This question already has answers here:
How to get highest count of associated model (Rails)?
(3 answers)
Closed 1 year ago.
I have a Postgres database with a many-to-many association table that's similar to what's down below.
id | item_id | item_tag_id
1 101 3
2 102 3
3 103 1
4 104 2
5 105 2
How can I get the top 5 most associated item_tag_id?
Basically, group by item and order by the count of rows (= count of items in a proper many-to-many design):
SELECT item_id, count(*)
FROM assoc_tbl
GROUP BY 1
ORDER BY 2 DESC
LIMIT 5;
There is a remaining corner-case: how to break ties for the top 5? Either define criteria (resulting in more ORDER BY expressions), or consider WITH TIES. See:
Get top row(s) with highest value, with ties
Can I do a max(count(*)) in SQL?
Related
I currently have two tables, users and coupons
id
first_name
1
Roberta
2
Oliver
3
Shayna
4
Fechin
id
discount
user_id
1
20%
1
2
40%
2
3
15%
3
4
30%
1
5
10%
1
6
70%
4
What I want to do is select from the coupons table until I've selected X users.
so If I chose X = 2 the resulting table would be
id
discount
user_id
1
20%
1
2
40%
2
4
30%
1
5
10%
1
I've tried using both dense_rank and row_number but they return the count of occurrences of each user_id not it's order.
SELECT id,
discount,
user_id,
dense_rank() OVER (PARTITION BY user_id)
FROM coupons
I'm guessing I need to do it in multiple subqueries (which is fine) where the first subquery would return something like
id
discount
user_id
order_of_occurence
1
20%
1
1
2
40%
2
2
3
15%
3
3
4
30%
1
1
5
10%
1
1
6
70%
4
4
which I can then use to filter by what I need.
PS: I'm using postgresql.
You've stated that you want to parameterize the query so that you can retrieve X users. I'm reading that as all coupons for the first X distinct user_ids in coupon id column order.
It appears your attempt was close. dense_rank() is the right idea. Since you want to look over the entire table you can't use partition by. And a sorting column is also required to determine the ranking.
with data as (
select *,
dense_rank() over (order by id) as dr
from coupons
)
select * from data where dr <= <X>;
I've got this table ratings:
id
user_id
type
value
0
0
Rest
4
1
0
Bar
3
2
0
Cine
2
3
0
Cafe
1
4
1
Rest
4
5
1
Bar
3
6
1
Cine
2
7
1
Cafe
5
8
2
Rest
4
9
2
Bar
3
10
3
Cine
2
11
3
Cafe
5
I want to have a table with a row for every pair (user_id, type) for the top 3 rated types through all users (ranked by sum(value) across the whole table).
Desired result:
user_id
type
value
0
Rest
4
0
Cafe
1
0
Bar
3
1
Rest
4
1
Cafe
5
1
Bar
3
2
Rest
4
3
Cafe
5
2
Bar
3
I was able to do this with two queries, one to get the top 3 and then another to get the rows where the type matches the top 3 types.
Does someone know how to fit this into a single query?
Get rows per user for the 3 highest ranking types, where types are ranked by the total sum of their value across the whole table.
So it's not exactly about the top 3 types per user, but about the top 3 types overall. Not all users will have rows for the top 3 types, even if there would be 3 or more types for the user.
Strategy:
Aggregate to get summed values per type (type_rnk).
Take only the top 3. (Break ties ...)
Join back to main table, eliminating any other types.
Order result by user_id, type_rnk DESC
SELECT r.user_id, r.type, r.value
FROM ratings r
JOIN (
SELECT type, sum(value) AS type_rnk
FROM ratings
GROUP BY 1
ORDER BY type_rnk DESC, type -- tiebreaker
LIMIT 3 -- strictly the top 3
) v USING (type)
ORDER BY user_id, type_rnk DESC;
db<>fiddle here
Since multiple types can have the same ranking, I added type to the sort order to break ties alphabetically by their name (as you did not specify otherwise).
Turns out, we don't need window functions - the ones with OVER and, optionally, PARTITION for this. (Since you asked in a comment).
I think you just want row_number(). Based on your results, you seem to want three rows per type, with the highest value:
select t.*
from (select t.*,
row_number() over (partition by type order by value desc) as seqnum
from t
) t
where seqnum <= 3;
Your description suggests that you might just want this per user, which is a slight tweak:
select t.*
from (select t.*,
row_number() over (partition by user order by value desc) as seqnum
from t
) t
where seqnum <= 3;
This question already has answers here:
How to use RANK() in SQL Server
(8 answers)
Closed 3 years ago.
How can I create a sequential value based on two rows within a table, for example, let's say I have a table containing an employee's ID and work state. I would expect the following values:
ID State Expected Value
-----------------------------
1 NY 1
1 PA 2
1 NY 1
2 NC 1
2 FL 2
2 MN 3
You can use dense_rank():
select t.*,
dense_rank() over (partition by id order by state) as expected
from t;
This question already has answers here:
Selecting COUNT(*) with DISTINCT
(7 answers)
Closed 4 years ago.
I need to query a database that has about 10-11 columns, including a column of id's and a column of role codes. Those are the 2 column that i'm interested in.
ID ROLE
1 a
2 a
2 b
2 c
3 a
4 a
4 b
I need to count how many role codes exist for each ID. (Basically like counting the number of times each id exists in the database)
Output should be something like this:
ID Count
1 1
2 3
3 1
4 2
Use count distinct:
SELECT ID, COUNT(DISTINCT ROLE)
FROM YOURTABLE
GROUP BY ID
I have the following table structure.
ITEM TOTAL
----------- -----------------
ID | TITLE ID |ITEMID|VALUE
1 A 1 2 6
2 B 2 1 4
3 C 3 3 3
4 D 4 3 8
5 E 5 1 2
6 F 6 5 4
7 4 5
8 2 8
9 2 7
10 1 3
11 2 2
12 3 6
I am using Apache Derby DB. I need to perform the average calculation in SQL. I need to show the list of item IDs and their average total of the last 3 records.
That is, for ITEM.ID 1, I will go to TOTAL table and select the last 3 records of the rows which are associated with the ITEMID 1. And take average of them. In Derby database, I am able to do this for a given item ID but I cannot make it without giving a specific ID. Let me show you what I've done it.
SELECT ITEM.ID, AVG(VALUE) FROM ITEM, TOTAL WHERE TOTAL.ITEMID = ITEM.ID GROUP BY ITEM.ID
This SQL gives the average of all items in a list. But this calculates for all values of the total tables. I need last 3 records only. So I changed the SQL to this:
SELECT AVG(VALUE) FROM (SELECT ROW_NUMBER() OVER() AS ROWNUM, TOTAL.* FROM TOTAL WHERE ITEMID = 1) AS TR WHERE ROWNUM > (SELECT COUNT(ID) FROM TOTAL WHERE ITEMID = 1) - 3
This works if I supply the item ID 1 or 2 etc. But I cannot do this for all items without giving an item ID.
I tried to do the same thing in ORACLE using partition and it worked. But derby does not support partitioning. There is WINDOW but I could not make use of it.
Oracle one
SELECT ITEMID, AVG(VALUE) FROM(SELECT ITEMID, VALUE, COUNT(*) OVER (PARTITION BY ITEMID) QTY, ROW_NUMBER() OVER (PARTITION BY ITEMID ORDER BY ID) IDX FROM TOTAL ORDER BY ITEMID, ID) WHERE IDX > QTY -3 GROUP BY ITEMID ORDER BY ITEMID
I need to use derby DB for its portability.
The desired output is this
RESULT
-----------------
ITEMID | AVERAGE
1 (9/3)
2 (17/3)
3 (17/3)
4 (5/1)
5 (4/1)
6 NULL
As you have noticed, Derby's support for the SQL 2003 "OLAP Operations" support is incomplete.
There was some initial work (see https://wiki.apache.org/db-derby/OLAPOperations), but that work was only partially completed.
I don't believe anyone is currently working on adding more functionality to Derby in this area.
So yes, Derby has a row_number function, but no, Derby does not (currently) have partition by.