在对象数组中合并两个对象

我在用洛达斯 groupBy .

Object.entries(_.groupBy([
    {
        "idItem": "5d656f10394d6524c821f1b1",
        "mark": 5,
        "date": "2018-11-27T00:00:00.000Z"
    },
    {
        "idItem": "5d656f10394d6524c821f1b1",
        "mark": 2,
        "date": "2018-11-27T00:00:00.000Z"
    },
    {
        "idItem": "5d656f10394d6524c821f1b1",
        "mark": 1,
        "date": "2018-12-27T00:00:00.000Z"
    },
    {
        "idItem": "5d656f10394d6524c821f1b1",
        "mark": 2,
        "date": "2018-12-27T00:00:00.000Z",
    }
], e => `${e.idItem}_${e.date}`)).reduce((acc, [key, value]) => {
  const mark = value.reduce((sum, el) => sum + el.mark, 0) / value.length
 const [idItem, date] = key.split('_')
  return [...acc, { idItem, date, mark }]
}, [])

试试这个:

const input = [
  { idItem: "5d656f10394d6524c821f1b1", mark: 5, date: "2018-11-27T00:00:00.000Z" },
  { idItem: "5d656f10394d6524c821f1b1", mark: 2, date: "2018-11-27T00:00:00.000Z" },
  { idItem: "5d656f10394d6524c821f1b1", mark: 1, date: "2018-12-27T00:00:00.000Z" },
  { idItem: "5d656f10394d6524c821f1b1", mark: 2, date: "2018-12-27T00:00:00.000Z" }
];

const ids = new Set(input.map(e => `${e.idItem} ${e.date}`));
const grouped = [...ids].map(id => input.filter(
  e => `${e.idItem} ${e.date}` === id
)).map(
  group => ({
    ...group[0],
    mark: group.reduce((acc, cur) => cur.mark + acc, 0) / group.length,
  })
);

您可以看到结果已注销 here

也许不是最有表现力的,但它起作用了。首先我们得到一组id和date的所有组合,然后过滤每个组合的输入。然后,我们将每个组的日期和id复制到结果中,并减少条目以获得平均值。

如果id和日期字符串没有逗号( , )在它们中,然后您可以使用这个解决方案,我首先生成一个对象,其中键是id和日期的串联,值是具有具有键的id和日期的元素数以及它们的标记之和的对象,然后这个对象将用于构造他想要阵列:

var arr = [
  { "idItem": "5d656f10394d6524c821f1b1", "mark": 5, "date": "2018-11-27T00:00:00.000Z" },
  { "idItem": "5d656f10394d6524c821f1b1", "mark": 2, "date": "2018-11-27T00:00:00.000Z" },
  { "idItem": "5d656f10394d6524c821f1b1", "mark": 1, "date": "2018-12-27T00:00:00.000Z" },
  { "idItem": "5d656f10394d6524c821f1b1", "mark": 2, "date": "2018-12-27T00:00:00.000Z" }
];

var obj = {};

arr.forEach((o) => {
  var k = o.idItem + ',' + o.date;
  if (obj.hasOwnProperty(k)) {
    obj[k].s += o.mark;
    obj[k].n += 1;
  } else {
    obj[k] = {s: o.mark, n: 1};
  }
});

var result = [];

Object.keys(obj).forEach((k) => {
  var p = k.split(',');
  result.push({idItem: p[0], date: p[1], mark: obj[k].s / obj[k].n})
});

console.log(result);

首先,我将声明我们将在本例中使用的数组:

var array = [
   {
      "idItem": "5d656f10394d6524c821f1b1",
      "mark": 5,
      "date": "2018-11-27T00:00:00.000Z"
   },
   {
      "idItem": "5d656f10394d6524c821f1b1",
      "mark": 2,
      "date": "2018-11-27T00:00:00.000Z"
   },
   {
      "idItem": "5d656f10394d6524c821f1b1",
      "mark": 1,
      "date": "2018-12-27T00:00:00.000Z"
   },
   {
      "idItem": "5d656f10394d6524c821f1b1",
      "mark": 2,
      "date": "2018-12-27T00:00:00.000Z",
   }
];

所以,现在的情况是:我有一个程序可以工作,但另一个实现可以 几乎 作品。我从一个有用的开始,它使用 alasql 图书馆:

要加载依赖项,请使用以下命令 <script> 标签:

<script src="https://cdn.jsdelivr.net/npm/alasql@0.4">

实施:

var newArray = alasql('SELECT idItem, AVG([mark]) AS [mark], date AS [date] 
FROM ? GROUP BY date',[array]);
console.log(JSON.stringify(newArray));

下一个程序只计算所有id的平均标记,不考虑日期(因此得到的平均值是2.5)。也许有人能想出如何正确运行这个程序并编辑我的答案:

var sum = {};
for(var i = 0; i < array.length; i++) {
   var ele = array[i];
   if (!sum[ele.idItem]) {
      sum[ele.idItem] = {};
      sum[ele.idItem]["sum"] = 0;
      sum[ele.idItem]["count"] = 0;
   }
   sum[ele.idItem]["sum"] += ele.mark;
   sum[ele.idItem]["count"]++;
}
var result = [];
for (var idItem in sum) {
    result.push({idItem: idItem, mark: sum[idItem]["sum"] / sum[idItem]["count"]});
}
console.log(JSON.stringify(result));