Related
I have a table will the data with exist data below:
Select Date, [Closing Balance] from StockClosing
Date | Closing Quantity
---------------------------
20200828 | 5
20200901 | 10
20200902 | 8
20200904 | 15
20200905 | 18
There are some missing date on the table, example 20200829 to 20200831 and 20200903.
Those closing quantity of the missing date will be follow as per previous day closing quantity.
I would like select the table result in a full range of date (show everyday) with the closing quantity. Expected result,
Date | Closing Quantity
---------------------------
20200828 | 5
20200829 | 5
20200830 | 5
20200831 | 5
20200901 | 10
20200902 | 8
20200903 | 8
20200904 | 15
20200905 | 18
Beside using cursor/for loop to insert the missing date and data 1 by 1, is that any SQL command can do it at once?
You have option to use recursive CTE.
For reference Click Here
;with cte as(
select max(date) date from YourTable
),cte1 as (
select min(date) date from YourTable
union all
select dateadd(day,1,cte1.date) date from cte1 where date<(select date from cte)
)select c.date,isnull(y.[Closing Quantity],
(select top 1 a.[Closing Quantity] from YourTable a where c.date>a.date order by a.date desc) )
as [Closing Quantity]
from cte1 c left join YourTable y on c.date=y.date
The easiest way to do this is to use LAST_VALUE along with the IGNORE NULLS option. Sadly, SQL Server does not support this. There is a workaround using analytic functions, but I would actually offer this simple option, which uses a correlated subquery to fill in the missing values:
WITH dates AS (
SELECT '20200828' AS Date UNION ALL
SELECT '20200829' UNION ALL
SELECT '20200830' UNION ALL
SELECT '20200831' UNION ALL
SELECT '20200901' UNION ALL
SELECT '20200902' UNION ALL
SELECT '20200903' UNION ALL
SELECT '20200904' UNION ALL
SELECT '20200905'
)
SELECT
d.Date,
(SELECT TOP 1 t2.closing FROM StockClosing t2
WHERE t2.Date <= d.Date AND t2.closing IS NOT NULL
ORDER BY t2.Date DESC) AS closing
FROM dates d
LEFT JOIN StockClosing t1
ON d.Date = t1.Date;
Demo
Account balance collection, that shows the account balance of a customer at a given day:
+---------------+---------+------------+
| customer_id | value | timestamp |
+---------------+---------+------------+
| 1 | -500 | 2019-10-12 |
| 1 | -300 | 2019-10-11 |
| 1 | -200 | 2019-10-10 |
| 1 | 0 | 2019-10-09 |
| 2 | 200 | 2019-09-10 |
| 1 | 600 | 2019-09-02 |
+---------------+---------+------------+
Notice, that customer #2 had no updates to his account balance in October.
I want to get the last account balance per customer per month. If there has been no account balance update for a customer in a given month, the last known account balance should be transferred to the current month. The result should look like that:
+---------------+---------+------------+
| customer_id | value | timestamp |
+---------------+---------+------------+
| 1 | -500 | 2019-10-12 |
| 2 | 200 | 2019-10-10 |
| 2 | 200 | 2019-09-10 |
| 1 | 600 | 2019-09-02 |
+---------------+---------+------------+
Since the account balance of customer #2 was not updated in October but in September, we create a copy of the row from September changing the date to October. Any ideas how to achieve this in BigQuery?
Below is for BigQuery Standard SQL
#standardSQL
WITH customers AS (
SELECT DISTINCT customer_id FROM `project.dataset.table`
), months AS (
SELECT month FROM (
SELECT DATE_TRUNC(MIN(timestamp), MONTH) min_month, DATE_TRUNC(MAX(timestamp), MONTH) max_month
FROM `project.dataset.table`
), UNNEST(GENERATE_DATE_ARRAY(min_month, max_month, INTERVAL 1 MONTH)) month
)
SELECT customer_id,
IFNULL(value, LEAD(value) OVER(win)) value,
IFNULL(timestamp, DATE_ADD(LEAD(timestamp) OVER(win), INTERVAL DATE_DIFF(month, LEAD(month) OVER(win), MONTH) MONTH)) timestamp
FROM months, customers
LEFT JOIN (
SELECT DATE_TRUNC(timestamp, MONTH) month, customer_id,
ARRAY_AGG(STRUCT(value, timestamp) ORDER BY timestamp DESC LIMIT 1)[OFFSET(0)].*
FROM `project.dataset.table`
GROUP BY month, customer_id
) USING(month, customer_id)
WINDOW win AS (PARTITION BY customer_id ORDER BY month DESC)
if to apply to sample data from your question - as it is in below example
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 customer_id, -500 value, DATE '2019-10-12' timestamp UNION ALL
SELECT 1, -300, '2019-10-11' UNION ALL
SELECT 1, -200, '2019-10-10' UNION ALL
SELECT 2, 200, '2019-09-10' UNION ALL
SELECT 2, 100, '2019-08-11' UNION ALL
SELECT 2, 50, '2019-07-12' UNION ALL
SELECT 1, 600, '2019-09-02'
), customers AS (
SELECT DISTINCT customer_id FROM `project.dataset.table`
), months AS (
SELECT month FROM (
SELECT DATE_TRUNC(MIN(timestamp), MONTH) min_month, DATE_TRUNC(MAX(timestamp), MONTH) max_month
FROM `project.dataset.table`
), UNNEST(GENERATE_DATE_ARRAY(min_month, max_month, INTERVAL 1 MONTH)) month
)
SELECT customer_id,
IFNULL(value, LEAD(value) OVER(win)) value,
IFNULL(timestamp, DATE_ADD(LEAD(timestamp) OVER(win), INTERVAL DATE_DIFF(month, LEAD(month) OVER(win), MONTH) MONTH)) timestamp
FROM months, customers
LEFT JOIN (
SELECT DATE_TRUNC(timestamp, MONTH) month, customer_id,
ARRAY_AGG(STRUCT(value, timestamp) ORDER BY timestamp DESC LIMIT 1)[OFFSET(0)].*
FROM `project.dataset.table`
GROUP BY month, customer_id
) USING(month, customer_id)
WINDOW win AS (PARTITION BY customer_id ORDER BY month DESC)
-- ORDER BY month DESC, customer_id
result is
Row customer_id value timestamp
1 1 -500 2019-10-12
2 2 200 2019-10-10
3 1 600 2019-09-02
4 2 200 2019-09-10
5 1 null null
6 2 100 2019-08-11
7 1 null null
8 2 50 2019-07-12
The following query should mostly answer your question by creating a 'month-end' record for each customer for every month and getting the most recent balance:
with
-- Generate a set of months
month_begins as (
select dt from unnest(generate_date_array('2019-01-01','2019-12-01', interval 1 month)) dt
),
-- Get the month ends
month_ends as (
select date_sub(date_add(dt, interval 1 month), interval 1 day) as month_end_date from month_begins
),
-- Cross Join and group so we get 1 customer record for every month to account for
-- situations where customer doesn't change balance in a month
user_month_ends as (
select
customer_id,
month_end_date
from `project.dataset.table`
cross join month_ends
group by 1,2
),
-- Fan out so for each month end, you get all balances prior to month end for each customer
values_prior_to_month_end as (
select
customer_id,
value,
timestamp,
month_end_date
from `project.dataset.table`
inner join user_month_ends using(customer_id)
where timestamp <= month_end_date
),
-- Order by most recent balance before month end, even if it was more than 1+ months ago
ordered as (
select
*,
row_number() over (partition by customer_id, month_end_date order by timestamp desc) as my_row
from values_prior_to_month_end
),
-- Finally, select only the most recent record for each customer per month
final as (
select
* except(my_row)
from ordered
where my_row = 1
)
select * from final
order by customer_id, month_end_date desc
A few caveats:
I did not order results to match your desired result set, and I also kept a month-end date to illustrate the concept. You can easily change the ordering and exclude unneeded fields.
In the month_begins CTE, I set a range of months into the future, so your result set will contain the most recent balance of 'future months'. To make this a bit prettier, consider changing '2019-12-01' to 'current_date()' and your query will always return to the end of the current month.
Your timestamp field looks to be dates, so I used date logic, but you should be able to apply the same principles to use timestamp logic if your underlying fields are actual timestamps.
In your result set, I'm not sure why your 2nd row (customer 2) would have a timestamp of '2019-10-10', that seems arbitrary as customer 2 has no 2nd balance record.
I purposefully split the logic into several CTEs so I could comment on each step easier, you could definitely perform several steps in the same code block for a more condensed query.
I have the following data set:
I want to create a new column that sums the last 7 days of sales. So the query result should look be the following:
Pls help
Thanks!
In standard SQL, you would use a window function -- assuming you have data for each day:
select t.*,
sum(sales) over (partition by itemid order by date rows between 6 preceding and current row) as sales_7
from t;
use sum() aggregate function and group by
select country,itemid,year,monthnumber,week sum(sales) as sales_last_7days from your_table
where date>=DATEADD(day, -7, getdate()) and date< getdate()
group by country,itemid,year,monthnumber,week
with window:
select (list other columns here), sum(sum(sales)) over
(partition by week
order by day
rows between 6 preceding and current row)
from table
group by date, week;
note that week doesen't change group by beacause a date is reffered to one week only, but it is needed in window.
Seems you are working with SQL Server if so, then you can use apply :
select t.*, t1.[last7day]
from table t outer apply
(select sum(t1.sales) as [last7day]
from table t1
where t.itemid = t1.itemid and
t1.date <= dateadd(day, -6, t.dt)
) t1;
If you don't have exactly one day for each row, for example if you have a list of transactions...
The below example completely confused me the first time I saw it, so I've tried to comment as much as I can to explain what's happening.
Suppose we have a table tbl with date column dt and amount column amt, and for each date in tbl we want to return a rolling sum of the amount from the current day and the past 6 days.
select distinct -- see note after code on what this distinct is doing.
dt
, ( -- Has to be in brackets to denote we're returning 1 value per row.
-- for each row of T1:
select sum(b.amt) -- the sum of amounts in T2. The where clause will restrict which rows in T2 will be summed.
from tbl T2
where T2.dt between T1.dt - 6 and T1.dt -- for each row in T1, give me all rows in T2 where the date is between 6 days before this T1 row's date and T1 row's date, giving us our rolling sum
-- WARNING: CHECK YOUR VERSION OF SQL FOR HOW TO SUBTRACT DAYS FROM A DATE, I'VE MADE IT (T1.dt - 6) FOR SIMPLICITY
-- we don't need a group by, because we're returning one value for each row in T1
)
from tbl T1
We have a main version of tbl, aliased T1. We then have a secondary table, aliased T2. For each row in T1, we're going to ask for a set of rows in T2 that we're going to sum before giving it to our main query.
To understand what's happening, run the code without the distinct. You'll notice that we have the same number of rows as in tbl, because the T2 statement is happening for every row in T1.
Notes:
If you have any days for which no rows exist in your table you will not get a calculation for this day. To be certain this doesn't happen, join your table to a table containing a distinct list of consecutive dates, and use this as your date column.
If you have nulls in your amount column the calculation will still work, but if the rolling average contains only nulls you will have null instead of 0 as your result. If that troubles you convert all your nulls to zero's before (or after) you use the query.
The beginning of the period will have a 'ramp up'. But this would be the same whatever method you use to do a rolling sum. If it bothers you, don't return the first 6 days.
Finally a worked example if you're playing along at home using SQL Server:
with tbl as (
-- a list of transactions from 1.10.2019 to 14.10.2019
select cast('2019-10-01' as date) dt, 1 amt
union select cast('2019-10-02' as date), 4
union select cast('2019-10-01' as date), 10
union select cast('2019-10-03' as date), 3
union select cast('2019-10-04' as date), 20
union select cast('2019-10-04' as date), 2
union select cast('2019-10-04' as date), 12
union select cast('2019-10-04' as date), 17
union select cast('2019-10-05' as date), null -- a whole week of null values because we all had the week off... I hope this data wasn't important
union select cast('2019-10-06' as date), null
union select cast('2019-10-07' as date), null
union select cast('2019-10-08' as date), null
union select cast('2019-10-09' as date), null
union select cast('2019-10-10' as date), null
union select cast('2019-10-10' as date), null
union select cast('2019-10-10' as date), null
union select cast('2019-10-11' as date), null
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-13' as date), 2
union select cast('2019-10-14' as date), 1000
)
select distinct
a.dt
, (
select sum(b.amt)
from tbl b
where b.dt between dateadd(dd, -6, a.dt) and a.dt
) past_7_days_amt
from tbl a
Returns:
+------------+-----------------+
| dt | past_7_days_amt |
+------------+-----------------+
| 2019-10-01 | 11 |
| 2019-10-02 | 15 |
| 2019-10-03 | 18 |
| 2019-10-04 | 69 |
| 2019-10-05 | 69 |
| 2019-10-06 | 69 |
| 2019-10-07 | 69 |
| 2019-10-08 | 58 |
| 2019-10-09 | 54 |
| 2019-10-10 | 51 |
| 2019-10-11 | NULL |
| 2019-10-12 | 1 |
| 2019-10-13 | 3 |
| 2019-10-14 | 1003 |
+------------+-----------------+
I have a table with the following info
|date | user_id | week_beg | month_beg|
SQL to create table with test values:
CREATE TABLE uniques
(
date DATE,
user_id INT,
week_beg DATE,
month_beg DATE
)
INSERT INTO uniques VALUES ('2013-01-01', 1, '2012-12-30', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-03', 3, '2012-12-30', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-06', 4, '2013-01-06', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-07', 4, '2013-01-06', '2013-01-01')
INPUT TABLE:
| date | user_id | week_beg | month_beg |
| 2013-01-01 | 1 | 2012-12-30 | 2013-01-01 |
| 2013-01-03 | 3 | 2012-12-30 | 2013-01-01 |
| 2013-01-06 | 4 | 2013-01-06 | 2013-01-01 |
| 2013-01-07 | 4 | 2013-01-06 | 2013-01-01 |
OUTPUT TABLE:
| date | time_series | cnt |
| 2013-01-01 | D | 1 |
| 2013-01-01 | W | 1 |
| 2013-01-01 | M | 1 |
| 2013-01-03 | D | 1 |
| 2013-01-03 | W | 2 |
| 2013-01-03 | M | 2 |
| 2013-01-06 | D | 1 |
| 2013-01-06 | W | 1 |
| 2013-01-06 | M | 3 |
| 2013-01-07 | D | 1 |
| 2013-01-07 | W | 1 |
| 2013-01-07 | M | 3 |
I want to calculate the number of distinct user_id's for a date:
For that date
For that week up to that date (Week to date)
For the month up to that date (Month to date)
1 is easy to calculate.
For 2 and 3 I am trying to use such queries:
SELECT
date,
'W' AS "time_series",
(COUNT DISTINCT user_id) COUNT (user_id) OVER (PARTITION BY week_beg) AS "cnt"
FROM user_subtitles
SELECT
date,
'M' AS "time_series",
(COUNT DISTINCT user_id) COUNT (user_id) OVER (PARTITION BY month_beg) AS "cnt"
FROM user_subtitles
Postgres does not allow window functions for DISTINCT calculation, so this approach does not work.
I have also tried out a GROUP BY approach, but it does not work as it gives me numbers for whole week/months.
Whats the best way to approach this problem?
Count all rows
SELECT date, '1_D' AS time_series, count(DISTINCT user_id) AS cnt
FROM uniques
GROUP BY 1
UNION ALL
SELECT DISTINCT ON (1)
date, '2_W', count(*) OVER (PARTITION BY week_beg ORDER BY date)
FROM uniques
UNION ALL
SELECT DISTINCT ON (1)
date, '3_M', count(*) OVER (PARTITION BY month_beg ORDER BY date)
FROM uniques
ORDER BY 1, time_series
Your columns week_beg and month_beg are 100 % redundant and can easily be replaced by
date_trunc('week', date + 1) - 1 and date_trunc('month', date) respectively.
Your week seems to start on Sunday (off by one), therefore the + 1 .. - 1.
The default frame of a window function with ORDER BY in the OVER clause uses is RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW. That's exactly what you need.
Use UNION ALL, not UNION.
Your unfortunate choice for time_series (D, W, M) does not sort well, I renamed to make the final ORDER BY easier.
This query can deal with multiple rows per day. Counts include all peers for a day.
More about DISTINCT ON:
Select first row in each GROUP BY group?
DISTINCT users per day
To count every user only once per day, use a CTE with DISTINCT ON:
WITH x AS (SELECT DISTINCT ON (1,2) date, user_id FROM uniques)
SELECT date, '1_D' AS time_series, count(user_id) AS cnt
FROM x
GROUP BY 1
UNION ALL
SELECT DISTINCT ON (1)
date, '2_W'
,count(*) OVER (PARTITION BY (date_trunc('week', date + 1)::date - 1)
ORDER BY date)
FROM x
UNION ALL
SELECT DISTINCT ON (1)
date, '3_M'
,count(*) OVER (PARTITION BY date_trunc('month', date) ORDER BY date)
FROM x
ORDER BY 1, 2
DISTINCT users over dynamic period of time
You can always resort to correlated subqueries. Tend to be slow with big tables!
Building on the previous queries:
WITH du AS (SELECT date, user_id FROM uniques GROUP BY 1,2)
,d AS (
SELECT date
,(date_trunc('week', date + 1)::date - 1) AS week_beg
,date_trunc('month', date)::date AS month_beg
FROM uniques
GROUP BY 1
)
SELECT date, '1_D' AS time_series, count(user_id) AS cnt
FROM du
GROUP BY 1
UNION ALL
SELECT date, '2_W', (SELECT count(DISTINCT user_id) FROM du
WHERE du.date BETWEEN d.week_beg AND d.date )
FROM d
GROUP BY date, week_beg
UNION ALL
SELECT date, '3_M', (SELECT count(DISTINCT user_id) FROM du
WHERE du.date BETWEEN d.month_beg AND d.date)
FROM d
GROUP BY date, month_beg
ORDER BY 1,2;
SQL Fiddle for all three solutions.
Faster with dense_rank()
#Clodoaldo came up with a major improvement: use the window function dense_rank(). Here is another idea for an optimized version. It should be even faster to exclude daily duplicates right away. The performance gain grows with the number of rows per day.
Building on a simplified and sanitized data model
- without the redundant columns
- day as column name instead of date
date is a reserved word in standard SQL and a basic type name in PostgreSQL and shouldn't be used as identifier.
CREATE TABLE uniques(
day date -- instead of "date"
,user_id int
);
Improved query:
WITH du AS (
SELECT DISTINCT ON (1, 2)
day, user_id
,date_trunc('week', day + 1)::date - 1 AS week_beg
,date_trunc('month', day)::date AS month_beg
FROM uniques
)
SELECT day, count(user_id) AS d, max(w) AS w, max(m) AS m
FROM (
SELECT user_id, day
,dense_rank() OVER(PARTITION BY week_beg ORDER BY user_id) AS w
,dense_rank() OVER(PARTITION BY month_beg ORDER BY user_id) AS m
FROM du
) s
GROUP BY day
ORDER BY day;
SQL Fiddle demonstrating the performance of 4 faster variants. It depends on your data distribution which is fastest for you.
All of them are about 10x as fast as the correlated subqueries version (which isn't bad for correlated subqueries).
Without correlated subqueries. SQL Fiddle
with u as (
select
"date", user_id,
date_trunc('week', "date" + 1)::date - 1 week_beg,
date_trunc('month', "date")::date month_beg
from uniques
)
select
"date", count(distinct user_id) D,
max(week_dr) W, max(month_dr) M
from (
select
user_id, "date",
dense_rank() over(partition by week_beg order by user_id) week_dr,
dense_rank() over(partition by month_beg order by user_id) month_dr
from u
) s
group by "date"
order by "date"
Try
SELECT
*
FROM
(
SELECT dates, count(user_id), 'D' as timesereis FROM users_data GROUP BY dates
UNION
SELECT max(dates), count(user_id), 'W' FROM users_data GROUP BY date_part('year',dates)+date_part('week',dates)
UNION
SELECT max(dates), count(user_id), 'M' FROM users_data GROUP BY date_part('year',dates)+date_part('week',dates)
) tEMP order by dates, timesereis
SQLFIDDLE
Try queries like this
SELECT count(distinct user_id), date_format(date, '%Y-%m-%d') as date_period
FROM uniques
GROUP By date_period
I have a postgres table with customer ID's, dates, and integers. I need to find the average of the top 3 records for each customer ID that have dates within the last year. I can do it with a single ID using the SQL below (id is the customer ID, weekending is the date, and maxattached is the integer).
One caveat: the maximum values are per month, meaning we're only looking at the highest value in a given month to create our dataset, thus why we're extracting month from the date.
SELECT
id,
round(avg(max),0)
FROM
(
select
id,
extract(month from weekending) as month,
extract(year from weekending) as year,
max(maxattached) as max
FROM
myTable
WHERE
weekending >= now() - interval '1 year' AND
id=110070 group by id,month,year
ORDER BY
max desc limit 3
) AS t
GROUP BY id;
How can I expand this query to include all ID's and a single averaged number for each one?
Here is some sample data:
ID | MaxAttached | Weekending
110070 | 5 | 2011-11-10
110070 | 6 | 2011-11-17
110071 | 4 | 2011-11-10
110071 | 7 | 2011-11-17
110070 | 3 | 2011-12-01
110071 | 8 | 2011-12-01
110070 | 5 | 2012-01-01
110071 | 9 | 2012-01-01
So, for this sample table, I would expect to receive the following results:
ID | MaxAttached
110070 | 5
110071 | 8
This averages the highest value in a given month for each ID (6,3,5 for 110070 and 7,8,9 for 110071)
Note: postgres version 8.1.15
First - get the max(maxattached) for every customer and month:
SELECT id,
max(maxattached) as max_att
FROM myTable
WHERE weekending >= now() - interval '1 year'
GROUP BY id, date_trunc('month',weekending);
Next - for every customer rank all his values:
SELECT id,
max_att,
row_number() OVER (PARTITION BY id ORDER BY max_att DESC) as max_att_rank
FROM <previous select here>;
Next - get the top 3 for every customer:
SELECT id,
max_att
FROM <previous select here>
WHERE max_att_rank <= 3;
Next - get the avg of the values for every customer:
SELECT id,
avg(max_att) as avg_att
FROM <previous select here>
GROUP BY id;
Next - just put all the queries together and rewrite/simplify them for your case.
UPDATE: Here is an SQLFiddle with your test data and the queries: SQLFiddle.
UPDATE2: Here is the query, that will work on 8.1 :
SELECT customer_id,
(SELECT round(avg(max_att),0)
FROM (SELECT max(maxattached) as max_att
FROM table1
WHERE weekending >= now() - interval '2 year'
AND id = ct.customer_id
GROUP BY date_trunc('month',weekending)
ORDER BY max_att DESC
LIMIT 3) sub
) as avg_att
FROM customer_table ct;
The idea - to take your initial query and run it for every customer (customer_table - table with all unique id for customers).
Here is SQLFiddle with this query: SQLFiddle.
Only tested on version 8.3 (8.1 is too old to be on SQLFiddle).
8.3 version
8.3 is the oldest version I've got access to, so I can't guarantee it'll work in 8.1
I'm using a temporary table to work out the best three records.
CREATE TABLE temp_highest_per_month as
select
id,
extract(month from weekending) as month,
extract(year from weekending) as year,
max(maxattached) as max_in_month,
0 as priority
FROM
myTable
WHERE
weekending >= now() - interval '1 year'
group by id,month,year;
UPDATE temp_highest_per_month t
SET priority =
(select count(*) from temp_highest_per_month t2
where t2.id = t.id and
(t.max_in_month < t2.max_in_month or
(t.max_in_month= t2.max_in_month and
t.year * 12 + t.month > t2.year * 12 + t.month)));
select id,round(avg(max_in_month),0)
from temp_highest_per_month
where priority <= 3
group by id;
The year & month are included in the working out the priority so that if two months have the same maximum, they'll still be included in the numbering correctly.
9.1 version
Similar to Igor's answer, but I used the With clause to split the steps.
with highest_per_month as
( select
id,
extract(month from weekending) as month,
extract(year from weekending) as year,
max(maxattached) as max_in_month
FROM
myTable
WHERE
weekending >= now() - interval '1 year'
group by id,month,year),
prioritised as
( select id, month, year, max_in_month,
row_number() over (partition by id, month, year
order by max_in_month desc)
as priority
from highest_per_month
)
select id, round(avg(max_in_month),0)
from prioritised
where priority <= 3
group by id;