数组上的元素数

Question

我有一个包含 30 个值的数组，这些值将随机分布在其他 3 个数组之间。我让它“几乎”正常工作，但是 3 个数组总是带有随机数量的元素，我需要其中 2 个有 8 个元素，其他有 14 个，我该如何定义呢？

            const arrAll = [s1, s2, s3, s4, s5, s6, s7, s8, s9, s10, i1, i2, i3, i4, i5, i6, i7, i8, i9, i10, e1, e2, e3, e4, e5, e6, e7, e8, e9, e10]

            const arrCard = []

            const arrCent = []

            const arrSec = []

            arrAll.map(l => {

                let rand = Math.floor(Math.random() * 9000)

                rand <= 3000 ? arrCard.push(l) : rand <= 6000 ? arrCent.push(l) : arrSec.push(l);

            })

Answer 1

一种解决方案是您将整个数组打乱，然后您只需根据需要选择每个数组中的元素数量。这是示例（shuffle 函数是从How can I shuffle an array? 复制粘贴的？）function shuffle(a) {&nbsp; &nbsp; for (let i = a.length - 1; i > 0; i--) {&nbsp; &nbsp; &nbsp; &nbsp; const j = Math.floor(Math.random() * (i + 1));&nbsp; &nbsp; &nbsp; &nbsp; [a[i], a[j]] = [a[j], a[i]];&nbsp; &nbsp; }&nbsp; &nbsp; return a;}function splitArrayRandomly(arr, sizes) {&nbsp; const shuffledArr = shuffle(arr);&nbsp; let pos = 0;&nbsp; const arrays = sizes.map(size => {&nbsp; &nbsp; const newArr = shuffledArr.slice(pos, pos + size);&nbsp; &nbsp; pos += size;&nbsp; &nbsp; return newArr;&nbsp; });&nbsp;&nbsp;&nbsp; return arrays;}const arr = [4, 7, 15, 22, 11, 6, 19];// This option create 3 arrays, first of size 4, second of size 2 and last with size 1const sizes = [4,2,1];console.log(splitArrayRandomly(arr, sizes));在您的情况下，您放入sizes这个数组[8, 8, 14]，它会返回三个具有这些大小的数组。然后，如果需要，您可以将它们放入您arrCard, arrCent, arrSec的变量中。

Answer 2

制作第二个数组，映射出数组的分布：const dist = [8, 8, 14];那将是 3 个数组——前两个数组的长度为 8，第三个数组的长度为 14。在演示中，该数组是第二个参数。接下来，随机打乱主数组。在演示中，主数组是第一个参数。为了随机打乱主数组，我们将使用Fisher-Yates Shuffle最后，for使用了两个循环。外部循环将创建一个空数组。它将迭代 dist 数组的长度。在演示中，这将是三倍。内部循环将.shift()* 从主数组中取出一个值（已随机重新排序（参见 #2）），然后.push()将其放入先前在外循环中定义的数组中（参见 #3.1）。每个内循环将迭代当前外循环迭代的索引位置指示的次数。示例（伪代码）/* On the second iteration of the outer loop, will be 8 */dist[1] = 8/* On the inner loop, it will take the values of the randomly reordered main&nbsp;&nbsp;array in the range of 9 thru 16 */["e9","i4","i10","s2","e8","i9","i3","i8"]*.shift()将永久获取索引 0 处的数组值。这样做是为了使主数组中的值不会在子数组中重复一旦内部循环的迭代完成，该数组将被.push()编辑到结果数组中。完成最后一个内部循环后，结果数组将作为二维数组返回。示例（伪代码）/* Returned as a two dimensional array */result = [[8 values], [8 values], [14 values]]/* Assign variables to each sub-array */const cardArr = result[0];const centArr = result[1];const sectArr = result[2];演示const main = ['s1', 's2', 's3', 's4', 's5', 's6', 's7', 's8', 's9', 's10', 'i1', 'i2', 'i3', 'i4', 'i5', 'i6', 'i7', 'i8', 'i9', 'i10', 'e1', 'e2', 'e3', 'e4', 'e5', 'e6', 'e7', 'e8', 'e9', 'e10'];const dist = [8, 8, 14];&nbsp;function randomDistribution(array, dist) {&nbsp; let total = array.length, split = dist.length;&nbsp; let result = [], temp, index;&nbsp; while(total) {&nbsp; &nbsp; index = Math.floor(Math.random() * total--);&nbsp; &nbsp; temp = array[total];&nbsp; &nbsp; array[total] = array[index];&nbsp; &nbsp; array[index] = temp;&nbsp; }&nbsp; for (let subIdx = 0; subIdx < split; subIdx++) {&nbsp; &nbsp; let subArr = [];&nbsp; &nbsp; for (let reIdx = 0; reIdx < dist[subIdx]; reIdx++) {&nbsp; &nbsp; &nbsp; subArr.push(array.shift());&nbsp; &nbsp; }&nbsp; &nbsp; result.push(subArr);&nbsp; }&nbsp; return result;}let twoDArray = randomDistribution(main, dist);console.log(JSON.stringify(twoDArray));const cardArray = twoDArray[0];const centArray = twoDArray[1];const sectArray = twoDArray[2];console.log(`cardArray: ${cardArray}`);console.log(`centArray: ${centArray}`);console.log(`sectArray: ${sectArray}`);.as-console-row-code {&nbsp; word-break: break-word;&nbsp; overflow-x: hidden;}

Answer 3

1）从输入数组中，通过生成随机索引来创建随机顺序，并限制数组的大小。2) 创建与大小数组相同长度的数组输出数组（初始化）。3) 维护o_idx（输出索引）以跟踪需要推送元素的输出数组。4）一旦获得r_idx（随机索引），然后将相应的值（arr[r_idx]）推送到相应的输出数组（output[o_idx]）。const randomOrderInSizes = (arr, sizes) => {&nbsp; const myRandom = () => Math.floor((Math.random() * 100) % arr.length);&nbsp; const temp = [];&nbsp; const output = sizes.map(() => []);&nbsp; let o_idx = 0;&nbsp; arr.forEach(() => {&nbsp; &nbsp; let r_idx = myRandom();&nbsp; &nbsp; while (temp.includes(r_idx)) {&nbsp; &nbsp; &nbsp; r_idx = myRandom();&nbsp; &nbsp; }&nbsp; &nbsp; temp.push(r_idx);&nbsp; &nbsp;&nbsp;&nbsp; &nbsp; if (sizes[o_idx] === output[o_idx].length) {&nbsp; &nbsp; &nbsp; o_idx += 1;&nbsp; &nbsp; }&nbsp; &nbsp; output[o_idx].push(arr[r_idx]);&nbsp; });&nbsp; return output;};const data = [21, 22, 23, 24, 25, 26, 27, 28, 29, 30,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 31, 32, 33, 34, 35, 36, 37, 38, 39, 40];&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;const required = randomOrderInSizes(data, [8, 8, 14]);console.log(required);