根据索引的嵌套数组的总和

Question

如果我有这些对应的数组，里面有 2 个嵌套数组（这可能有 2 个或更多）：const nums = [    [4, 23, 20, 23, 6, 8, 4, 0],      // Each array consists of 8 items    [7, 5, 2, 2, 0, 0, 0, 0] ];如何根据它们的索引与它们执行加法？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

这支持 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 2

您可以使用 reduce 和内部循环。需要注意的一些事情是不同的数组长度和不是数字的值。

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))