javascript - typescript /lodash : how to get results/objects with multiple common properties

Question

我正在寻找具有多个属性的 groupBy 函数的 lodash/typerscript 解决方案。这是我到目前为止所尝试过的:

JSON:

[
    {
        "jobId": 1001,
        "id": 1,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1001,
        "id": 2,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1002,
        "id": 1,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1002,
        "id": 1,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 2,
        "balance": 100
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 2,
        "balance": 100
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 2,
        "balance": 100
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 2,
        "balance": 100
    }
]

typescript 代码:

  // Group of same ids
  let mergeResults = _.groupBy(this.BSSDealresults, function (n) {
     return n.timeSegmentId;
  });
  const objKeys = Object.keys(mergeResults);
  const AggregateResults = [];
  objKeys.forEach(item => {
    if (mergeResults[item].length > 1) {

      let agg_balance = 0;

      mergeResults[item].forEach(element => {
        agg_balance += element.balance;
      });
      AggregateResults.push({
        id: mergeResults[item][0].id,
        jobId: mergeResults[item][0].jobId,
        balance: agg_balance
      });
    } else {
      AggregateResults.push(mergeResults[item][0]);
    }
  })

此解决方案提供仅按相同“id”分组的余额聚合，因为我应用了基于属性“id”的 lodash“_groupBy”函数。

而我正在寻找基于多个属性的聚合结果，例如id、jobId 和 timeSegId。

在 lodash/typescript 中有什么方法可以让我们使用多个属性来应用 grouby 吗？

预期输出应该是:

[
    {
        "jobId": 1001,
        "id": 1,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1001,
        "id": 2,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1002,
        "id": 1,
        "timeSegmentId": 1,
        "balance": 200
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 1,
        "balance": 100
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 2,
        "balance": 100
    },
    {
        "jobId": 1003,
        "id": 1,
        "timeSegmentId": 2,
        "balance": 300
    }
]

Answer 1

您可以通过从 3 个 ID 中创建一个字符串来为 _.groupBy() 生成 key 。然后你就可以_.map() groups BSSDealresults 对象，使用 object spread 克隆第一个元素(或 Object.assign() ，并获得与 _.sumBy() 的合并余额:

const BSSDealresults = [{"jobId":1001,"id":1,"timeSegmentId":1,"balance":100},{"jobId":1001,"id":2,"timeSegmentId":1,"balance":100},{"jobId":1002,"id":1,"timeSegmentId":1,"balance":100},{"jobId":1002,"id":1,"timeSegmentId":1,"balance":100},{"jobId":1003,"id":1,"timeSegmentId":1,"balance":100},{"jobId":1003,"id":1,"timeSegmentId":2,"balance":100}];

const mergeResults = _.groupBy(BSSDealresults, ({ id, jobId, timeSegmentId }) => `${id}-${jobId}-${timeSegmentId}`);

const AggregateResults = _.map(mergeResults, group => ({
  ...group[0],
  balance: _.sumBy(group, 'balance')
}));

console.log(AggregateResults);

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