Question

我有一个包含日期和值的对象列表。我想创建一个新的对象，该对象将对象中值的总和相加。到目前为止，我有一个解决方案，但对代码不满意，想知道您是否有更好的解决方案？

此外，我想按顺序显示最近的一年。

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

// Create new object that sum the total of value
let totalDividendsPerYear = {}
Object.entries(dividendsYears).forEach((entry, index) => {
    let sum = 0

    // Adds up dividends value per year
    entry[1].forEach((item) => (
        sum += Number(item.value)
    ))

    totalDividendsPerYear[entry[0]] = sum
});

// Reverse array to sort the recent date to the oldest date
const sortRecentDate = Object.assign([], totalDividendsPerYear).reverse();

预期结果：

{
    1978: 0.00789,
    1977: 0.0043100000000000005,
    1976: 0.00186
}

Answer 1

您可以使用单个Object.keys和Array.reduce来做到这一点：

const data = { 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"} ] }

let result = Object.keys(data).reduce((r,k) => {
  r[k] = data[k].reduce((acc,c) => (+acc.value || acc || 0) + (+c.value))
  return r
}, {})

console.log(result)

您也可以在一行中牺牲一些可读性：

const data = { 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"} ] }

let result = Object.keys(data).reduce((r,k) => 
   (r[k] = data[k].reduce((a,c) => (+a.value || a || 0) + (+c.value))) && r, {})

console.log(result)

Answer 2

这是一个单一的解决方案：

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 result = Object.entries(dividendsYears).reduce((a,b) => { a[b[0]] = b[1].map(i => parseFloat(i.value)).reduce((k,j) => k + j); return a; },{});

document.write(JSON.stringify(result));

Answer 3

我想我得到一个较短的：;）

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 result = Object.entries(dividendsYears).reduce((r,x)=>{r[x[0]]=x[1].reduce((a,c)=>a+parseFloat(c.value),0); return r},{});

console.log( result )

如何遍历具有数组并求和值的对象？

3 个答案: