one to many relation between table columns. Grouping and finding combinations - sql

Question

In sample table t0 :
OrderID | ProductID
0001 1254
0001 1252
0002 0038
0003 1254
0003 1252
0003 1432
0004 0038
0004 1254
0004 1252
I need to find the most popular combination of two ProductIDs under one OrderID. The purpose is to decide which products are more likely to be sold together in one order e.g phone - handsfree. I think the logic is to group by OrderID, calculate every possible combination of productID pairs, count them per OrderID and select the TOP 2, but i realy can't tell if it is doable..

Answer 1

A "self-join" may be used but ensuring that one of the product ids is greater then than the other so that we get get "pairs" of products per order. Then it is simple to count:
Demo
CREATE TABLE OrderDetail
([OrderID] int, [ProductID] int)
;
INSERT INTO OrderDetail
([OrderID], [ProductID])
VALUES
(0001, 1254), (0001, 1252), (0002, 0038), (0003, 1254), (0003, 1252), (0003, 1432), (0004, 0038), (0004, 1254), (0004, 1252)
;
Query 1:
select -- top(2)
od1.ProductID, od2.ProductID, count(*) count_of
from OrderDetail od1
inner join OrderDetail od2 on od1.OrderID = od2.OrderID and od2.ProductID > od1.ProductID
group by
od1.ProductID, od2.ProductID
order by
count_of DESC
Results:
| ProductID | ProductID | count_of |
|-----------|-----------|----------|
| 1252 | 1254 | 3 |
| 1252 | 1432 | 1 |
| 1254 | 1432 | 1 |
| 38 | 1252 | 1 |
| 38 | 1254 | 1 |
----
With respect to displaying the "top 2" or whatever. You are likely to get "equal top" results so I would suggest you need to use dense_rank() and you may even want to "unpivot" the result so you have a single column of productids with their associated rank. How often you perform this and/or store this I leave to you.
with ProductPairs as (
select
p1, p2, count_pair
, dense_rank() over(order by count_pair DESC) as ranked
from (
select
od1.ProductID p1, od2.ProductID p2, count(*) count_pair
from OrderDetail od1
inner join OrderDetail od2 on od1.OrderID = od2.OrderID and od2.ProductID > od1.ProductID
group by
od1.ProductID, od2.ProductID
) d
)
, RankedProducts as (
select p1 as ProductID, ranked, count_pair
from ProductPairs
union all
select p2 as ProductID, ranked, count_pair
from ProductPairs
)
select *
from RankedProducts
where ranked <= 2
order by ranked, ProductID

Answer 2

WITH products as (
SELECT DISTINCT ProductID
FROM orders
), permutation as (
SELECT p1.ProductID as pidA,
p2.ProductID as pidB
FROM products p1
JOIN products p2
ON p1.ProductID < p2.ProductID
), check_frequency as (
SELECT pidA, pidB, COUNT (o2.orderID) total_orders
FROM permutations p
LEFT JOIN orders o1
ON p.pidA = o1.ProductID
LEFT JOIN orders o2
ON p.pidB = o2.ProductID
AND o1.orderID = o2.orderID
GROUP BY pidA, pidB
)
SELECT TOP 2 *
FROM check_frequency
ORDER BY total_orders DESC

Answer 3

The following query calculates the number of two-way combinations
among all orders in Orderline:
SELECT SUM(numprods * (numprods - 1)/2) as numcombo2
FROM ( SELECT orderid, COUNT(DISTINCT productid) as numprods
FROM orderline ol
GROUP BY orderid ) o
Notice that this query counts distinct products rather than order lines, so
orders with the same product on multiple lines do not affect the count.
The number of two-way combinations is 185,791. This is useful because the
number of combinations pretty much determines how quickly the query generating
them runs. A single order with a large number of products can seriously
degrade performance. For instance, if one order contains a thousand
products, there would be about five hundred thousand two-way combinations
in just that one order—versus 185,791 in all the orders data. As the number of
products in the largest order increases, the number of combinations increases
much faster.subject to the conditions:
The two products in the pair are different
No two combinations have the same two products.
The approach for calculating the combinations is to do a self-join on the Orderline
table, with duplicate product pairs removed. The goal is to get all pairs of
products
The first condition is easily met by filtering out any pairs where the two products
are equal. The second condition is also easily met, by requiring that the
first product id be smaller than the second product id. The following query
generates all the combinations in a subquery and counts the number of orders
containing each one:
SELECT p1, p2, COUNT(*) as numorders
FROM (SELECT op1.orderid, op1.productid as p1, op2.productid as p2
FROM (SELECT DISTINCT orderid, productid FROM orderline) op1 JOIN
(SELECT DISTINCT orderid, productid FROM orderline) op2
ON op1.orderid = op2.orderid AND
op1.productid < op2.productid
) combinations
GROUP BY p1, p2
source Data Analysis Using SQL and Excel

Answer 4

Try using the following commnand:
SELECT T1.orderID,T1.productId,T2.productID,Count(*) as Occurence
FROM TBL T1 INNER JOIN TBL T2
ON T1.orderid = T2.orderid
WHERE t1.productid > T2.productId
GROUP BY T1.orderID,T1.productId,T2.productID
ORDER BY Occurence DESC
SQL fiddle