JavaScript中按参与者分组对象数组并聚合排班数据

这个需求本质是做数据结构的反转：把原结构中「排班关联多个参与人」的一对多关系，转换成「参与人关联多个排班」的一对多关系，用中间哈希表做映射的方式实现效率最高，不需要多层嵌套循环查找。

实现代码 #

// 原始接口返回数据，直接替换成接口实际返回的数组即可
const rawData = [
   {
      "participant_details":[
         {"participant_id":2,"participant_name":"Bond James"},
         {"participant_id":3,"participant_name":"Barkley Charles"}
      ],
      "schedule_details":{
         "schedule_id":17,"schedule_name":"bug test","job_type":"Registered Nurse",
         "schedule_type":"Shared/Grouped Schedule","number_of_shifts":10,
         "start_date":"2022-05-30 23:00:00","end_date":"2022-06-03 06:00:00"
      }
   },
   {
      "participant_details":[{"participant_id":3,"participant_name":"Barkley Charles"}],
      "schedule_details":{
         "schedule_id":18,"schedule_name":"June Chuk Schedule","job_type":"Coordinator - Operations",
         "schedule_type":"Individual Schedule","number_of_shifts":6,
         "start_date":"2022-06-02 09:00:00","end_date":"2022-06-07 15:00:00"
      }
   },
   {
      "participant_details":[
         {"participant_id":2,"participant_name":"Bond James"},
         {"participant_id":3,"participant_name":"Barkley Charles"}
      ],
      "schedule_details":{
         "schedule_id":19,"schedule_name":"Grouped Re-assignment test","job_type":"People & Culture",
         "schedule_type":"Shared/Grouped Schedule","number_of_shifts":6,
         "start_date":"2022-06-04 19:00:00","end_date":"2022-06-10 02:00:00"
      }
   }
]

function groupScheduleByParticipant(sourceData) {
  const participantMap = {};
  for (const item of sourceData) {
    const schedule = item.schedule_details;
    for (const p of item.participant_details) {
      // 首次遍历到该参与人时初始化结构
      if (!participantMap[p.participant_id]) {
        participantMap[p.participant_id] = {
          participant_id: p.participant_id,
          participant_name: p.participant_name,
          schedule_details: []
        };
      }
      // 追加当前排班到对应参与人的列表中
      participantMap[p.participant_id].schedule_details.push(schedule);
    }
  }
  return Object.values(participantMap);
}

// 调用即可得到预期格式结果
const targetResult = groupScheduleByParticipant(rawData);

可选优化 #

如果原始数据可能出现「同一个参与人重复关联同一条排班」的异常情况，可以在push排班之前加一行去重判断，避免重复数据：

// 去重逻辑：判断当前排班是否已经在用户的排班列表里
const isExisted = participantMap[p.participant_id].schedule_details
  .some(s => s.schedule_id === schedule.schedule_id);
if (!isExisted) {
  participantMap[p.participant_id].schedule_details.push(schedule);
}

如果处理的数据量特别大，可以把中间存储的普通对象换成ES6的Map结构，查找性能会有小幅提升，核心逻辑不需要改动。

内容的提问来源于stack exchange，提问作者suo