Mongodb $group 和 sum + 添加所有大于

假设我的文档如下所示:

[
  {
    name: "Example 1",
    year: "2012"
  },
  {
    name: "Example 2",
    year: "2012"
  },
  {
    name: "Example 3",
    year: "2013"
  },
  {
    name: "Example 4",
    year: "2014"
  }
]

使用aggregation，有没有办法达到group by year and sum the document count，但额外增加了add the sum of all later years？

我想要的result美元是:

[
    {
        _id: "2012",
        count: 4 // years 2012-2014
    },
    {
        _id: "2013",
        count: 2 // years 2013-2014
    },
    {
        _id: "2014",
        count: 1 // only year 2014
    }
]

现在，我使用的是普通的$group+$sum，它让我分别计算出每一年的计数，然后用JavaScript对它们进行排序.我希望有一种更简单的方法来消除额外的JS代码:

yearCounts: [           
    { $group: { _id: "$year", count: { $sum: 1 } } }
]

const yearCounts: { _id: string, count: number }[] = aggregationResult[0].yearCounts || [];
const yearCountsSummed = yearCounts.map((yearCount: { _id: string, count: number }) => {
    const yearsUntil = yearCounts.filter(year => year._id >= yearCount._id);
    const countSummed = yearsUntil.map(yearCount => yearCount.count).reduce((a, b) => a + b) || 0;
    return countSummed;
});