Having the following table:
+--------+-------+-------+-------+
| categ. | elem. | atr_1 | atr_2 |
+--------+-------+-------+-------+
| 1 | 1 | 2 | 1 |
| 1 | 2 | 2 | 2 |
| 2 | 3 | 1 | 3 |
| 2 | 4 | 1 | 3 |
+--------+-------+-------+-------+
...I'm trying to obtain the resulting table showing the best element per category:
+--------+--------+
| categ. | elem. |
+--------+--------+
| 1 | 2 |
| 2 | 3 |
+- ------+--------+
In order to determine which element is the 'best' per category the system needs to check which element has the max(atr_1) per category. If more than one element is retrieved will look at max(atr_2) of the retrieved elements. If more than one element is retrieved one of the resulting ones will be randomly assigned to the category.
I'm not able to figure out how to aggregate and use the conditional statements in order to compose the required query. Any suggestion?
I'm using standard SQL in Google BigQuery.
Thanks in advance
The BigQuery'ish way to solve this would just use aggregation:
select (array_agg(t order by atr_1 desc, atr_2 desc limit 1))[ordinal(1)].* except (atr_1, atr_2)
from t
group by categ;
We can use ROW_NUMBER here:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY category ORDER BY atr_1 DESC, atr_2 DESC) rn
FROM yourTable
)
SELECT category, element
FROM cte
WHERE rn = 1;
Below is for BigQuery Standard SQL
#standardSQL
SELECT AS VALUE
ARRAY_AGG(
STRUCT(categ, elem) ORDER BY atr_1 DESC, atr_2 DESC LIMIT 1
)[OFFSET(0)]
FROM `project.dataset.table`
GROUP BY categ
if to apply to sample data from your question - output is
Row categ elem
1 1 2
2 2 3
Related
I'm implementing a view to store leaderboard data of the top 10 users that is computed using an expensive COUNT(*). I'm planning on the view to look something like this:
id SERIAL PRIMARY KEY
user_id TEXT
type TEXT
rank INTEGER
count INTEGER
-- adding an index to user_id
-- adding a two-column unique index to user_id and type
I'm having trouble with seeing how this view should be created to properly account for the rank and type. Essentially, I have a big table (~30 million rows) like this:
+----+---------+---------+----------------------------+
| id | user_id | type | created_at |
+----+---------+---------+----------------------------+
| 1 | 1 | Diamond | 2021-05-11 17:35:18.399517 |
| 2 | 1 | Diamond | 2021-05-12 17:35:17.399517 |
| 3 | 1 | Diamond | 2021-05-12 17:35:18.399517 |
| 4 | 2 | Diamond | 2021-05-13 17:35:18.399517 |
| 5 | 1 | Clay | 2021-05-14 17:35:18.399517 |
| 6 | 1 | Clay | 2021-05-15 17:35:18.399517 |
+----+---------+---------+----------------------------+
With the table above, I'm trying to achieve something like this:
+----+---------+---------+------+-------+
| id | user_id | type | rank | count |
+----+---------+---------+------+-------+
| 1 | 1 | Diamond | 1 | 3 |
| 2 | 2 | Diamond | 2 | 1 |
| 3 | 1 | Clay | 1 | 2 |
| 4 | 1 | Weekly | 1 | 5 | -- 3 diamonds + 2 clay obtained between Mon-Sun
| 5 | 2 | Weekly | 2 | 1 |
+----+---------+---------+------+-------+
By Weekly I am counting the time from the last Sunday to the upcoming Sunday.
Is this doable using only SQL, or is some kind of script needed? If doable, how would this be done? It's worth mentioning that there are thousands of different types, so not having to manually specify type would be preferred.
If there's anything unclear, please let me know and I'll do my best to clarify. Thanks!
The "weekly" rows are produced in a different way compared to the "user" rows (I called them two different "categories"). To get the result you want you can combine two queries using UNION ALL.
For example:
select 'u' as category, user_id, type,
rank() over(partition by type order by count(*) desc) as rk,
count(*) as cnt
from scores
group by user_id, type
union all
select 'w', user_id, 'Weekly',
rank() over(order by count(*) desc),
count(*) as cnt
from scores
group by user_id
order by category, type desc, rk
Result:
category user_id type rk cnt
--------- -------- -------- --- ---
u 1 Diamond 1 3
u 2 Diamond 2 1
u 1 Clay 1 2
w 1 Weekly 1 5
w 2 Weekly 2 1
See running example at DB Fiddle.
Note: For the sake of simplicity I left the filtering by timestamp out of the query. If you really needed to include only the rows of the last 7 days (or other period of time), it would be a matter of adding a WHERE clause in both subqueries.
I think this is what you were talking about, right?
WITH scores_plus_weekly AS ((
SELECT id, user_id, 'Weekly' AS type, created_at
FROM scores
WHERE created_at BETWEEN '2021-05-10' AND '2021-05-17'
)
UNION (
SELECT * FROM scores
))
SELECT
row_number() OVER (ORDER BY CASE "type" WHEN 'Diamond' THEN 0 WHEN 'Clay' THEN 1 ELSE 2 END, count(*) DESC) as "id",
user_id,
"type",
row_number() OVER (PARTITION BY count(*) DESC) as "rank",
count(*)
FROM scores_plus_weekly
GROUP BY user_id, "type"
ORDER BY "id";
I'm sure this is not the only way, but I thought the result wasn't too complex. This query first combines the original database with all scores from this week. For the sake of consistency I picked a date range that matches your entire example set. It then groups by user_id and type to get the counts for each combination. The row_numbers will give you the overall rank and the rank per type. A big part of this query consists of sorting by type, so if you're joining another table that contains the order or priority of the types, the CASE can probably be simplified.
Then, lastly, this entire query can be caught in a view using the CREATE VIEW score_ranks AS , followed by your query.
I working on a query for SQL Server 2016. I have order by serial_no and group by pay_type and I would like to add row number same example below
row_no | pay_type | serial_no
1 | A | 4000118445
2 | A | 4000118458
3 | A | 4000118461
4 | A | 4000118473
5 | A | 4000118486
1 | B | 4000118499
2 | B | 4000118506
3 | B | 4000118519
4 | B | 4000118521
1 | A | 4000118534
2 | A | 4000118547
3 | A | 4000118550
1 | B | 4000118562
2 | B | 4000118565
3 | B | 4000118570
4 | B | 4000118572
Help me please..
SELECT
ROW_NUMBER() OVER(PARTITION BY paytype ORDER BY serial_no) as row_no,
paytype, serial_no
FROM table
ORDER BY serial_no
You can assign groups to adjacent pay types that are the same and then use row_number(). For this purpose, the difference of row numbers is a good way to determine the groups:
select row_number() over (partition by pay_type, seqnum - seqnum_2 order by serial_no) as row_no,
t.*
from (select t.*,
row_number() over (order by serial_no) as seqnum,
row_number() over (partition by pay_type order by serial_no) as seqnum_2
from t
) t;
This type of problem is one example of a gaps-and-islands problem. Why does the difference of row numbers work? I find that the simplest way to understand is to look at the results of the subquery.
Here is a db<>fiddle.
add this to your select list
ROW_NUMBER() OVER ( ORDER BY (SELECT 1) )
since you already sorting by your stuff, so you don't need to sorting in your windowing function so consuming less CPU,
I have a table that has a number column and an attribute column like this:
1.
+-----+-----+
| num | att |
-------------
| 1 | a |
| 1 | b |
| 1 | a |
| 2 | a |
| 2 | b |
| 2 | b |
+------------
I want to make the number unique, and the attribute to be whichever attribute occured most often for that number, like this (This is the end-product im interrested in) :
2.
+-----+-----+
| num | att |
-------------
| 1 | a |
| 2 | b |
+------------
I have been working on this for a while and managed to write myself a query that looks up how many times an attribute occurs for a given number like this:
3.
+-----+-----+-----+
| num | att |count|
------------------+
| 1 | a | 1 |
| 1 | b | 2 |
| 2 | a | 1 |
| 2 | b | 2 |
+-----------------+
But I can't think of a way to only select those rows from the above table where the count is the highest (for each number of course).
So basically what I am asking is given table 3, how do I select only the rows with the highest count for each number (Of course an answer describing providing a way to get from table 1 to table 2 directly also works as an answer :) )
You can use aggregation and window functions:
select num, att
from (
select num, att, row_number() over(partition by num order by count(*) desc, att) rn
from mytable
group by num, att
) t
where rn = 1
For each num, this brings the most frequent att; if there are ties, the smaller att is retained.
Oracle has an aggregation function that does this, stats_mode().:
select num, stats_mode(att)
from t
group by num;
In statistics, the most common value is called the mode -- hence the name of the function.
Here is a db<>fiddle.
You can use group by and count as below
select id, col, count(col) as count
from
df_b_sql
group by id, col
i have a table in a postgres DB which has the following structure:
id | date | groupme1 | groupme2 | value
----------------------------------------
1 |
2 |
3 |
Now i want to achieve the following:
Grouping the table after groupme1 and groupme2
Get the value for every group
But only the last entry for each group-compination (odered after date)
Example:
id | date | groupme1 | groupme2 | value
---------------------------------------
| | A | 1 | 4
| | A | 2 | 7
| | A | 3 | 3
| | B | 1 | 9
My current approach looks like this:
SELECT a.*
FROM table AS a
JOIN (SELECT max(id) AS id
FROM table
GROUP BY groupme1, groupme2) AS b
ON a.id = b.id
The Problems of this approach:
it asumes that higher dates have a higher id
it takes long
Is there a faster and better way of doing this? Can windowing function help with this?
I think you just want window functions:
select t.*
from (select t.*,
row_number() over (partition by groupme1, groupme2 order by date desc) as seqnum
from t
) t
where seqnum = 1;
Or, a better way to do this in Postgres uses distinct on:
select distinct on (groupme1, groupme2) t.*
from t
order by groupme1, groupme2, date desc;
I'm not looking for the answer as much as what to search for as I think this is possible. I have a query where the result can be as such:
| ID | CODE | RANK |
I want to base rank off of the code so my I get these results
| 1 | A | 1 |
| 1 | B | 1 |
| 2 | A | 1 |
| 2 | C | 1 |
| 3 | B | 2 |
| 3 | C | 2 |
| 4 | C | 3 |
Basically, based on the group of IDs, if any of the CODEs = a certain value I want to adjust the rank so then I can order by rank first and then other columns. Never sure how to phrase things in SQL.
I tried
CASE WHEN CODE = 'A' THEN 1 WHEN CODE = 'B' THEN 2 ELSE 3 END rank
ORDER BY rank DESC
But I want to keep the ids together, I don't want them broken apart, I was thinking of doing all ranks the same based on the highest if I can't solve it another way?
Thoughts of a SQL function to look at?
You could use the MIN() OVER() analytic function to get the minimum rank value per group, and just order by that;
WITH cte AS (
SELECT id, code,
MIN(CASE WHEN code='A' THEN 1 WHEN code='B' THEN 2 ELSE 3 END)
OVER (PARTITION BY id) rank
FROM mytable
)
SELECT * FROM cte
ORDER BY rank, id, code
An SQLfiddle to test with.