I have a dataset with different clients, and their sales count. Over time, some clients get added and deleted from the data. How do I make sure that when I look at the sales counts, that I am only using a selection of the clients that were in the data set all the time? Ie if I have a client that doesn't have a record for 2018-03, then I don't want that client to be part of the entire query. If a clients does not have a record in 2020-03, then I also do not want this client to be part of the entire query.
For example, the following query:
select DATE_PART (y, sold_date)as year, DATE_PART (mm, sold_date) as month, count(distinct(client))
from sales_data
where sold_date > '2018-01-01'
group by year, month
order by year,month
Yields
year month count
2018 1 78
2018 2 83
2018 3 80
2018 4 83
2018 5 84
2018 6 81
2018 7 83
2018 8 90
2018 9 89
2018 10 95
2018 11 94
2018 12 97
2019 1 102
2019 2 103
2019 3 102
2019 4 105
2019 5 103
2019 6 104
2019 7 104
2019 8 106
2019 9 106
2019 10 108
2019 11 109
2019 12 104
2020 1 104
2020 2 102
2020 3 103
2020 4 98
2020 5 97
2020 6 79
So I want to only use the clients that are in all months, they should not be more than 78, because there can not be more users than the minimal month (2018-1).
FYI, I am using Amazon Redshift here but I am OK with a query that's rdbms agnostic or works for SQL-Server/Oracle/MySQL/PostgreSQL, I am just interested in a pattern on how to solve this issue effectively.
If I'm understanding what you want correctly, and if this is just a one-off query, you could use a correlated subquery in the where clause:
SELECT
DATE_PART(y, s.sold_date) AS year,
DATE_PART(mm, s.sold_date) AS month,
COUNT(DISTINCT s.client)
FROM
sales_data AS s
WHERE
EXISTS (
SELECT sd.client FROM sales_data AS sd WHERE DATE_PART(y,
sd.sold_date) = 2018 AND DATE_PART(mm, sd.sold_date) = 1 AND
sd.client = s.client
) AND
s.sold_date > '2018-01-01'
GROUP BY
year,
month
ORDER
DATE_PART(y, s.sold_date),
DATE_PART(mm, s.sold_date)
presence in all months can be done with 2-step aggregation:
group sales data by customer ID having all months
group sales data joined to (1) by year, month
like this (=12 can be a dynamic expression, depending on the amount of history you have)
with
stable_customers as (
select customer_id
from sales_data
group by 1
having count(distinct date_trunc('month' from sold_date)=12
)
select
DATE_PART (y, sold_date) as year
,DATE_PART (mm, sold_date) as month,
,count(1)
from sales_date
join stable_customers
using (customer_id)
where sold_date > '2018-01-01'
group by year, month
order by year,month
Use window functions. Unfortunately, SQL Server does not support count(distinct) as a window function. Fortunately, there is a simple work-around using dense_rank():
select year, month, count(distinct client)
from (select sd.*, year, month,
(dense_rank() over (order by year, month) +
dense_rank() over (order by year desc, month desc)
) as num_months,
(dense_rank() over (partition by client order by year, month) +
dense_rank() over (partition by client order by year desc, month desc)
) as num_months_client
from sales_data sd cross apply
(values (year(sold_date), month(sold_date))) v(year, month)
where sd.sold_date > '2018-01-01'
) sd
where num_months_client = num_months
group by year, month
order by year, month;
Note: This looks at all months that are in the data. If all clients are missing 2019-03, then that months is not considered at all.
Related
In my data set, each customer has some orders on different dates.
For each customer each month, I want to check his/her last order in the previous month in which city.
For example, it is my data for one of the customers.
customer
year
month
day
order id
city id
1544
2022
2
6
413
9
1544
2022
2
17
39
10
1544
2022
3
5
115
21
1544
2022
5
29
2153
4
1544
2022
5
30
955
9
the result should be the same as this:
customer
year
month
city of last order of prev month(prevCity)
1544
2022
2
null or 9
1544
2022
3
10
1544
2022
5
21
(the first row of the above table is not my question now. )
I write my query using last_value the same as this:
select customer,
year,
month,
last_value(City) over (partition by customer, year, month order by created_at desc
ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING) as prevCity
from table1
but the result is false!
How can I correct this?
Using the window function lag() over() in concert with the WITH TIES clause
Select top 1 with ties
customer
,year
,month
,LastCityID = lag([city id],1) over (partition by customer order by year, month,day)
From YourTable
order by row_number() over (partition by customer,year,month order by year, month,day)
Or an Nudge More Perforamt
with cte as (
Select *
,LastCityID = lag([city id],1) over (partition by customer order by year, month,day)
,RN = row_number() over (partition by customer,year,month order by year, month,day)
From YourTable
)
Select customer
,year
,month
,LastCityID
From cte
Where RN =1
Results
customer year month LastCityID
1544 2022 2 NULL
1544 2022 3 10
1544 2022 5 21
I have a table as following (using bigquery):
id
year
month
day
rating
111
2020
11
30
4
111
2020
12
01
4
112
2020
11
30
5
113
2020
11
30
5
Is there a way in which I can select ids that have ratings that are consecutively (two or more consecutive records) low (low as in both records' ratings less than 4.5)?
For example, my desired output is:
id
year
month
day
rating
111
2020
11
30
4
111
2020
12
01
4
If you want all rows, then you need to look at both the previous rating and the next rating:
SELECT t.*
FROM (SELECT t.*,
LAG(rating) OVER (PARTITION BY id ORDER BY year, month, day ASC) AS prev_rating,
LEAD(rating) OVER (PARTITION BY id ORDER BY year, month, day ASC) AS next_rating,
FROM dataset.table t
) t
WHERE (rating < 4.5 and prev_rating < 4.5) OR
(rating < 4.5 and next_rating < 4.5)
Below is for BigQuery Standard SQL
select * except(grp, seq_len)
from (
select *, sum(1) over(partition by grp) seq_len
from (
select *,
countif(rating >= 4.5) over(partition by id order by year, month, day) grp
from `project.dataset.table`
)
where rating < 4.5
)
where seq_len > 1
I'm writing a simple query on Amazon Redshift as follows:
SELECT EXTRACT(year FROM created_at) AS year,
EXTRACT(month FROM created_at) AS month,
member_id,
COUNT(*) as pageviews
FROM TABLE
GROUP BY year,
month,
member_id
ORDER BY year,
month,
member_id
This gives me the following result as an example:
year month member_id pageviews
2015 1 100 29
2015 2 100 22
2015 3 100 178
2015 4 100 34
2015 1 200 56
2015 3 200 16
Here's the result I would like to have:
year month member_id pageviews
2015 1 100 29
2015 2 100 22
2015 3 100 178
2015 4 100 34
2015 1 200 56
2015 2 200 0
2015 3 200 16
2015 4 200 0
In the result above, notice the additional rows with zero pageviews.
How do I get this result? Any help would be much appreciated.
Use a cross join to generate the rows and then a left join to bring in the data:
SELECT EXTRACT(year FROM created_at) AS year,
EXTRACT(month FROM created_at) AS month,
m.member_id,
COUNT(t.member_id) as pageviews
FROM (SELECT DISTINCT EXTRACT(year FROM created_at) AS year, EXTRACT(month FROM created_at) AS month FROM TABLE) ym CROSS JOIN
(SELECT DISTINCT member_id FROM TABLE) m LEFT JOIN
TABLE t
ON EXTRACT(year FROM created_at) AS month = ym.year AND
EXTRACT(month FROM created_at) AS month = ym.month AND
t.member_id = m.member_id
GROUP BY ym.year, ym.month, m.member_id
ORDER BY ym.year, ym.month, m.member_id;
This assumes that all year/month combinations are included in the table.
If you have other tables that are better sources for members and the dates, try them -- that may be faster than SELECT DISTINCT.
I have a table like this.
Year Month TenDays Pay
========================================
2015 8 2 12
2016 8 1 43
2016 8 2 11
2016 9 1 22
2016 9 2 33
2016 9 3 4
2016 9 3 25
I want to have SQL query that calculate sum of 'Pay' as 'TotalTenDays' group by 'year' and 'Month' and 'TenDays'
and also calculate sum of 'Pay' as 'TotalMonth' group by 'year' and 'Month'.
I can do that with "union all" but I am searching for a way without using union and 'with cte as()'.
Is it Possible?
Expected table must be like this:
Year Month TenDays TotalTenDays TotalMonth
====================================================================
2015 8 2 12 12
2016 8 1 43 54
2016 8 2 11 54
2016 9 1 22 84
2016 9 2 33 84
2016 9 3 29 84
The answer depends on the database dialect.
The first 4 columns are standard SQL GROUP BY logic, i.e. GROUP BY Year, Month, TenDays with a SUM(Pay) AS TotalTenDays result column.
The TotalMonth column is best done with a windowing function using the OVER clause, but that's only if the SQL dialect supports it.
E.g. for SQL Server, you can do this:
SELECT Year, Month, TenDays
, SUM(Pay) AS TotalTenDays
, SUM(SUM(Pay)) OVER (PARTITION BY Year, Month) AS TotalMonth
FROM MyTable
GROUP BY Year, Month, TenDays
ORDER BY Year, Month, TenDays
See SQL Fiddle for running query using MS SQL Server 2017.
If the SQL dialect doesn't support windowing functions, then suggestion in comment by Jonathan Leffler is a good alternative:
You need to create two sub-queries that do the aggregations, and then join the results of those two sub-queries. Each sub-query will have a different GROUP BY clause.
SELECT a.Year, a.Month, a.TenDays, a.TotalTenDays, b.TotalMonth
FROM ( SELECT Year, Month, TenDays
, SUM(Pay) AS TotalTenDays
FROM MyTable
GROUP BY Year, Month, TenDays
) a
JOIN ( SELECT Year, Month
, SUM(Pay) AS TotalMonth
FROM MyTable
GROUP BY Year, Month
) b ON b.Year = a.Year
AND b.Month = a.Month
ORDER BY a.Year, a.Month, a.TenDays
See SQL Fiddle for running query using MySQL 5.6.
I have this table in SQL Server:
Year Month Quantity
----------------------------
2015 January 10
2015 February 20
2015 March 30
2014 November 40
2014 August 50
How can I identify the different years and months adding two more columns that group the same years with a number and then different months in sequential way like the example
Year Month Quantity Group Subgroup
------------------------------------------------
2015 January 10 1 1
2015 February 20 1 2
2015 March 30 1 3
2014 November 40 2 1
2014 August 50 2 2
You can use DENSE_RANK to calculate the groups for you:
SELECT t1.*, DENSE_RANK() OVER (ORDER BY Year DESC) AS [Group],
DENSE_RANK() OVER (PARTITION BY Year ORDER BY DATEPART(month, Month + ' 01 2010')) AS [SubGroup]
FROM t1
ORDER BY 4, 5
See this fiddle.
To associate group and subgroup with a number you can do this:
WITH RankedTable AS (
SELECT year, month, quantity,
ROW_NUMBER() OVER (partition by year order by Month) AS rn
FROM yourtable)
SELECT year, month, quantity,
SUM (CASE WHEN rn = 1 THEN 1 ELSE 0 END) OVER (ORDER BY YEAR) as year_group,
rn AS subgroup
FROM RankedTable
Here ROW_NUMBER() OVER clause calculates rank of a month within a year.
And SUM() ... OVER calculates running SUM for the months with rank 1.
SQL Fiddle