Question

如果我在这些具有2个嵌套数组（可能有2个或更多）的嵌套数组中对应数组：

const nums = [
    [4, 23, 20, 23, 6, 8, 4, 0],      // Each array consists of 8 items
    [7, 5, 2, 2, 0, 0, 0, 0] 
];

如何根据他们的索引与他们进行加法运算？

预期结果：

// 11 is from 4 (array1 - index1) + 7 (array2 - index1)
// and so on.
[11, 28, 22, 25, 6, 8, 4, 0]

我所做的是：

// This will work but it will only be applicable for 2 arrays as what if there will be 2 or more making it dynamic 

const total = Array.from({ length: 8 }, (_, i) => nums[0][i] + nums[1][i]);

Answer 1

一个可能的解决方案是将第一个内部数组的每个元素Array.map()变为同一列中元素的总和。为了获得同一列中元素的汇总，我们可以在map()内使用Array.reduce()：

const nums = [
  [4, 23, 20, 23, 6, 8, 4, 0],
  [7, 5, 2, 2, 0, 0, 0, 0],
  [1, 3, 4, 7, 1, 1, 1, 1],
];

let [first, ...rest] = nums;
let res = first.map((e, i) => rest.reduce((sum, x) => sum + x[i], e));

console.log(res);

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

Answer 2

此功能支持n个嵌套数组

const nums = [
  [4, 23, 20, 23, 6, 8, 4, 0],
  [7, 5, 2, 2, 0, 0, 0, 0],
  [2, 1, 2, 5, 7, 8, 9, 4]
];


const total = nums.reduce((a, b) => a.map((c, i) => c + b[i]));

console.log(total);

Answer 3

您可以使用嵌套的forEach()循环

const nums = [
    [4, 23, 20, 23, 6, 8, 4, 0],
    [7, 5, 2, 2, 0, 0, 0, 0] 
];

function sum(arr){
  let max = Math.max(...arr.map(x => x.length));
  let res = Array(max).fill(0)
  res.forEach((x,i) => {
    nums.forEach(a => {
      
      res[i] = res[i] +  (a[i] || 0)
    })
  })
  return res;
}

console.log(sum(nums))

Answer 4

您可以使用reduce和一个内部循环。要注意的一些事情是不同的数组长度和值（不是数字）。

const nums = [
    [4, 23, 20, 23, 6, 8, 4, 0],      // Each array consists of 8 items
    [7, 5, 2, 2, 0, 0, 0, 0] 
];
const otherNums = [
    [4, 23, 20, 23, 6, 8, 4, 0, 9, 55],      // Each array consists of 8 items
    [7, 5, 2, 2, 0, 0, 0, 0, "cat", null, 78],
    [7, 5, 2, 2, 0, 0, 0, 0, "dog", null, 78],
    [7, 5, 2, 2, 0, 0, 0, 0, "elephant", null, 78] 
];

const sumArraysByIndex = nums => nums.reduce((sums, array) => {
  for (const index in array) {
    if (sums[index] === undefined) sums[index] = 0
    if (isNaN(array[index])) return sums
    sums[index] += array[index]
  }
  return sums
}, [])

console.log(sumArraysByIndex(nums))
console.log(sumArraysByIndex(otherNums))

Answer 5

获取最小长度subarray，然后使用索引返回该长度的数组来创建该长度的数组。

const nums = [
    [4, 23, 20, 23, 6, 8, 4, 0],
    [7, 5, 2, 2, 0, 0, 0, 0]
];

const [arr1, arr2] = nums;
const min = Math.min(nums[0].length, nums[1].length);

const output = Array.from({length: min}, (_, i) => arr1[i] + arr2[i]);

console.log(output);

根据其索引嵌套数组的总和

5 个答案: