I have a postgresql table
cubing=# SELECT * FROM times;
count | name | time
-------+---------+--------
4 | sean | 32.97
5 | Austin | 15.64
6 | Kirk | 117.02
I retrieve all from it with SELECT * FROM times ORDER BY time ASC. But now I want to give the user the option to search for a specific value (say, WHERE name = Austin) and have it tell them what rank they are in the table. Right now, I have SELECT name,time, RANK () OVER ( ORDER BY time ASC) rank_number FROM times. From how I understand it, that is giving me the rank of the entire table. I would like the rank, name, and time of who I am searching for. I am afraid if I added a where clause to my last SELECT statement with the name Austin, it would only find where the name equals Austin and rank those, rather than the rank of Austin in the rest of the table.
thanks for reading
I think the behavior you want here is to first rank your current data, then query it with some WHERE filter:
WITH cte AS (
SELECT *, RANK() OVER (ORDER BY time) rank_number
FROM times
)
SELECT count, name, time
FROM cte
WHERE name = 'Austin';
The point here is that at the time we do a query searching for Austin, the ranks for each row in your original table have already been generated.
Edit:
If you're running this query from an application, it would probably be best to avoid CTE syntax. Instead, just inline the CTE as a subquery:
SELECT count, name, time, rank_number
FROM
(
SELECT *, RANK() OVER (ORDER BY time) rank_number
FROM times
) t
WHERE name = 'Austin';
Related
I have the following sample table
And the output should be the first date (for each id) when cum_rev reaches the 100 mark.
I tried the following, because I taught with group bz trick and the where condition i will only get the first occurrence of value higher than 100.
SELECT id
,pd
,cum_rev
FROM (
SELECT id
,pd
,rev
,SUM(rev) OVER (
PARTITION BY id
ORDER BY pd
) AS cum_rev
FROM tab1
)
WHERE cum_rev >= 100
GROUP BY id
But it is not working, and I get the following error. And also when I add an alias is not helping
ERROR: subquery in FROM must have an alias LINE 4: FROM (
^ HINT: For example, FROM (SELECT ...) [AS] foo.
So the desired output is:
2 2015-04-02 135.70
3 2015-07-03 102.36
Do I need another approach? Can anyone help?
Thanks
demo:db<>fiddle
SELECT
id, total
FROM (
SELECT
*,
SUM(rev) OVER (PARTITION BY id ORDER BY pd) - rev as prev_total,
SUM(rev) OVER (PARTITION BY id ORDER BY pd) as total
FROM tab1
) s
WHERE total >= 100 AND prev_total < 100
You can use the cumulative SUM() window function for each id group (partition). To find the first which goes over a threshold you need to check the previous value for being under the threshold while the current one meets it.
PS: You got the error because your subquery is missing an alias. In my example its just s
I've used WINDOW functions before but only when working with data that has a fixed cadence/interval. I am likely missing something simple in aggregation but I've never had a scenario where I'm not working with fixed intervals.
I have a table the records samples at arbitrary timestamps. A sample is only recorded when it is a delta from the previous sample and the sample rate is completely irregular due to a large number of conditions. The table is very simple:
id (int)
happened_at (timestamp)
sensor_id (int)
new_value (float)
I'm trying to construct a query that will include a count of all of the samples before the happened_at of a given result row. So given an ultra simple 2 row sample data set:
id|happened_at |sensor_id| new_value
1 |2019-06-07:21:41|134679 | 123.331
2 |2019-06-07:19:00|134679 | 100.009
I'd like the result set to look like this:
happened_at |sensor_id | new_value | sample_count
2019-06-07:21:41|134679 |123.331 |2
2019-06-07:19:00|134679 |123.331 |1
I've tried:
SELECT *,
(SELECT count(sample_history.id) OVER (PARTITION BY score_history.sensor_id
ORDER BY sample_history.happened_at DESC))
FROM sensor_history
ORDER by happened_at DESC
and the duh not going to work.
(SELECT count(*)
FROM sample_history
WHERE sample_history.happened_at <= sample_timestamp)
Insights greatly appreciated.
Get rid of the SELECT (sub-query) when using the window function.
SELECT *,
count(*) OVER (PARTITION BY sensor_id ORDER BY happened_at DESC)
FROM sensor_history
ORDER BY happened_at DESC
I'm trying to get a leaderboard of summed user scores from a list of user score entries. A single user can have more than one entry in this table.
I have the following table:
rewards
=======
user_id | amount
I want to add up all of the amount values for given users and then rank them on a global leaderboard. Here's the query I'm trying to run:
SELECT user_id, SUM(amount) AS score, rank() OVER (PARTITION BY user_id) FROM rewards;
I'm getting the following error:
ERROR: column "rewards.user_id" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: SELECT user_id, SUM(amount) AS score, rank() OVER (PARTITION...
Isn't user_id already in an "aggregate function" because I'm trying to partition on it? The PostgreSQL manual shows the following entry which I feel is a direct parallel of mine, so I'm not sure why mine's not working:
SELECT depname, empno, salary, avg(salary) OVER (PARTITION BY depname) FROM empsalary;
They're not grouping by depname, so how come theirs works?
For example, for the following data:
user_id | score
===============
1 | 2
1 | 3
2 | 5
3 | 1
I would expect the following output (I have made a "tie" between users 1 and 2):
user_id | SUM(score) | rank
===========================
1 | 5 | 1
2 | 5 | 1
3 | 1 | 3
So user 1 has a total score of 5 and is ranked #1, user 2 is tied with a score of 5 and thus is also rank #1, and user 3 is ranked #3 with a score of 1.
You need to GROUP BY user_id since it's not being aggregated. Then you can rank by SUM(score) descending as you want;
SQL Fiddle Demo
SELECT user_id, SUM(score), RANK() OVER (ORDER BY SUM(score) DESC)
FROM rewards
GROUP BY user_id;
user_id | sum | rank
---------+-----+------
1 | 5 | 1
2 | 5 | 1
3 | 1 | 3
There is a difference between window functions and aggregate functions. Some functions can be used both as a window function and an aggregate function, which can cause confusion. Window functions can be recognized by the OVER clause in the query.
The query in your case then becomes, split in doing first an aggregate on user_id followed by a window function on the total_amount.
SELECT user_id, total_amount, RANK() OVER (ORDER BY total_amount DESC)
FROM (
SELECT user_id, SUM(amount) total_amount
FROM table
GROUP BY user_id
) q
ORDER BY total_amount DESC
If you have
SELECT user_id, SUM(amount) ....
^^^
agreagted function (not window function)
....
FROM .....
You need
GROUP BY user_id
I'm trying to rank a subset of data within a table but I think I am doing something wrong. I cannot find much information about the rank() feature for postgres, maybe I'm looking in the wrong place. Either way:
I'd like to know the rank of an id that falls within a cluster of a table based on a date. My query is as follows:
select cluster_id,feed_id,pub_date,rank
from (select feed_id,pub_date,cluster_id,rank()
over (order by pub_date asc) from url_info)
as bar where cluster_id = 9876 and feed_id = 1234;
I'm modeling this after the following stackoverflow post: postgres rank
The reason I think I am doing something wrong is that there are only 39 rows in url_info that are in cluster_id 9876 and this query ran for 10 minutes and never came back. (actually re-ran it for quite a while and it returned no results, yet there is a row in cluster 9876 for id 1234) I'm expecting this will tell me something like "id 1234 was 5th for the criteria given). It will return a relative rank according to my query constraints, correct?
This is postgres 8.4 btw.
By placing the rank() function in the subselect and not specifying a PARTITION BY in the over clause or any predicate in that subselect, your query is asking to produce a rank over the entire url_info table ordered by pub_date. This is likely why it ran so long as to rank over all of url_info, Pg must sort the entire table by pub_date, which will take a while if the table is very large.
It appears you want to generate a rank for just the set of records selected by the where clause, in which case, all you need do is eliminate the subselect and the rank function is implicitly over the set of records matching that predicate.
select
cluster_id
,feed_id
,pub_date
,rank() over (order by pub_date asc) as rank
from url_info
where cluster_id = 9876 and feed_id = 1234;
If what you really wanted was the rank within the cluster, regardless of the feed_id, you can rank in a subselect which filters to that cluster:
select ranked.*
from (
select
cluster_id
,feed_id
,pub_date
,rank() over (order by pub_date asc) as rank
from url_info
where cluster_id = 9876
) as ranked
where feed_id = 1234;
Sharing another example of DENSE_RANK() of PostgreSQL.
Find top 3 students sample query.
Reference taken from this blog:
Create a table with sample data:
CREATE TABLE tbl_Students
(
StudID INT
,StudName CHARACTER VARYING
,TotalMark INT
);
INSERT INTO tbl_Students
VALUES
(1,'Anvesh',88),(2,'Neevan',78)
,(3,'Roy',90),(4,'Mahi',88)
,(5,'Maria',81),(6,'Jenny',90);
Using DENSE_RANK(), Calculate RANK of students:
;WITH cteStud AS
(
SELECT
StudName
,Totalmark
,DENSE_RANK() OVER (ORDER BY TotalMark DESC) AS StudRank
FROM tbl_Students
)
SELECT
StudName
,Totalmark
,StudRank
FROM cteStud
WHERE StudRank <= 3;
The Result:
studname | totalmark | studrank
----------+-----------+----------
Roy | 90 | 1
Jenny | 90 | 1
Anvesh | 88 | 2
Mahi | 88 | 2
Maria | 81 | 3
(5 rows)
For Postgresql 8.x, I have an answers table containing (id, user_id, question_id, choice) where choice is a string value. I need a query that will return a set of records (all columns returned) for all unique choice values. What I'm looking for is a single representative record for each unique choice. I also want to have an aggregate votes column that is a count() of the number of records matching each unique choice accompanying each record. I want to force choice to lowercase for this comparison to be made (HeLLo and Hello should be considered equal). I can't GROUP BY lower(choice) because I want all columns in the result-set. Grouping by all columns causes all records to return, including all duplicates.
1. Closest I've gotten
select lower(choice), count(choice) as votes from answers where question_id = 21 group by lower(choice) order by votes desc;
The issue with this is it will not return all columns.
lower | votes
-----------------------------------------------+-------
dancing in the moonlight | 8
pumped up kicks | 7
party rock anthem | 6
sexy and i know it | 5
moves like jagger | 4
2. Trying with all columns
select *, count(choice) as votes from answers where question_id = 21 group by lower(choice) order by votes desc;
Because I am not specifying every column from the SELECT in my GROUP BY, this throws an error telling me to do so.
3. Specifying all columns in the GROUP BY
select *, count(choice) as votes from answers where question_id = 21 group by lower(choice), id, user_id, question_id, choice order by votes desc;
This simply dumps the table with votes column as 1 for all records.
How can I get the vote count and unique representative records from 1., but with all columns from the table returned?
Join grouped results back with primary table, then show only one row for each (question,answer) combination.
similar to this:
WITH top5 AS (
select question_id, lower(choice) as choice, count(*) as votes
from answers
where question_id = 21
group by question_id , lower(choice)
order by count(*) desc
limit 5
)
SELECT DISTINCT ON(question_id,choice) *
FROM top5
JOIN answers USING(question_id,lower(choice))
ORDER BY question_id, lower(choice), answers.id;
Here's what I ended up with:
SELECT answers.*, cc.votes as votes FROM answers join (
select max(id) as id, count(id) as votes
from answers
group by trim(lower(choice))
) cc
on answers.id = cc.id ORDER BY votes desc, lower(response) asc