Question

这里我更新了我的示例代码，

var distanceSumOfprev = 0;
          var distancesArr = [1000, 2000, 1000, 2000, 1000, 2000];
          for (var i = 1; i < distancesArr.length; i++) {
            distanceSumOfprev += distancesArr[i]
            console.log(distanceSumOfprev);
          }

这就是我目前所得到的

  distanceSumOfprev = [2000, 3000, 5000, 6000, 8000];

预期结果

distanceSumOfprev = [2000, 4000, 6000, 7000, 9000];

Answer 1

var distancesArr = [1000, 2000, 1000, 2000, 1000, 2000];
// since we need to add each value with value at previous index, distanceSumOfprev is initialized with value corresponding to first index.
var distanceSumOfprev = distancesArr[0];
for (var i = 0; i < distancesArr.length - 1; i++) {
  //distanceSumOfprev will be having the sum of all values upto the index pointing to current loop variable.
  distanceSumOfprev += distancesArr[i + 1];
  console.log(distanceSumOfprev);
}

请检查

Answer 2

您可以使用最后一笔的闭包并映射数组的其余部分。

var array = [1000, 2000, 1000, 2000, 1000, 2000],
    result = array.map(function (s) {
        return function(a) {
            return s += a;
        };
    }(array.shift()));

console.log(result);

ES6

var array = [1000, 2000, 1000, 2000, 1000, 2000],
    result = array.map((s => a => s += a)(array.shift()));

console.log(result);

Answer 3

您正在寻找的是被称为cumsum，它是累积总和的缩写以下代码演示了如何在原始数组的副本上执行此操作：

＆＃13;

var distancesArr = [1000, 2000, 1000, 2000, 1000, 2000];
var cumSumArr = distancesArr.slice();
for (var i = 1; i < distancesArr.length; i++) {
    cumSumArr[i] += cumSumArr[i-1];
}
console.log(distancesArr);
console.log(cumSumArr);

＆＃13;

Answer 4

我认为你的意思是输出数组应该以{{1}}开头。然后，我们所要做的就是在现有循环内再添加一个For i = 2 To 20 If Not Rows(i).Hidden 'do stuff End If End With循环向后3000

＆＃13;

for

＆＃13;

由于这里的小调整，您可以获得所需的输出： [3000 *，4000,6000,7000,9000]

Answer 5

您可以使用reduce()并返回新数组。

var distancesArr = [1000, 2000, 1000, 2000, 1000, 2000];

var result = distancesArr.reduce((function(c) {
  return (r, e, i) => (i == 0 ? c = e : r.push(c += e), r)
})(0), [])

console.log(result)

Answer 6

＆＃13;

var distanceSumOfprev = 0;
var distancesArr = [1000, 2000, 1000, 2000, 1000, 2000];
for (var i = 1; i < distancesArr.length; i++) {
    distanceSumOfprev = i===1?distancesArr[i]:distancesArr[i]+distanceSumOfprev+distancesArr[0]
    console.log(distanceSumOfprev);
}

＆＃13;

如何在一个数组中存储以前索引值的总和？

6 个答案: