I have a table in the form of
USER |VALUE |DATE
-------------------------
user1 | 1337 | 2019-11-01
user1 | 1338 | 2019-03-28
user2 | 1234 | 2019-04-23
user2 | 4567 | 2019-05-05
and want to get the maximum value of every user with the associated date. When I do something like
SELECT max(VALUE) FROM table
GROUP BY USER
I have the problem that I can neither aggregate nor group by the DATE. How can I tell PostgreSQL that I just wanna have the date associated with the row where the max. value is?
Thanks!
Click: demo:db<>fiddle
You can use DISTINCT ON: This gives you the first record of an ordered group. In your case the groups is your user column. This you have to order by value DESC to get the max value to be the first record. This will be taken - including the associated date value.
SELECT DISTINCT ON (user)
*
FROM
mytable
ORDER BY user, value DESC
If you want other aggregations in the query, you can also use array_agg() to mimic a "first" function:
select user, max(value), count(*),
(array_agg(date order by value desc))[1] as date_at_max
from t
group by user;
How can I get all IDs that have more than 10 entries on one day?
Here is the sample data:
ID | Time
__________________________
4 | 2019-02-14 17:22:43
__________________________
2 | 2019-04-27 07:51:09
__________________________
83 | 2018-01-07 08:38:37
__________________________
I am having a hard time using count and going through and finding all of the ones on the same day. The Hour:Min:Sec is what is causing problems for me.
For MySql it would be:
select distinct id from tablename
group by id, date(time)
having count(*) > 10
The date() function rejects the time part of the column, so the grouping is done only by the date part.
For SqlServer you would use:
convert(date, time)
I want to find the minimum value of a column in a certain date range of a table.
so lets say I have a table like the following,
Date | Value
---------------
01-26 | 2
01-26 | 1
01-27 | 2
01-27 | 4
01-28 | 3
01-28 | 5
How can I apply the MIN() function to the subgroup of the Value column so that the result might be
Date | MIN(Value)
---------------
01-26 | 1
01-27 | 2
01-28 | 3
I thought about GROUP BY .. or such but couldn't figure out how to get the results into a table.
Using UNION and JOIN isn't quite scalable because the query could be using a date range of a month
Group by should work:
Select date, min( value )
From table1
Group by date
Maybe too simple, but seems like this would work
Select Min(col1), datecol from yourtable group by datecol;
HTH
Assume I have a table with only two columns: id, maturity. maturity is some date in the future and is representative of until when a specific entry will be available. Thus it's different for different entries but is not necessarily unique. And with time number of entries which have not reached this maturity date changes.
I need to count a number of entries from such a table that were available on a specific date (thus entries that have not reached their maturity). So I basically need to join this two queries:
SELECT generate_series as date FROM generate_series('2015-10-01'::date, now()::date, '1 day');
SELECT COUNT(id) FROM mytable WHERE mytable.maturity > now()::date;
where instead of now()::date I need to put entry from the generated series. I'm sure this has to be simple enough, but I can't quite get around it. I need the resulting solution to remain a query, thus it seems that I can't use for loops.
Sample table entries:
id | maturity
---+-------------------
1 | 2015-10-03
2 | 2015-10-05
3 | 2015-10-11
4 | 2015-10-11
Expected output:
date | count
------------+-------------------
2015-10-01 | 4
2015-10-02 | 4
2015-10-03 | 3
2015-10-04 | 3
2015-10-05 | 2
2015-10-06 | 2
NOTE: This count doesn't constantly decrease, since new entries are added and this count increases.
You have to use fields of outer query in WHERE clause of a sub-query. This can be done if the subquery is in the SELECT clause of the outer query:
SELECT generate_series,
(SELECT COUNT(id)
FROM mytable
WHERE mytable.maturity > generate_series)
FROM generate_series('2015-10-01'::date, now()::date, '1 day');
More info: http://www.techonthenet.com/sql_server/subqueries.php
I think you want to group your data by the maturity Date.
Check this:
select maturity,count(*) as count
from your_table group by maturity;
I have a table (in Postgres 9.1) that looks something like this:
CREATE TABLE actions (
user_id: INTEGER,
date: DATE,
action: VARCHAR(255),
count: INTEGER
)
For example:
user_id | date | action | count
---------------+------------+--------------+-------
1 | 2013-01-01 | Email | 1
1 | 2013-01-02 | Call | 3
1 | 2013-01-03 | Email | 3
1 | 2013-01-04 | Call | 2
1 | 2013-01-04 | Voicemail | 2
1 | 2013-01-04 | Email | 2
2 | 2013-01-04 | Email | 2
I would like to be able to view a user's total actions over time for a specific set of actions; for example, Calls + Emails:
user_id | date | count
-----------+-------------+---------
1 | 2013-01-01 | 1
1 | 2013-01-02 | 4
1 | 2013-01-03 | 7
1 | 2013-01-04 | 11
2 | 2013-01-04 | 2
The monstrosity that I've created so far looks like this:
SELECT
date, user_id, SUM(count) OVER (PARTITION BY user_id ORDER BY date) AS count
FROM
actions
WHERE
action IN ('Call', 'Email')
GROUP BY
user_id, date, count;
Which works for single actions, but seems to break for multiple actions when they happen on the same day, for example instead of the expected 11 on 2013-01-04, we get 9:
date | user_id | count
------------+--------------+-------
2013-01-01 | 1 | 1
2013-01-02 | 1 | 4
2013-01-03 | 1 | 7
2013-01-04 | 1 | 9 <-- should be 11?
2013-01-04 | 2 | 2
Is it possible to tweak my query to resolve this issue? I tried removing the grouping on count, but Postgres doesn't seem to like that:
column "actions.count" must appear in the GROUP BY clause
or be used in an aggregate function
LINE 2: date, user_id, SUM(count) OVER (PARTITION BY user...
^
This query produces the result you are looking for:
SELECT DISTINCT
date, user_id, SUM(count) OVER (PARTITION BY user_id ORDER BY date) AS count
FROM actions
WHERE
action IN ('Call', 'Email');
The default window is already what you want, according to the official docs and the "DISTINCT" eliminates duplicate rows when both Emails and Calls happen on the same day.
See SQL Fiddle.
The table has a column named "count", and the expresion in the SELECT clause is aliased as "count", it is ambiguous.
Read documentation: http://www.postgresql.org/docs/9.0/static/sql-select.html#SQL-GROUPBY
In case of ambiguity, a GROUP BY name will be interpreted as an
input-column name rather than an output column name.
That means, that your query does not group by "count" evaluated in the SELECT clause, but rather it groups by "count" values taken from the table.
This query gives expected results, see SQL Fiddle
SELECT date, user_id, count
from (
Select date, user_id,
SUM(count) OVER (PARTITION BY user_id ORDER BY date) AS count
FROM actions
WHERE
action IN ('Call', 'Email')
) alias
GROUP BY
user_id, date, count;
Asserts
It is unclear whether you want to sort by user_id or date
It is also unclear whether you want to include dates in the result list, for which there is no row in the base table. In this case, refer to this closely related answer:
PostgreSQL: running count of rows for a query 'by minute'
Repair names
First off, I am using this test table instead of your problematic table:
CREATE TEMP TABLE actions (
user_id integer,
thedate date,
action text,
ct integer
);
Your use of reserved words and function names as identifiers (column names) is part of the problem.
Repair query
Combine aggregate and window functions
Since aggregate functions are applied first, your original query lumps the two rows found for user_id = 1 and thedate = '2013-01-04' into one. You have to multiply by count(*) to get the actual running count.
You can do this without subquery, since you can combine aggregate functions and window functions. Aggregate functions are applied first. You can even have a window functions over the result of aggregate functions.
SELECT thedate
, user_id
, sum(ct * count(*)) OVER (PARTITION BY user_id
ORDER BY thedate) AS running_ct
FROM actions
WHERE action IN ('Call', 'Email')
GROUP BY user_id, thedate, ct
ORDER BY user_id, thedate;
Or simplify to:
...
, sum(sum(ct)) OVER (PARTITION BY user_id
ORDER BY thedate) AS running_ct
...
This should also be the fastest of the solutions presented.
Here, the inner sum() is an aggregate function, while the outer sum() is a window function - over the result of the aggregate function.
Or use DISTINCT
Another way would to use DISTINCT or DISTINCT ON, since that is applied after window functions:
DISTINCT - this is possible, since running_ct is guaranteed to be the same in this case anyway, since all peers are summed at once for the default frame definition of window functions.
SELECT DISTINCT
thedate
, user_id
, sum(ct) OVER (PARTITION BY user_id ORDER BY thedate) AS running_ct
FROM actions
WHERE action IN ('Call', 'Email')
ORDER BY thedate, user_id;
Or simplify with DISTINCT ON:
SELECT DISTINCT ON (thedate, user_id)
...
->SQLfiddle demonstrating all variants.