More efficient way to SELECT rows from PARTITION BY - sql

Suppose I have the following table:
+----+-------------+-------------+
| id | step_number | employee_id |
+----+-------------+-------------+
| 1 | 1 | 3 |
| 1 | 2 | 3 |
| 1 | 3 | 4 |
| 2 | 2 | 3 |
| 2 | 3 | 4 |
| 2 | 4 | 5 |
+----+-------------+-------------+
My desired results are:
+----+-------------+-------------+
| id | step_number | employee_id |
+----+-------------+-------------+
| 1 | 1 | 3 |
| 2 | 2 | 3 |
+----+-------------+-------------+
My current solution is:
SELECT
*
FROM
(SELECT
id,
step_number,
MIN(step_number) OVER (PARTITION BY id) AS min_step_number,
employee_id
FROM
table_name) AS t
WHERE
t.step_number = t.min_step_number
Is there a more efficient way I could be doing this?
I'm currently using postgresql, version 12.

In Postgres, I would recommend using distinct on to adress this greatest-n-per-group problem:
select distinct on (id) t.*
from mytbale t
order by id, step_number
This Postgres extension to the SQL standard has usually better performance than the standard approach using window functions (and, as a bonus, the syntax is neater).
Note that this assumes unicity of (id, step_number) tuples: otherwise, the results might be different than those of your query (which allows ties, while distinct on does not).

Related

Merging multiple "state-change" time series

Given a number of tables like the following, representing state-changes at time t of an entity identified by id:
| A | | B |
| t | id | a | | t | id | b |
| - | -- | - | | - | -- | - |
| 0 | 1 | 1 | | 0 | 1 | 3 |
| 1 | 1 | 2 | | 2 | 1 | 2 |
| 5 | 1 | 3 | | 3 | 1 | 1 |
where t is in reality a DateTime field with millisecond precision (making discretisation infeasible), how would I go about creating the following output?
| output |
| t | id | a | b |
| - | -- | - | - |
| 0 | 1 | 1 | 3 |
| 1 | 1 | 2 | 3 |
| 2 | 1 | 2 | 2 |
| 3 | 1 | 2 | 1 |
| 5 | 1 | 3 | 1 |
The idea is that for any given input timestamp, the entire state of a selected entity can be extracted by selecting one row from the resulting table. So the latest state of each variable corresponding to any time needs to be present in each row.
I've tried various JOIN statements, but I seem to be getting nowhere.
Note that in my use case:
rows also need to be joined by entity id
there may be more than two source tables to be merged
I'm running PostgreSQL, but I will eventually translate the query to SQLAlchemy, so a pure SQLAlchemy solution would be even better
I've created a db<>fiddle with the example data.
I think you want a full join and some other manipulations. The ideal would be:
select t, id,
last_value(a.a ignore nulls) over (partition by id order by t) as a,
last_value(b.b ignore nulls) over (partition by id order by t) as b
from a full join
b
using (t, id);
But . . . Postgres doesn't support ignore nulls. So an alternative method is:
select t, id,
max(a) over (partition by id, grp_a) as a,
max(b) over (partition by id, grp_b) as b
from (select *,
count(a.a) over (partition by id order by t) as grp_a,
count(b.b) over (partition by id order by t) as grp_b
from a full join
b
using (t, id)
) ab;

How to select values, where each one depends on a previously aggregated state?

I have the following table:
|-----|-----|
| i d | val |
|-----|-----|
| 1 | 1 |
|-----|-----|
| 2 | 4 |
|-----|-----|
| 3 | 3 |
|-----|-----|
| 4 | 7 |
|-----|-----|
Can I get the following output:
|-----|
| sum |
|-----|
| 1 |
|-----|
| 5 |
|-----|
| 8 |
|-----|
| 1 5 |
|-----|
using a single SQLite3 SELECT-query? I know it could be easily achieved using variables, but SQLite3 lacks those. Maybe some recursive query? Thanks.
No.
In a relational database table rows do not have any order. If you specify an order for the rows, then it's possible to write a query.
Now, you could add an extra column to sort the rows. For example:
| val | sort
|-----|-----
| 1 | 10
| 4 | 20
| 3 | 30
| 7 | 40
The query could be:
select
sum(val) over(order by sort)
from my_table
For the updated question, you can write:
select
sum(val) over(order by id)
from my_table
By using the order of the id column and if you want only the sum column, you can do this:
select (select sum(val) from tablename where id <= t.id) sum
from tablename t

Window functions limited by value in separate column

I have a "responses" table in my postgres database that looks like
| id | question_id |
| 1 | 1 |
| 2 | 2 |
| 3 | 1 |
| 4 | 2 |
| 5 | 2 |
I want to produce a table with the response and question id, as well as the id of the previous response with that same question id, as such
| id | question_id | lag_resp_id |
| 1 | 1 | |
| 2 | 2 | |
| 3 | 1 | 1 |
| 4 | 2 | 2 |
| 5 | 2 | 4 |
Obviously pulling "lag(responses.id) over (order by responses.id)" will pull the previous response id regardless of question_id. I attempted the below subquery, but I know it is wrong since I am basically making a table of all lag ids for each question id in the subquery.
select
responses.question_id,
responses.id as response_id,
(select
lag(r2.id, 1) over (order by r2.id)
from
responses as r2
where
r2.question_id = responses.question_id
)
from
responses
I don't know if I'm on the right track with the subquery, or if I need to do something more advanced (which may involve "partition by", which I do not know how to use).
Any help would be hugely appreciated.
Use partition by. There is no need for a correlated subquery here.
select id,question_id,
lag(id) over (partition by question_id order by id) lag_resp_id
from responses

Filtering using aggregation functions

I would like to filter my table by MIN() function but still keep columns which cant be grouped.
I have table:
+----+----------+----------------------+
| ID | distance | geom |
+----+----------+----------------------+
| 1 | 2 | DSDGSAsd23423DSFF |
| 2 | 11.2 | SXSADVERG678BNDVS4 |
| 2 | 2 | XCZFETEFD567687SDF |
| 3 | 24 | SADASDSVG3423FD |
| 3 | 10 | SDFSDFSDF343DFDGF |
| 4 | 34 | SFDHGHJ546GHJHJHJ |
| 5 | 22 | SDFSGTHHGHGFHUKJYU45 |
| 6 | 78 | SDFDGDHKIKUI45 |
| 6 | 15 | DSGDHHJGHJKHGKHJKJ65 |
+----+----------+----------------------+
This is what I would like to achieve:
+----+----------+----------------------+
| ID | distance | geom |
+----+----------+----------------------+
| 1 | 2 | DSDGSAsd23423DSFF |
| 2 | 2 | XCZFETEFD567687SDF |
| 3 | 10 | SDFSDFSDF343DFDGF |
| 4 | 34 | SFDHGHJ546GHJHJHJ |
| 5 | 22 | SDFSGTHHGHGFHUKJYU45 |
| 6 | 15 | DSGDHHJGHJKHGKHJKJ65 |
+----+----------+----------------------+
it is possible when I use MIN() on distance column and grouping by ID but then I loose my geom which is essential.
The query looks like this:
SELECT "ID", MIN(distance) AS distance FROM somefile GROUP BY "ID"
the result is:
+----+----------+
| ID | distance |
+----+----------+
| 1 | 2 |
| 2 | 2 |
| 3 | 10 |
| 4 | 34 |
| 5 | 22 |
| 6 | 15 |
+----+----------+
but this is not what I want.
Any suggestions?
One common approach to this is to find the minimum values in a derived table that you join with:
SELECT somefile."ID", somefile.distance, somefile.geom
FROM somefile
JOIN (
SELECT "ID", MIN(distance) AS distance FROM somefile GROUP BY "ID"
) t ON t.distance = somefile.distance AND t.ID = somefile.ID;
Sample SQL Fiddle
You need a window function to do this:
SELECT "ID", distance, geom
FROM (
SELECT "ID", distance, geom, rank() OVER (PARTITION BY "ID" ORDER BY distance) AS rnk
FROM somefile) sub
WHERE rnk = 1;
This effectively orders the entire set of rows first by the "ID" value, then by the distance and returns the record for each "ID" where the distance is minimal - no need to do a GROUP BY.
select a.*,b.geom from
(SELECT ID, MIN(distance) AS distance FROM somefile GROUP BY ID) as a
inner join somefile as b on a.id=b.id and a.distance=b.distance
You can use "distinct on" clause of the PostgreSQL.
select distinct on(id) id, distance, geom
from table_name
order by distance;
I think this is what you are exactly looking for.
For more details on how "distinct on" works, refer the documentation and the example.
But, remember, using "distinct on" does not comply to SQL standards.

selecting data with highest field value in a field

I have a table, and I'd like to select rows with the highest value. For example:
----------------
| user | index |
----------------
| 1 | 1 |
| 2 | 1 |
| 2 | 2 |
| 3 | 4 |
| 3 | 7 |
| 4 | 1 |
| 5 | 1 |
----------------
Expected result:
----------------
| user | index |
----------------
| 1 | 1 |
| 2 | 2 |
| 3 | 7 |
| 4 | 1 |
| 5 | 1 |
----------------
How may I do so? I assume it can be done by some oracle function I am not aware of?
Thanks in advance :-)
You can use MAX() function for that with grouping user column like this:
SELECT "user"
,MAX("index") AS "index"
FROM Table1
GROUP BY "user"
ORDER BY "user";
Result:
| USER | INDEX |
----------------
| 1 | 1 |
| 2 | 2 |
| 3 | 7 |
| 4 | 1 |
| 5 | 1 |
See this SQLFiddle
if you have more than one column
select user , index
from (
select u.* , row_number() over (partition by user order by index desc) as rnk
from some_table u)
where rnk = 1
user is a reserved word - you should use a different name for the column.
select user,max(index) index from tbl
group by user;
Alternatively, you can use analytic functions:
select user,index, max(index) over (partition by user order by 1 ) highest from YOURTABLE
Note: Try NOT to use words like user, index, date etc.. as your column names, as they are reserved words for Oracle. If you will use, then use them with quotation marks, eg. "index", "date"...