Question

我有以下对象数组：

[{
   id: 1,
   amount: 2000,
   date: "2018-01-31T00:00:00.000Z"
},{
   id: 2,
   amount: 3000,
   date: "2017-07-31T00:00:00.000Z"
},{
   id: 3,
   amount: 6000,
   date: "2018-01-31T00:00:00.000Z"
},{
   id: 4,
   amount: 7000,
   date: "2017-01-31T00:00:00.000Z"
},{
   id: 5,
   amount: 5000,
   date: "2017-03-31T00:00:00.000Z"
},{
   id: 6,
   amount: 3000,
   date: "2018-02-22T00:00:00.000Z"
},{
   id: 7,
   amount: 4500,
   date: "2017-01-31T00:00:00.000Z"
}]

我正在调用以下命令按年份和日期对对象进行分组：

_(data)
.groupBy(o => new Date(o.date).getFullYear() + '-' + new Date(o.date).getMonth())
.map(o1 => _(o1).map(o2 => o2.amount).sum())

上面的代码给我一些总和，如[xx，yy，aaa，bbb，...]

现在我需要确保数组中的这些值将被排序（2018-2的总和将是第一个，2017-1的总和将在最后）。

如上所述，当结果将包含已排序对象的数组时也会很好，其中每个对象都将包含句点键“year-month”以检测当前值是什么。预期的输出将是这样的：

[
  {period: "2018-2", amount:3000}, // sum of 2018-2
  {period: "2018-1", amount: 8000 }, // sum of 2018-1
  {period: "2017-7", amount: 3000} // sum of 2017-7
  {period: "2017-3", amount: 5000} // sum of 2017-3
  {period: "2017-1" amount: 11500} // sum of 2017-1
]

可以编辑此链以获得所需的结果吗？谢谢！

Answer 1

您可以将函数reduce与函数Object.values

一起使用

Just vanilla javascript

＆＃13;

var data = [{   id: 1,   amount: 2000,   date: "2018-01-31T00:00:00.000Z"},{   id: 2,   amount: 3000,   date: "2017-07-31T00:00:00.000Z"},{   id: 3,   amount: 6000,   date: "2018-01-31T00:00:00.000Z"},{   id: 4,   amount: 7000,   date: "2017-01-31T00:00:00.000Z"},{   id: 5,   amount: 5000,   date: "2017-03-31T00:00:00.000Z"},{   id: 6,   amount: 3000,   date: "2018-02-22T00:00:00.000Z"},{   id: 7,   amount: 4500,   date: "2017-01-31T00:00:00.000Z"}];

var result = Object.values(data.reduce((a, c) => {
  var [year, month] = c.date.split('-');
  var key = `${year}-${+month}`;
  (a[key] || (a[key] = {period: key, amount: 0})).amount += c.amount;
  return a;
}, {})).sort((a, b) => b.period.localeCompare(a.period));

console.log(result);

＆＃13;

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＆＃13;

Answer 2

使用lodash，您可以先groupBy，然后对条目进行排序，然后对map这些条目进行排序，以便为每个时段获取period和amount之和的对象。

const data = [{"id":1,"amount":2000,"date":"2018-01-31T00:00:00.000Z"},{"id":2,"amount":3000,"date":"2017-07-31T00:00:00.000Z"},{"id":3,"amount":6000,"date":"2018-01-31T00:00:00.000Z"},{"id":4,"amount":7000,"date":"2017-01-31T00:00:00.000Z"},{"id":5,"amount":5000,"date":"2017-03-31T00:00:00.000Z"},{"id":6,"amount":3000,"date":"2018-02-22T00:00:00.000Z"},{"id":7,"amount":4500,"date":"2017-01-31T00:00:00.000Z"}]

const result = _.chain(data)
  .groupBy(({date}) => {
    const d = new Date(date)
    return d.getFullYear() + '-' + (d.getMonth() + 1)
  })
  .entries()
  .sort(([a], [b]) => b.localeCompare(a))
  .map(([period, arr]) =>  ({period, amount: _.sumBy(arr, ({amount}) => amount)}))

console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>

Answer 3

您可以获取分组数据并按升序排序（使用lodash的唯一可能方式）并反转订单。

var data = [{ id: 1, amount: 2000, date: "2018-01-31T00:00:00.000Z" }, { id: 2, amount: 3000, date: "2017-07-31T00:00:00.000Z" },{ id: 3, amount: 6000, date: "2018-01-31T00:00:00.000Z" }, { id: 4, amount: 7000, date: "2017-01-31T00:00:00.000Z" }, { id: 5, amount: 5000, date: "2017-03-31T00:00:00.000Z" }, { id: 6, amount: 3000, date: "2018-02-22T00:00:00.000Z" }, { id: 7, amount: 4500, date: "2017-01-31T00:00:00.000Z" }],
    result = _(data)
        .groupBy(o => o.date.slice(0, 7))
        .map((array, sort) => ({ sort, date: sort.split('-').map(Number).join('-'), amount: _.sumBy(array, 'amount') }))
        .sortBy('sort')
        .reverse()
        .map(({ date, amount }) => ({ date, amount }))
        .value();


console.log(result);

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<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.15.0/lodash.min.js"></script>

按照Lodash

3 个答案: