案不正确；看评论。它被留在这里供以后参考，因为这种想法并不罕见。]
[1,2,3,4,5,6].sort(function() {
return .5 - Math.random();
});
实际无偏混洗算法是Fisher-Yates（aka Knuth）混洗。
参见https://github.com/coolaj86/knuth-shuffle
您可以在此处看到出色的可视化效果（以及与此链接相关的原始文章）
function shuffle(array) {
var currentIndex = array.length, temporaryValue, randomIndex;
// While there remain elements to shuffle...
while (0 !== currentIndex) {
// Pick a remaining element...
randomIndex = Math.floor(Math.random() * currentIndex);
currentIndex -= 1;
// And swap it with the current element.
temporaryValue = array[currentIndex];
array[currentIndex] = array[randomIndex];
array[randomIndex] = temporaryValue;
}
return array;
}
// Used like so
var arr = [2, 11, 37, 42];
shuffle(arr);
console.log(arr);有关使用的算法的更多信息。
