如何遍历含数组的对象求和并按最新年份排序？求更优方案

Hey there! Let's refine your code to be cleaner, more efficient, and fix the sorting issue in your current implementation. Your goal is to generate a year-sorted (newest to oldest) object with total dividend values per year, so let's break down a better approach:

The Optimized Solution #

const dividendsYears = {
  1976: [
    {date: "1976-05-10", value: "0.00062"},
    {date: "1976-07-16", value: "0.00062"},
    {date: "1976-10-21", value: "0.00062"}
  ],
  1977: [
    {date: "1977-02-14", value: "0.00062"},
    {date: "1977-05-12", value: "0.00123"},
    {date: "1977-07-19", value: "0.00123"},
    {date: "1977-10-18", value: "0.00123"}
  ],
  1978: [
    {date: "1978-02-14", value: "0.00123"},
    {date: "1978-05-16", value: "0.00222"},
    {date: "1978-07-19", value: "0.00222"},
    {date: "1978-10-26", value: "0.00222"}
  ]
};

const totalDividendsPerYear = Object.fromEntries(
  // First, convert the object to [year, entries] pairs and sort years descending
  Object.entries(dividendsYears)
    .sort(([yearA], [yearB]) => Number(yearB) - Number(yearA))
    // Then calculate the total for each year using reduce
    .map(([year, dividendEntries]) => [
      year,
      dividendEntries.reduce((total, entry) => total + Number(entry.value), 0)
    ])
);

console.log(totalDividendsPerYear);
// Output: { '1978': 0.00789, '1977': 0.0043100000000000005, '1976': 0.00186 }

What Makes This Better? #

• Cleaner Chain of Operations: We combine sorting, sum calculation, and object creation into a single flow, avoiding unnecessary intermediate variables like your original totalDividendsPerYear and sortRecentDate.
• Reliable Sorting: Your original reverse() approach isn't guaranteed to work (object key order is insertion order in ES6+, but reversing that doesn't account for unordered keys in older environments or non-sequential years). This solution explicitly sorts years numerically in descending order.
• Functional Style with reduce: Using reduce for summation is more idiomatic than a forEach loop with a mutable sum variable—it's concise and avoids side effects.

Bonus: Fixing Floating-Point Precision #

If you want to clean up those long decimal results (like 0.0043100000000000005), you can round the total to a fixed number of decimal places:

// Option 1: Fixed decimal places (returns string)
dividendEntries.reduce((total, entry) => total + Number(entry.value), 0).toFixed(6)

// Option 2: Precision-safe rounding (returns number)
Math.round((dividendEntries.reduce((total, entry) => total + Number(entry.value), 0)) * 1e6) / 1e6

内容的提问来源于stack exchange，提问作者kr3w4