Let's say I have a customer table like so:
id | start_date | created_at
-----------------------------
1 | 2020-1-15 | 2020-1-15
1 | 2020-1-16 | 2020-1-15
1 | 2020-1-16 | 2020-1-16
2 | 2020-1-15 | 2020-1-15
2 | 2020-1-16 | 2020-1-15
I want to get 1 row per customer id that has the max(start_date) and if it's the same date will use the max(created_at).
Result should look like this:
id | start_date | created_at
-----------------------------
1 | 2020-1-16 | 2020-1-16
2 | 2020-1-16 | 2020-1-15
I'm having a hard time with window functions as I thought a partition by id would work but I have 2 dates.
Maybe I use a group by?
please try this one, you could use order by two columns
SELECT * FROM (
SELECT id, start_date, Created_At, ROW_NUMBER()OVER(PARTITION BY id ORDER BY start_date DESC, Created_At DESC) AS R
FROM #date
) A
WHERE A.R = 1
Related
I am trying to write a query to return supplier ID (sup_id), order date and the order ID of the first order (based on earliest time).
+--------+--------+------------+--------+-----------------+
|orderid | sup_id | items | sales | order_ts |
+--------+--------+------------+--------+-----------------+
|1111132 | 3 | 1 | 27,0 | 24/04/17 13:00 |
|1111137 | 3 | 2 | 69,0 | 02/02/17 16:30 |
|1111147 | 1 | 1 | 87,0 | 25/04/17 08:25 |
|1111153 | 1 | 3 | 82,0 | 05/11/17 10:30 |
|1111155 | 2 | 1 | 29,0 | 03/07/17 02:30 |
|1111160 | 2 | 2 | 44,0 | 30/01/17 20:45 |
|....... | ... | ... | ... | ... ... |
+--------+--------+------------+--------+-----------------+
Output I am looking for:
+--------+--------+------------+
| sup_id | date | order_id |
+--------+--------+------------+
|....... | ... | ... |
+--------+--------+------------+
I tried using a subquery in the join clause as below but didn't know how to join it without having selected order_id.
SELECT sup_id, date(order_ts), order_id
FROM sales s
JOIN
(
SELECT sup_id, date(order_ts) as date, min(time(order_date))
FROM sales
GROUP BY merchant_id, date
) m
on ...
Kindly assist.
You can use not exists:
select *
from sales
where not exists (
-- find sales for same supplier, earlier date, same day
select *
from sales as older
where older.sup_id = sales.sup_id
and older.order_ts < sales.order_ts
and older.order_ts >= cast(sales.order_ts as date)
)
The query below might not be the fastest in the world, but it should give you all information you need.
select order_id, sup_id, items, sales, order_ts
from sales s
where order_ts <= (
select min(order_ts)
from sales m
where m.sup_id = s.sup_id
)
select sup_id, min(order_ts), min(order_id) from sales
where order_ts = '2022-15-03'
group by sup_id
Assumed orderid is an identity / auto increment column
Posting here in case someone with more knowledge than may be able to help me with some direction.
I have a table like this:
| Row | date |user id | score |
-----------------------------------
| 1 | 20201120 | 1 | 26 |
-----------------------------------
| 2 | 20201121 | 1 | 14 |
-----------------------------------
| 3 | 20201125 | 1 | 0 |
-----------------------------------
| 4 | 20201114 | 2 | 32 |
-----------------------------------
| 5 | 20201116 | 2 | 0 |
-----------------------------------
| 6 | 20201120 | 2 | 23 |
-----------------------------------
However, from this, I need to have a record for each user for each day where if a day is missing for a user, then the last score recorded should be maintained then I would have something like this:
| Row | date |user id | score |
-----------------------------------
| 1 | 20201120 | 1 | 26 |
-----------------------------------
| 2 | 20201121 | 1 | 14 |
-----------------------------------
| 3 | 20201122 | 1 | 14 |
-----------------------------------
| 4 | 20201123 | 1 | 14 |
-----------------------------------
| 5 | 20201124 | 1 | 14 |
-----------------------------------
| 6 | 20201125 | 1 | 0 |
-----------------------------------
| 7 | 20201114 | 2 | 32 |
-----------------------------------
| 8 | 20201115 | 2 | 32 |
-----------------------------------
| 9 | 20201116 | 2 | 0 |
-----------------------------------
| 10 | 20201117 | 2 | 0 |
-----------------------------------
| 11 | 20201118 | 2 | 0 |
-----------------------------------
| 12 | 20201119 | 2 | 0 |
-----------------------------------
| 13 | 20201120 | 2 | 23 |
-----------------------------------
I'm trying to to this in BigQuery using StandardSQL. I have an idea of how to keep the same score across following empty dates, but I really don't know how to add new rows for missing dates for each user. Also, just to keep in mind, this example only has 2 users, but in my data I have more than 1500.
My end goal would be to show something like the average of the score per day. For background, because of our logic, if the score wasn't recorded in a specific day, this means that the user is still in the last score recorded which is why I need a score for every user every day.
I'd really appreciate any help I could get! I've been trying different options without success
Below is for BigQuery Standard SQL
#standardSQL
select date, user_id,
last_value(score ignore nulls) over(partition by user_id order by date) as score
from (
select user_id, format_date('%Y%m%d', day) date,
from (
select user_id, min(parse_date('%Y%m%d', date)) min_date, max(parse_date('%Y%m%d', date)) max_date
from `project.dataset.table`
group by user_id
) a, unnest(generate_date_array(min_date, max_date)) day
)
left join `project.dataset.table` b
using(date, user_id)
-- order by user_id, date
if applied to sample data from your question - output is
One option uses generate_date_array() to create the series of dates of each user, then brings the table with a left join.
select d.date, d.user_id,
last_value(t.score ignore nulls) over(partition by d.user_id order by d.date) as score
from (
select t.user_id, d.date
from mytable t
cross join unnest(generate_date_array(min(date), max(date), interval 1 day)) d(date)
group by t.user_id
) d
left join mytable t on t.user_id = d.user_id and t.date = d.date
I think the most efficient method is to use generate_date_array() but in a very particular way:
with t as (
select t.*,
date_add(lead(date) over (partition by user_id order by date), interval -1 day) as next_date
from t
)
select row_number() over (order by t.user_id, dte) as id,
t.user_id, dte, t.score
from t cross join join
unnest(generate_date_array(date,
coalesce(next_date, date)
interval 1 day
)
) dte;
I'm trying to rank the rows in the following table that looks like this:
| ID | Key | Date | Row|
*****************************
| P175 | 5 | 2017-01| 2 |
| P175 | 5 | 2017-02| 2 |
| P175 | 5 | 2017-03| 2 |
| P175 | 12 | 2017-03| 1 |
| P175 | 12 | 2017-04| 1 |
| P175 | 12 | 2017-05| 1 |
This person has two Keys at once during 2017-03, but I want the formula to put '1' for the rows where Key=12 since it reflects the most recent records.
I want the same formula to also work for the people who don't have overlapping Keys, putting '1' for the most recent records:
| ID | Key | Date | Row|
*****************************
| P170 | 8 | 2017-01| 2 |
| P170 | 8 | 2017-02| 2 |
| P170 | 8 | 2017-03| 2 |
| P170 | 6 | 2017-04| 1 |
| P170 | 6 | 2017-05| 1 |
I've tried variations of ROW_NUMBER() OVER PARTITION BY and DENSE_RANK but cannot figure out the correct formula. Thanks for your help.
First calculate the max date for the key. Then use dense_rank():
select t.*,
dense_rank() over (partition by id order by max_date desc, key) as row
from (select t.*, max(date) over (partition by id, key) as max_date
from t
) t;
If the ranges for each key did not overlap, you could do this with a cumulative count distinct:
select t.*, count(distinct key) over (partition by id order by date desc) as rank
from t;
However, this would not work in the first case. I just find it interesting that this does almost the same thing as the first query.
I guess you are looking for something like this
select personid, mykey, month,
dense_rank() over (partition by personid order by mykey desc) rown
from personkeys
order by month
see the example
http://sqlfiddle.com/#!15/cf751/8
I have a table that accommodates data that is logically groupable by multiple properties (foreign key for example). Data is sequential over continuous time interval; i.e. it is a time series data. What I am trying to achieve is to select only latest values for each group of groups.
Here is example data:
+-----------------------------------------+
| code | value | date | relation_id |
+-----------------------------------------+
| A | 1 | 01.01.2016 | 1 |
| A | 2 | 02.01.2016 | 1 |
| A | 3 | 03.01.2016 | 1 |
| A | 4 | 01.01.2016 | 2 |
| A | 5 | 02.01.2016 | 2 |
| A | 6 | 03.01.2016 | 2 |
| B | 1 | 01.01.2016 | 1 |
| B | 2 | 02.01.2016 | 1 |
| B | 3 | 03.01.2016 | 1 |
| B | 4 | 01.01.2016 | 2 |
| B | 5 | 02.01.2016 | 2 |
| B | 6 | 03.01.2016 | 2 |
+-----------------------------------------+
And here is example of desired output:
+-----------------------------------------+
| code | value | date | relation_id |
+-----------------------------------------+
| A | 3 | 03.01.2016 | 1 |
| A | 6 | 03.01.2016 | 2 |
| B | 3 | 03.01.2016 | 1 |
| B | 6 | 03.01.2016 | 2 |
+-----------------------------------------+
To put this in perspective — for every related object I want to select each code with latest date.
Here is a select I came with. I've used ROW_NUMBER OVER (PARTITION BY...) approach:
SELECT indicators.code, indicators.dimension, indicators.unit, x.value, x.date, x.ticker, x.name
FROM (
SELECT
ROW_NUMBER() OVER (PARTITION BY indicator_id ORDER BY date DESC) AS r,
t.indicator_id, t.value, t.date, t.company_id, companies.sic_id,
companies.ticker, companies.name
FROM fundamentals t
INNER JOIN companies on companies.id = t.company_id
WHERE companies.sic_id = 89
) x
INNER JOIN indicators on indicators.id = x.indicator_id
WHERE x.r <= (SELECT count(*) FROM companies where sic_id = 89)
It works but the problem is that it is painfully slow; when working with about 5% of production data which equals to roughly 3 million fundamentals records this select take about 10 seconds to finish. My guess is that happens due to subselect selecting huge amounts of records first.
Is there any way to speed this query up or am I digging in wrong direction trying to do it the way I do?
Postgres offers the convenient distinct on for this purpose:
select distinct on (relation_id, code) t.*
from t
order by relation_id, code, date desc;
So your query uses different column names than your sample data, so it's hard to tell, but it looks like you just want to group by everything except for date? Assuming you don't have multiple most recent dates, something like this should work. Basically don't use the window function, use a proper group by, and your engine should optimize the query better.
SELECT mytable.code,
mytable.value,
mytable.date,
mytable.relation_id
FROM mytable
JOIN (
SELECT code,
max(date) as date,
relation_id
FROM mytable
GROUP BY code, relation_id
) Q1
ON Q1.code = mytable.code
AND Q1.date = mytable.date
AND Q1.relation_id = mytable.relation_id
Other option:
SELECT DISTINCT Code,
Relation_ID,
FIRST_VALUE(Value) OVER (PARTITION BY Code, Relation_ID ORDER BY Date DESC) Value,
FIRST_VALUE(Date) OVER (PARTITION BY Code, Relation_ID ORDER BY Date DESC) Date
FROM mytable
This will return top value for what ever you partition by, and for whatever you order by.
I believe we can try something like this
SELECT CODE,Relation_ID,Date,MAX(value)value FROM mytable
GROUP BY CODE,Relation_ID,Date
In the table below
+-------+-----------------------+
| id | timestamp |
+-------+-----------------------+
| 1 | 2010-06-10 14:35:30 |
| 2 | 2010-06-10 15:27:35 |
| 3 | 2010-06-10 16:39:36 |
| 4 | 2010-06-11 14:55:30 |
| 5 | 2010-06-11 18:45:31 |
| 6 | 2010-06-12 20:25:31 |
+-------+-----------------------+
I want to be able to count the dates (time is ignored). So the output should be like below:
+-------+-----------------------+
| id | type | count |
+-------+-----------------------+
| 1 | 2010-06-10 | 3 |
| 2 | 2010-06-11 | 2 |
| 3 | 2010-06-12 | 1 |
+-------+-----------------------+
What would be the query for this?
This works if you can live without the id column in the result:
SELECT DATE(timestamp) AS type, COUNT(*) AS `count`
FROM sometable
GROUP BY DATE(timestamp)
ORDER BY DATE(timestamp)
SELECT
DATE(timestamp),
COUNT(*)
FROM
My_Table
GROUP BY
DATE(timestampe)
This doesn't give you a row number for each row. I don't know if (why?) that's important.
select date(timestamp) as type, count(*)
from your_table
group by type;
This might work...
select DATE(timestamp), count(timestamp)
from _table
group by timestamp
order by count(timestamp) desc
SELECT count( DATE_FORMAT( timestamp, '%Y-%m-%d/' ) ) , DATE_FORMAT( timestamp, '%Y-%m-%d' )
FROM tablename
group by DATE_FORMAT( timestamp, '%Y-%m-%d' );
You can not include the id in the select or the count will be off.