1

我正在尝试计算唯一访问者的数量。我首先按总检查它，而没有按任何时间框架将其分开。

主表（大数据表示例）：

+-----------+----+-------+
|theDateTime|vD  | vis   |
+----------------+-------+
|2018-10-03 |123 |abc    |
|2018-10-04 |123 |abc    |
|2018-10-04 |123 |pqr    |
|2018-10-05 |123 |xyz    |
+-----------+----+-------+

上述的总不同计数将为 3 但当我按天分组时abc计数两次。先是3号，然后是2号。我只想计算第一个。

我的总查询：

select
  d.eId AS vD
  , COUNT(DISTINCT visitorId) AS vis
 from decisions  
 WHERE d.eId = 123 
 AND timestamp BETWEEN unix_timestamp('2018-10-03 00:00:00')*1000 AND 
 unix_timestamp('2018-10-06 12:17:00')*1000
 GROUP BY d.eId
 ORDER BY vId

我的结果：

+----+---------+
| vD | vis     |
+----+---------+
|123 | 3       |
+----+---------+

我的每日查询：

select DISTINCT
cast(from_unixtime(timestamp DIV 1000) AS date) AS theDateTime
, d.eId AS vD
, COUNT(DISTINCT visitorId) AS vis
from decisions  
WHERE timestamp BETWEEN unix_timestamp('2018-10-03 00:00:00')*1000 AND 
unix_timestamp('2018-10-06 12:17:00')*1000
AND d.eId IN (11550123588)
GROUP BY cast(from_unixtime(timestamp DIV 1000) AS date), 
d.vD
ORDER BY vD, theDateTime

我的结果：

+-----------+----+-------+
|theDateTime|vD  | vis   |
+----------------+-------+
|2018-10-03 |123 |   1   |
|2018-10-04 |123 |   2   |
|2018-10-05 |123 |   1   |
+-----------+----+-------+

总数为1122585。WH这超过总和

我知道这是因为以防万一访客在不同的一天重复，当我按天分组时，他被计算了两次。如果他已经在第 1 天被统计，我有没有办法在第 2 天不统计访客？

请帮忙！

4

2 回答 2

0

如果我理解正确，您只需要不同的数据视图。

val df = Seq(("2018-10-03",123,"abc"),
("2018-10-04",123,"abc"),
("2018-10-05",123,"pqr"),
("2018-10-06",123,"xyz")).toDF("theDateTime","vD","vis").withColumn("theDateTime", $"theDateTime".cast("timestamp"));

df.show

import org.apache.spark.sql.functions._
val df1 = df.groupBy("vis").pivot("vD").agg(min("theDateTime")).sort($"123")
df1.show

+---+-------------------+
|vis|                123|
+---+-------------------+
|abc|2018-10-03 00:00:00|
|pqr|2018-10-05 00:00:00|
|xyz|2018-10-06 00:00:00|
+---+-------------------+

现在，如果您按“123”分组，您将能够获得每天的唯一计数。这有帮助吗？

于 2018-10-14T01:45:42.600 回答
0

如果我理解正确，您可以在 SQL 中使用子查询执行此操作：

select min_dt, count(distinct visitorId) AS vis
from (select eid, vis, min(thedatetime) as min_dt
      from decisions d
      where d.eid = 123 and . . .
      group by vis, eid
     ) d
group by min_dt
于 2018-10-14T03:56:12.770 回答