Question

这是一个完美创建数学序列的代码。唯一不同的是，它只对序列中的每个数字进行四舍五入，并最终分别计算最后一个数字。

目前的输出是：[0, 6, 12, 18, 24]

现在的问题是，我想要一个好的方法来创建另一个函数，该函数求和数组中的第一项和第二项，并像这样修改现有项：

因此，首先，代码创建了以下代码：[0, 6, 12, 18, 24]

然后我想将0和6相加并得到：[6, 12, 18, 24]

然后6和12：[18, 20, 22]

然后：[38, 22]

和：[60]

注意：每次我想为sn（数组中所有数字的总和）定义不同的值

编辑：换句话说，在新功能中，我必须再次重新分配这些行：

let a1 = 0;  // first item of the array
let sn = 60; // sum of all numbers in the array
let total_Slides = 5;

var n = total_Slides;  // total array or sequence length
var last = (sn - (n/2 * a1)) * 2/n  // equation to calculate the last number if we want the sum of whole sequence will be equel to sn
var d = (last - a1) / (n - 1)

代码如下：

let reducer = (accumulator, currentValue) => accumulator + currentValue;
let reducere = (accumulator, currentValue) => accumulator + currentValue;

let a1 = 0;  // first item of the array
let sn = 60; // sum of all numbers in the array
let total_Slides = 5;

var n = total_Slides;  // total array or sequence length
var last = (sn - (n/2 * a1)) * 2/n  // equation to calculate the last number if we want the sum of whole sequence will be equel to sn
var d = (last - a1) / (n - 1) 

var next = [];
var sum = 0;

for (let i = 1; i < total_Slides; i++){
      sum += a1 + (d * ( i - 1));  
      var pusher = int_part(sum);
      next.push(pusher);
      sum -= pusher
    }  
  
  let sumer = sn - (next.reduce(reducer));
  next.push(sumer) 
  let sumerr = next.reduce(reducere);
  console.log(sumerr)
  console.log(next)
 
function int_part(y){
  return Math.trunc(y);

}

function float_part(d){
  let int_part = Math.trunc(d);
  
  return Number((d-int_part).toFixed(2)); 
}

Answer 1

您可以获取前两个元素，并将其添加到广告中，然后将此值作为新数组的第一项并连接索引2以后的所有项。

var array = [0, 6, 12, 18, 24];

while (array.length > 1) {
    array = [array[0] + array[1]].concat(array.slice(2));
    console.log(...array);
}

Answer 2

您可以使用.splice()替换数组中的当前元素。

const numbers = [0, 6, 12, 18, 24];

while (numbers.length >= 2) {
  let [a, b] = numbers;
  numbers.splice(0, 2, a + b);
}

console.log(numbers);

修改现有的数学序列数组

2 个答案: