我如何添加一个计数来对 SQL Hive 中的 null 个值进行排名？

Question

This is what I have right now:

| time  | car_id | order | in_order |
|-------|--------|-------|----------|
| 12:31 | 32     | null  | 0        |
| 12:33 | 32     | null  | 0        |
| 12:35 | 32     | null  | 0        |
| 12:37 | 32     | 123   | 1        |
| 12:38 | 32     | 123   | 1        |
| 12:39 | 32     | 123   | 1        |
| 12:41 | 32     | 123   | 1        |
| 12:43 | 32     | 123   | 1        |
| 12:45 | 32     | null  | 0        |
| 12:47 | 32     | null  | 0        |
| 12:49 | 32     | 321   | 1        |
| 12:51 | 32     | 321   | 1        |

I'm trying to rank orders, including those who have null values, in this case by car_id. This is the result I'm looking for:

| time  | car_id | order | in_order | row |
|-------|--------|-------|----------|-----|
| 12:31 | 32     | null  | 0        | 1   |
| 12:33 | 32     | null  | 0        | 1   |
| 12:35 | 32     | null  | 0        | 1   |
| 12:37 | 32     | 123   | 1        | 2   |
| 12:38 | 32     | 123   | 1        | 2   |
| 12:39 | 32     | 123   | 1        | 2   |
| 12:41 | 32     | 123   | 1        | 2   |
| 12:43 | 32     | 123   | 1        | 2   |
| 12:45 | 32     | null  | 0        | 3   |
| 12:47 | 32     | null  | 0        | 3   |
| 12:49 | 32     | 321   | 1        | 4   |
| 12:51 | 32     | 321   | 1        | 4   |

I just don't know how to manage a count for the null values. Thanks!

Answer 1

You can count the number of non-NULL values before each row and then use dense_rank() :

select t.*,
       dense_rank() over (partition by car_id order by grp) as row
from (select t.*,
             count(order) over (partition by car_id order by time) as grp
      from t
     ) t;