How to calculate average of values without including the last value (sql)? - sql

Question

I have a table. I partition it by the id and want to calculate average of the values previous to the current, without including the current value. Here is a sample table:
+----+-------+------------+
| id | Value | Date |
+----+-------+------------+
| 1 | 51 | 2020-11-26 |
| 1 | 45 | 2020-11-25 |
| 1 | 47 | 2020-11-24 |
| 2 | 32 | 2020-11-26 |
| 2 | 51 | 2020-11-25 |
| 2 | 45 | 2020-11-24 |
| 3 | 47 | 2020-11-26 |
| 3 | 32 | 2020-11-25 |
| 3 | 35 | 2020-11-24 |
+----+-------+------------+
In this case, it means calculating the average of values for dates BEFORE 2020-11-26. This is the expected result
+----+-------+
| id | Value |
+----+-------+
| 1 | 46 |
| 2 | 48 |
| 3 | 33.5 |
+----+-------+
I have calculated it using ROWS N PRECEDING but it appears that this way I average N preceding + last row, and I want to exclude the last row (which is the most recent date in my case).
Here is my query:
SELECT ID,
(avg(Value) OVER(
PARTITION BY ID
ORDER BY Date
ROWS 9 PRECEDING )) as avg9
FROM t1

Answer 1

Then define your window in full using both the start and ends with BETWEEN:
SELECT ID,
(AVG(Value) OVER (PARTITION BY ID ORDER BY Date ROWS BETWEEN 9 PRECEDING AND 1 PRECEDING)) AS avg9
FROM t1;

Answer 2

Why not just filter:
select id, avg(value)
from t1
where date < '2020-11-26'
group by id;
If you want the date to be flexible -- say the most recent value for each date, then:
select id, avg(value)
from (select t1.*,
max(date) over (partition by id) as max_date
from t1
) t1
where date < max_date
group by id;

Answer 3

Do a row_number() over (Partition by id ORDER BY [Date] DESC). This will give a rank = 1 to the row with latest date. Wrap it within a CTE and then calculate avg for each partition where RANK > 1. Please check syntax.
;with a as
(
select id, value, Date, row_number() over (partition by id order by date
desc) as RN
)
select id, avg(Value) from a group by id where r.RN > 1