记一次逻辑代码的实现(数组内数据按照指定时间差进行分组)

业务场景

有如下数据：

　　id　　　　　　　　intime　　　　　　　outtime
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26
1190771865,2019-11-26 16:42:46,2019-11-26 16:42:46
1190771865,2019-11-26 17:23:11,2019-11-26 17:23:11
1190771865,2019-11-26 13:27:26,2019-11-26 13:27:26

需求：

　　针对以上数据进行重组，重组规则为：

　　　　对以上数据进行intime升序排序，后一条数据与前一条数据的intime进行比较

　　　　1、如果第二条与第一条数据的差值大于120min，则直接舍弃第一条数据

　　　　2、后一条数据与前一条数据差值小于120，则保留上一条数据的intime，将这一条的intime当做上一条的outtime，继续往后遍历，知道遍历到最后一条数据

　　　　3、如果后一条数据与前一条数据的差值大于120min，则将该条数据当做新的一条数据，继续循环上面的规则

代码实现:

1、将上面数据处理成为一个array，即(aaa,Array(id,intime,outtime))
注：在这之前已经将每条数据中的进出时间转换为了时间戳
mergedDataTmp.map(x => (x._1, .distinct.filter(x => x._2<= x._2)))
      .mapPartitions(iter => {
        iter.map(x => {
          var count = 0
          var iterNum = 0
          val tList = new ListBuffer[(String, (String, String, String))]()
          val vs = x._2.sortWith((a, b) => a._2 < b._2).toIterator
          val vsList = vs.toList
          val vsLength = vsList.length
          var tmpV = ""
          for (t <- vsList) {

            iterNum += 1
            if (count == 0) {
              tList += ((x._1, t))
              count += 1
            } else {
              val compareTime = if (!tList.isEmpty) {
                (DateUtil.dateToTimeStamp(t._2) - DateUtil.dateToTimeStamp(tList.last._2._2)) / 1000 >= 120 * 60
              } else {
                false
              }
              if (compareTime && count == 1) {
                // (如果后一条记录的进时间)-(前一条记录的进时间)>=120min
                tList.remove(tList.length - 1)
                tList += ((x._1, t))
              } else if (compareTime && count > 1) {
                // (如果后一条记录的进时间)-(前一条记录的进时间)>=120min
                val lastRecord = tList.last
                tList(tList.length - 1) = (x._1, (t._1, lastRecord._2._2, tmpV, t._3))
                tList += ((x._1, t))
                count = 1
              } else {
                // 如果后一条记录的进时间 - 前一条记录的进时间<120min
                count += 1
                if (iterNum == vsLength) {
                  val lastRecord = tList.last
                  tList(tList.length - 1) = (x._1, (t._1,lastRecord._2._2, t._3))
                }
                tmpV = t._2
              }
            }
          }
          tList
        })
      }).flatMap(x => x)