我有一个这样的数组:

var arr1 = ["a", "b", "c", "d"];

我如何随机化/打乱它?


当前回答

重建整个阵列,逐个将每个元素放在一个随机位置。

[1,2,3].reduce((a,x,i)=>{a.splice(Math.floor(Math.random()*(i+1)),0,x);return a},[])

变量ia=[1,2,3];var it=1000;var f=(a,x,i)=>{a.splice(Math.floor(Math.random()*(i+1)),0,x);返回a};var a=新数组(it).fill(ia).map(x=>x.reduce(f,[]));var r=新数组(ia.length).fill(0).map((x,i)=>a.reduce((i2,x2)=>x2[i]+i2,0)/it)console.log(“这些值应该相当相等:”,r);

其他回答

使用Ramda的功能解决方案。

const {map, compose, sortBy, prop} = require('ramda')

const shuffle = compose(
  map(prop('v')),
  sortBy(prop('i')),
  map(v => ({v, i: Math.random()}))
)

shuffle([1,2,3,4,5,6,7])

有趣的是,没有非变异递归答案:

var shuffle=arr=>{常量重复=(arr,currentIndex)=>{console.log(“什么?”,JSON.stringify(arr))如果(当前索引==0){返回arr;}const randomIndex=数学地板(Math.random()*当前索引);常量swap=arr[currentIndex];arr[currentIndex]=arr[randomIndex];arr[randomIndex]=掉期;返回重复(arr,当前索引-1);}返回递归(arr.map(x=>x),arr.length-1);};var arr=[1,2,3,4,5,[6];console.log(shuffle(arr));控制台日志(arr);

Fisher Yates的另一个实现,使用严格模式:

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;
}

无序排列到位

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纯,迭代

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
};

可靠性和性能测试

本页上的一些解决方案不可靠(它们只是部分随机化了阵列)。其他解决方案的效率明显较低。使用testShuffleArrayFun(见下文),我们可以测试阵列洗牌功能的可靠性和性能。

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}`)
}

其他解决方案

其他解决方案只是为了好玩。

ES6纯,递归

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纯使用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纯使用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]
    )
}

Fisher Yates的这种变体稍微更有效,因为它避免了元素与自身的交换:

function shuffle(array) {
  var elementsRemaining = array.length, temp, randomIndex;
  while (elementsRemaining > 1) {
    randomIndex = Math.floor(Math.random() * elementsRemaining--);
    if (randomIndex != elementsRemaining) {
      temp = array[elementsRemaining];
      array[elementsRemaining] = array[randomIndex];
      array[randomIndex] = temp;
    }
  }
  return array;
}