这是我的表:
+--------+----+----------+----+
| station|temp| dateS|hour|
+--------+----+----------+----+
|Roma | 2.2|2018-10-02| 1|
|Roma | 1.5|2018-10-02| 2|
|Roma | 1.4|2018-10-02| 3|
|Roma | 1.4|2018-10-02| 4|
|Milano | 0.6|2018-11-02| 12|
|Milano | 1.0|2018-11-02| 13|
|Napoli | 0.3|2018-12-02| 20|
|Napoli | 0.0|2018-12-02| 21|
|Napoli | 1.8|2018-12-02| 4|
|Napoli | 2.0|2018-12-03| 5|
|Napoli | 1.8|2018-12-03| 6|
+--------+----+----------+----+
我想连续4小时记录在记录中的站(显然是连续4个小时)。例如,仅显示罗姆人,因为同一天连续4个小时(2018-10-02)...
我尝试编写查询,但没有...
我尝试使用铅窗口功能。我通过在站和日期进行分区并按小时订购来安排数据框,并计算了上一个小时的差异。如果我们连续4个小时考虑,则在"差异"列中应该有三个1。为了找出这一点,我已经根据站和日期收集了所有差异,并检查了它是否包含" 1 1 1"。同一代码如下所示。我希望这很有帮助。
//Creating Test Data
val df = Seq(("Roma",2.2,"2018-10-02",1 )
, ("Roma",1.5,"2018-10-02",2 )
, ("Roma",1.4,"2018-10-02",3 )
, ("Roma",1.4,"2018-10-02",4 )
, ("Milano",0.6,"2018-11-02",12 )
, ("Milano",1.0,"2018-11-02",13 )
, ("Napoli",0.3,"2018-12-02",20 )
, ("Napoli",0.0,"2018-12-02",21 )
, ("Napoli",1.8,"2018-12-02",4 )
, ("Napoli",2.0,"2018-12-03",5 )
, ("Napoli",1.8,"2018-12-03",6))
.toDF("station", "temp", "dateS", "hour")
val filterDF = df.withColumn("hour_lead", lead($"hour", 1)
.over(Window.partitionBy("station","dateS")
.orderBy(col("hour")))
.filter($"hour_lead".isNotNull)
.withColumn("hour_diff", $"hour_lead" - $"hour")
.groupBy("station","dateS")
.agg(collect_list($"hour_diff".cast("string")).as("hour_diff_list"))
.withColumn("hour_diff_list_str",
concat(lit(" "),
concat_ws(" ", $"hour_diff_list"),
lit(" ")))
.filter($"hour_diff_list_str".contains(" 1 1 1 "))
filterDF.show(false)
+-------+----------+--------------+------------------+
|station|dateS |hour_diff_list|hour_diff_list_str|
+-------+----------+--------------+------------------+
|Roma |2018-10-02|[1, 1, 1] | 1 1 1 |
+-------+----------+--------------+------------------+