如何在对象数组中找到特定键的总和[重复]答案

【问题标题】：How to find the sum of a specific key in array of objects [duplicate]如何在对象数组中找到特定键的总和[重复]
【发布时间】：2021-08-17 05:14:50
【问题描述】：

我有一个具有以下结构的对象数组：

[
 {
     _id: 60b18e0a7ba49519c1ee63b4,
     course: 'sfsdf',
     earnings: 100,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 1,
     year: 2019
   },
   {
     _id: 60b18e1a7ba49519c1ee63b5,
     course: 'sfsdf',
     earnings: 200,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 2,
     year: 2019
   },
   {
     _id: 60b18e227ba49519c1ee63b6,
     course: 'sfsdf',
     earnings: 300,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 3,
     year: 2019
   },
   {
     _id: 60b18e297ba49519c1ee63b7,
     course: 'sfsdf',
     earnings: 400,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 4,
     year: 2019
   },
   {
     _id: 60b18b907ba49519c1ee63a5,
     course: 'html-css',
     earnings: 1000,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 3,
     year: 2021
   },
   {
     _id: 60b18c457ba49519c1ee63a9,
     course: 'html-css',
     earnings: 500,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 2,
     year: 2020
   }
]

我想以某种方式操作上面的数组，使结果看起来像：

[
    { label: 2019, earnings: 1000 },
    { label: 2021, earnings: 1000 },
    { label: 2020, earnings: 500 },
]

其中label 是唯一的年份，earnings 是该年份所有收入的总和。

问题说明：

2019 年有 4 个对象，每年的收益分别为 100、200、300、400，总计为 1000（100 + 200 + 300 + 400），因此结果数组有 { label: 2019, earnings: 1000 }
2021 年有 1 个对象，收入为 1000，因此总收入为 1000，因此结果数组有 { label: 2021, earnings: 1000 }

2020 年也是如此。

【问题讨论】：

标签： javascript arrays mapreduce javascript-objects reduce

【解决方案1】：

您可以使用findIndex 和forEach

const response = [
 {
     _id: "60b18e0a7ba49519c1ee63b4",
     course: 'sfsdf',
     earnings: 100,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 1,
     year: 2019
   },
   {
     _id: "60b18e1a7ba49519c1ee63b5",
     course: 'sfsdf',
     earnings: 200,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 2,
     year: 2019
   },
   {
     _id: "60b18e227ba49519c1ee63b6",
     course: 'sfsdf',
     earnings: 300,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 3,
     year: 2019
   },
   {
     _id: "60b18e297ba49519c1ee63b7",
     course: 'sfsdf',
     earnings: 400,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 4,
     year: 2019
   },
   {
     _id: "60b18b907ba49519c1ee63a5",
     course: 'html-css',
     earnings: 1000,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 3,
     year: 2021
   },
   {
     _id: "60b18c457ba49519c1ee63a9",
     course: 'html-css',
     earnings: 500,
     creator: '607497d5f4bcc83fa1f3cedf',
     month: 2,
     year: 2020
   }
];

const result = [];
response.forEach(({earnings, year}) => {
  const index = result.findIndex(({label}) => label === year);
  if (index === -1) result.push({label: year, earnings});
  else result[index].earnings += earnings;
})

console.log(result)

【讨论】：

【解决方案2】：

const result = arr.reduce((acc, obj) => {
    const idx = acc.findIndex(o => o.label === obj.year);
    if (idx !== -1) acc[idx].earnings += obj.earnings;
    else acc.push({ label: obj.year, earnings: obj.earnings });
    return acc;
}, [])

console.log(result)

【讨论】：