selecting data with highest field value in a field - sql

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"...

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;

More efficient way to SELECT rows from PARTITION BY

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).

calculating sum of rows with identical id

Let's imagine a table with two columns ex:
| Value | ID |
+-------+----+
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
| 1 | 2 |
| 2 | 2 |
| 2 | 2 |
What I am trying to do is to calculate the sum of those with similar id and display them in different table like:
| Sum | ID |
+-----+----+
| 9 | 1 |
| 5 | 2 |
and so on.
I could find a sum of a known id by
SELECT SUM(VALUE) FROM MYTABLE WHERE ID = 1;
However not sure on how to find sum of different id's separately, could you give an idea on how to proceed?
Select SUM(VALUE),ID FROM MYTABLE GROUP BY ID
Use GROUP BY clause:
SELECT SUM(VALUE) Sum, ID FROM MYTABLE GROUP BY ID;
SELECT SUM(VALUE),ID FROM MYTABLE Group By ID

How to query and get only consecutive rows with the same column(s) value(s)

I have this activities table generated by the public_activity gem. Which has this two fields key owner_id part from others.
I'm trying to query the activities in chunks base on the owner_id and the key columns.
So if we have this data:
| id | owner_id | key |
| ---- | ---------- | -------------- |
| 1 | 1 | product.create |
| 2 | 1 | product.create |
| 3 | 1 | product.update |
| 4 | 2 | product.update |
| 5 | 2 | product.destroy |
| 6 | 2 | product.destroy |
| 7 | 2 | product.destroy |
| 8 | 1 | product.create |
| 9 | 1 | product.create |
The chunks would the matching rows.
Represented here by ids:
[1,2]
[3]
[4]
[5,6,7]
[8,9]
Chunks will be loaded one at a time via ajax, so the query will be preceded with a condition where(id > the_last_activity_fetched_id)
Any thoughts on how to build the necessary query are really welcome.
You can achieve this using array_agg:
select key,owner_id , array_agg(id) chunks from
(select key,owner_id,id from my_table order by id) A
group by key,owner_id
order by chunks
Here is an example: SQLFIDDLE
Using array_agg()
select owner_id,key, array_agg(id) chunks
from t
group by 1,2
order by 3
Demo
or
string_agg()
select owner_id,key,string_agg(id::text,',') chunks
from t
group by 1,2
order by 3
Demo
As per OP's comment
select owner_id,key
, array_agg(id) chunks
from t
group by 1,2 order by 3 desc limit 1

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.