Question

我在Hive中有一个如下表。

我想为seconds相同的列计算id中的时差，并在time_diff列中获取值。

Table

+-----+---------+------------------------+
| id  |  event  |            eventdate   |
+-----+---------+------------------------+
| 1   | sent    | 2017-11-23 03:49:50.0  |
| 1   | sent    | 2017-11-23 03:49:59.0  |
| 2   | sent    | 2017-11-23 04:49:59.0  |
| 1   | click   | 2017-11-24 03:49:50.0  |
+-----+---------+------------------------+

我已经完成了以下操作

SELECT *, coalesce(unix_timestamp(eventdate) - unix_timestamp(LAG(eventdate) OVER(PARTITION BY ID ORDER BY eventdate)),0) time_diff FROM Table;

Result

+-----+---------+------------------------+-----------+
| id  |  event  |            eventdate   |time_diff  |
+-----+---------+------------------------+-----------+
| 1   | sent    | 2017-11-23 03:49:50.0  | 0         |
| 1   | sent    | 2017-11-23 03:49:59.0  | 9         |
| 2   | sent    | 2017-11-23 04:49:59.0  | 0         |
| 1   | click   | 2017-11-24 03:49:50.0  | 86391     |
+-----+---------+------------------------+-----------+

我得到了想要的东西，但有一个例外。在id列中1为event和sent为time_diff的结果中，有两个值0和9 。在应用滞后函数后，我希望所有sent事件在0列中都有time_diff。

Expected result：

+-----+---------+------------------------+-----------+
| id  |  event  |            eventdate   |time_diff  |
+-----+---------+------------------------+-----------+
| 1   | sent    | 2017-11-23 03:49:50.0  | 0         |
| 1   | sent    | 2017-11-23 03:49:59.0  | 0         |
| 2   | sent    | 2017-11-23 04:49:59.0  | 0         |
| 1   | click   | 2017-11-24 03:49:50.0  | 86391     |
+-----+---------+------------------------+-----------+

如何获得预期的结果？

Answer 1

您可以使用case表达式：

SELECT *,
       (case when event = 'sent' then 0
             else coalesce(unix_timestamp(eventdate) - unix_timestamp(LAG(eventdate) OVER(PARTITION BY ID ORDER BY eventdate)), 0)
        end) as time_diff 
FROM Table;

在Hive / SQL中使用滞后功能时列中的默认值

1 个答案: