我有一个这样的数组:
var arr1 = ["a", "b", "c", "d"];
我如何随机化/打乱它?
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;
}