GROUP BY one column, then GROUP BY another column - sql

I have a database table t with a sales table:
ID
TYPE
AGE
1
B
20
1
BP
20
1
BP
20
1
P
20
2
B
30
2
BP
30
2
BP
30
3
P
40
If a person buys a bundle it appears the bundle sale (TYPE B) and the different bundle products (TYPE BP), all with the same ID. So a bundle with 2 products appears 3 times (1x TYPE B and 2x TYPE BP) and has the same ID.
A person can also buy any other product in that single sale (TYPE P), which has also the same ID.
I need to calculate the average/min/max age of the customers but the multiple entries per sale tamper with the correct calculation.
The real average age is
(20 + 30 + 40) / 3 = 30
and not
(20+20+20+20 + 30+30+30 + 40) / 8 = 26,25
But I don't know how I can reduce the sales to a single row entry AND get the 4 needed values?
Do I need to GROUP BY twice (first by ID, then by AGE?) and if yes, how can I do it?
My code so far:
SELECT
AVERAGE(AGE)
, MIN(AGE)
, MAX(AGE)
, MEDIAN(AGE)
FROM t
but that does count every row.

Assuming the age is the same for all rows with the same ID (which in itself indicates a normalisation problem), you can use nest aggregation:
select avg(min(age)) from sales
group by id
AVG(MIN(AGE))
-------------
30
SQL Fiddle
The example in the documentation is very similar; and is explained as:
This calculation evaluates the inner aggregate (MAX(salary)) for each group defined by the GROUP BY clause (department_id), and aggregates the results again.
So for your version:
This calculation evaluates the inner aggregate (MIN(age)) for each group defined by the GROUP BY clause (id), and aggregates the results again.
It doesn't really matter whether the inner aggregate is min or max - again, assuming they are all the same - it's just to get a single value per ID, which can then be averaged.
You can do the same for the other values in your original query:
select
avg(min(age)) as avg_age,
min(min(age)) as min_age,
max(min(age)) as max_age,
median(min(age)) as med_age
from sales
group by id;
AVG_AGE MIN_AGE MAX_AGE MED_AGE
------- ------- ------- -------
30 20 40 30
Or if you prefer you could get the one-age-per-ID values once ina CTE or subquery and apply the second layer of aggregation to that:
select
avg(age) as avg_age,
min(age) as min_age,
max(age) as max_age,
median(age) as med_age
from (
select min(age) as age
from sales
group by id
);
which gets the same result.
SQL Fiddle

Related

How to consecutively count everything greater than or equal to itself in SQL?

Let's say if I have a table that contains Equipment IDs of equipments for each Equipment Type and Equipment Age, how can I do a Count Distinct of Equipment IDs that have at least that Equipment Age.
For example, let's say this is all the data we have:
equipment_type
equipment_id
equipment_age
Screwdriver
A123
1
Screwdriver
A234
2
Screwdriver
A345
2
Screwdriver
A456
2
Screwdriver
A567
3
I would like the output to be:
equipment_type
equipment_age
count_of_equipment_at_least_this_age
Screwdriver
1
5
Screwdriver
2
4
Screwdriver
3
1
Reason is there are 5 screwdrivers that are at least 1 day old, 4 screwdrivers at least 2 days old and only 1 screwdriver at least 3 days old.
So far I was only able to do count of equipments that falls within each equipment_age (like this query shown below), but not "at least that equipment_age".
SELECT
equipment_type,
equipment_age,
COUNT(DISTINCT equipment_id) as count_of_equipments
FROM equipment_table
GROUP BY 1, 2
Consider below join-less solution
select distinct
equipment_type,
equipment_age,
count(*) over equipment_at_least_this_age as count_of_equipment_at_least_this_age
from equipment_table
window equipment_at_least_this_age as (
partition by equipment_type
order by equipment_age
range between current row and unbounded following
)
if applied to sample data in your question - output is
Use a self join approach:
SELECT
e1.equipment_type,
e1.equipment_age,
COUNT(*) AS count_of_equipments
FROM equipment_table e1
INNER JOIN equipment_table e2
ON e2.equipment_type = e1.equipment_type AND
e2.equipment_age >= e1.equipment_age
GROUP BY 1, 2
ORDER BY 1, 2;
GROUP BY restricts the scope of COUNT to the rows in the group, i.e. it will not let you reach other rows (rows with equipment_age greater than that of the current group). So you need a subquery or windowing functions to get those. One way:
SELECT
equipment_type,
equipment_age,
(Select COUNT(*)
from equipment_table cnt
where cnt.equipment_type = a.equipment_type
AND cnt.equipment_age >= a.equipment_age
) as count_of_equipments
FROM equipment_table a
GROUP BY 1, 2, 3
I am not sure if your environment supports this syntax, though. If not, let us know we will find another way.

Finding all instances where a foreign key appears multiple times grouped by month

I am not too familiar with SQL, and I have been tasked with something that I quite frankly have no clue how to go about it.
I am just going to simplify the tables to the point where only the necessary fields are taken into consideration.
The tables look as follows.
Submission(course(string), student(foreign_key), date-submitted)
Student(id)
What I need to do is produce a table of active students per month, per course with a total. An active student being anyone who has more than 4 submissions in the month. I am only looking at specific courses, so I will need to hard code the values that I need, for the sake of the example "CourseA" and "CourseB"
The result should be as follows
month | courseA | CourseB | Total
------------------------------------------
03/2020 50 27 77
02/2020 25 12 37
01/2020 43 20 63
Any help would be greatly apreciated
You can do this with two levels of aggregation: first by month, course and student (while filtering on students having more than 4 submissions), then by month (while pivoting the dataset):
select
month_submitted,
count(*) filter(where course = 'courseA') active_students_in_courseA,
count(*) filter(where course = 'courseB') active_students_in_courseB,
count(*) total
from (
select
date_trunc('month', date_submitted) month_submitted,
course,
student_id,
count(*) no_submissions
from submission
where course in ('courseA', 'courseB')
group by 1, 2, 3
having count(*) > 4
) t
group by 1
You could do subqueries using the WITH keyword like this:
WITH monthsA AS (
SELECT to_char(date-submitted, "MM/YYYY") as month, course, COUNT(*) as students
FROM Submission
WHERE course = 'courseA'
GROUP BY 1, 2
), monthsB AS (
SELECT to_char(date-submitted, "MM/YYYY") as month, course, COUNT(*) AS students
FROM Submission
WHERE course = 'courseB'
GROUP BY 1, 2
)
SELECT ma.month,
COALESE(ma.students, 0) AS courseA,
COALESCE(mb.students) AS courseB,
COALESCE(ma.students, 0) + COALESCE(mb.students, 0) AS Total
FROM monthsA ma
LEFT JOIN monthsB mb ON ma.month = mb.month
ORDER BY 1 DESC

Calculate percentages of columns in Oracle SQL

I have three columns, all consisting of 1's and 0's. For each of these columns, how can I calculate the percentage of people (one person is one row/ id) who have a 1 in the first column and a 1 in the second or third column in oracle SQL?
For instance:
id marketing_campaign personal_campaign sales
1 1 0 0
2 1 1 0
1 0 1 1
4 0 0 1
So in this case, of all the people who were subjected to a marketing_campaign, 50 percent were subjected to a personal campaign as well, but zero percent is present in sales (no one bought anything).
Ultimately, I want to find out the order in which people get to the sales moment. Do they first go from marketing campaign to a personal campaign and then to sales, or do they buy anyway regardless of these channels.
This is a fictional example, so I realize that in this example there are many other ways to do this, but I hope anyone can help!
The outcome that I'm looking for is something like this:
percentage marketing_campaign/ personal campaign = 50 %
percentage marketing_campaign/sales = 0%
etc (for all the three column combinations)
Use count, sum and case expressions, together with basic arithmetic operators +,/,*
COUNT(*) gives a total count of people in the table
SUM(column) gives a sum of 1 in given column
case expressions make possible to implement more complex conditions
The common pattern is X / COUNT(*) * 100 which is used to calculate a percent of given value ( val / total * 100% )
An example:
SELECT
-- percentage of people that have 1 in marketing_campaign column
SUM( marketing_campaign ) / COUNT(*) * 100 As marketing_campaign_percent,
-- percentage of people that have 1 in sales column
SUM( sales ) / COUNT(*) * 100 As sales_percent,
-- complex condition:
-- percentage of people (one person is one row/ id) who have a 1
-- in the first column and a 1 in the second or third column
COUNT(
CASE WHEN marketing_campaign = 1
AND ( personal_campaign = 1 OR sales = 1 )
THEN 1 END
) / COUNT(*) * 100 As complex_condition_percent
FROM table;
You can get your percentages like this :
SELECT COUNT(*),
ROUND(100*(SUM(personal_campaign) / sum(count(*)) over ()),2) perc_personal_campaign,
ROUND(100*(SUM(sales) / sum(count(*)) over ()),2) perc_sales
FROM (
SELECT ID,
CASE
WHEN SUM(personal_campaign) > 0 THEN 1
ELSE 0
end AS personal_campaign,
CASE
WHEN SUM(sales) > 0 THEN 1
ELSE 0
end AS sales
FROM the_table
WHERE ID IN
(SELECT ID FROM the_table WHERE marketing_campaign = 1)
GROUP BY ID
)
I have a bit overcomplicated things because your data is still unclear to me. The subquery ensures that all duplicates are cleaned up and that you only have for each person a 1 or 0 in marketing_campaign and sales
About your second question :
Ultimately, I want to find out the order in which people get to the
sales moment. Do they first go from marketing campaign to a personal
campaign and then to sales, or do they buy anyway regardless of these
channels.
This is impossible to do in this state because you don't have in your table, either :
a unique row identifier that would keep the order in which the rows were inserted
a timestamp column that would tell when the rows were inserted.
Without this, the order of rows returned from your table will be unpredictable, or if you prefer, pure random.

SQL - Count Results of 2 Columns

I have the following table which contains ID's and UserId's.
ID UserID
1111 11
1111 300
1111 51
1122 11
1122 22
1122 3333
1122 45
I'm trying to count the distinct number of 'IDs' so that I get a total, but I also need to get a total of ID's that have also seen the that particular ID as well... To get the ID's, I've had to perform a subquery within another table to get ID's, I then pass this into the main query... Now I just want the results to be displayed as follows.
So I get a Total No for ID and a Total Number for Users ID - Also would like to add another column to get average as well for each ID
TotalID Total_UserID Average
2 7 3.5
If Possible I would also like to get an average as well, but not sure how to calculate that. So I would need to count all the 'UserID's for an ID add them altogether and then find the AVG. (Any Advice on that caluclation would be appreciated.)
Current Query.
SELECT DISTINCT(a.ID)
,COUNT(b.UserID)
FROM a
INNER JOIN b ON someID = someID
WHERE a.ID IN ( SELECT ID FROM c WHERE GROUPID = 9999)
GROUP BY a.ID
Which then Lists all the IDs and COUNT's all the USERID.. I would like a total of both columns. I've tried warpping the query in a
SELECT COUNT(*) FROM (
but this only counts the ID's which is great, but how do I count the USERID column as well
You seem to want this:
SELECT COUNT(DISTINCT a.ID), COUNT(b.UserID),
COUNT(b.UserID) * 1.0 / COUNT(DISTINCT a.ID)
FROM a INNER JOIN
b
ON someID = someID
WHERE a.ID IN ( SELECT ID FROM c WHERE GROUPID = 9999);
Note: DISTINCT is not a function. It applies to the whole row, so it is misleading to put an expression in parentheses after it.
Also, the GROUP BY is unnecessary.
The 1.0 is because SQL Server does integer arithmetic and this is a simple way to convert a number to a decimal form.
You can use
SELECT COUNT(DISTINCT a.ID) ...
to count all distinct values
Read details here
I believe you want this:
select TotalID,
Total_UserID,
sum(Total_UserID+TotalID) as Total,
Total_UserID/TotalID as Average
from (
SELECT (DISTINCT a.ID) as TotalID
,COUNT(b.UserID) as Total_UserID
FROM a
INNER JOIN b ON someID = someID
WHERE a.ID IN ( SELECT ID FROM c WHERE GROUPID = 9999)
) x

How can I SELECT the max row in a table SQL?

I have a little problem.
My table is:
Bill Product ID Units Sold
----|-----------|------------
1 | 10 | 25
1 | 20 | 30
2 | 30 | 11
3 | 40 | 40
3 | 20 | 20
I want to SELECT the product which has sold the most units; in this sample case, it should be the product with ID 20, showing 50 units.
I have tried this:
SELECT
SUM(pv."Units sold")
FROM
"Products" pv
GROUP BY
pv.Product ID;
But this shows all the products, how can I select only the product with the most units sold?
Leaving aside for the moment the possibility of having multiple products with the same number of units sold, you can always sort your results by the sum, highest first, and take the first row:
SELECT pv."Product ID", SUM(pv."Units sold")
FROM "Products" pv
GROUP BY pv."Product ID"
ORDER BY SUM(pv."Units sold") DESC
LIMIT 1
I'm not quite sure whether the double-quote syntax for column and table names will work - exact syntax will depend on your specific RDBMS.
Now, if you do want to get multiple rows when more than one product has the same sum, then the SQL will become a bit more complicated:
SELECT pv.`Product ID`, SUM(pv.`Units sold`)
FROM `Products` pv
GROUP BY pv.`Product ID`
HAVING SUM(pv.`Units sold`) = (
select max(sums)
from (
SELECT SUM(pv2.`Units sold`) as "sums"
FROM `Products` pv2
GROUP BY pv2.`Product ID`
) as subq
)
Here's the sqlfiddle
SELECT SUM(pv."Units sold") as `sum`
FROM "Products" pv
group by pv.Product ID
ORDER BY sum DESC
LIMIT 1
limit 1 + order by
The Best and effective way to this is Max function
Here's The General Syntax of Max function
SELECT MAX(ID) AS id
FROM Products;
and in your Case
SELECT MAX(Units Sold) from products
Here is the Complete Reference to MIN and MAX functions in Query
Click Here