如何统计SQL中所有组合出现的次数？

Question

是否有任何选项可以在不使用临时表或过程的情况下获得一个 SQL 查询中所有元素的计数组合？

考虑以下三个表：

• products (id, Product_name)

• transactions (id, date)

• transaction_has_product (id, Product_id, transaction_id)

示例数据

• 产品

  <前><代码>1 AAA 2 血脑屏障 3CCC

• 交易

  <前><代码>1 some_date 2 某个日期

• transaction_has_products

  <前><代码>1 1 1 2 2 1 3 3 1 4 1 2 5 2 2

结果应该是：

AAA, BBB = 2   
AAA, CCC = 1   
BBB, CCC = 1   
AAA, BBB, CCC = 1

Answer 1

这并不容易，因为与其他行相比，最后一行中的匹配产品数量不同。您也许可以使用某种 GROUP_CONCAT() 运算符（在 MySQL 中可用；可在其他 DBMS 中实现，例如 Informix 和可能的 PostgreSQL）来完成此操作，但我对此没有信心。

成对匹配

SELECT p1.product_name AS name1, p2.product_name AS name2, COUNT(*)
  FROM (SELECT p.product_name, h.transaction_id
          FROM products AS p
          JOIN transactions_has_products AS h ON h.product_id = p.product_id
       ) AS p1
  JOIN (SELECT p.product_name, h.transaction_id
          FROM products AS p
          JOIN transactions_has_products AS h ON h.product_id = p.product_id
       ) AS p2
    ON p1.transaction_id = p2.transaction_id
   AND p1.product_name   < p2.product_name
 GROUP BY p1.name, p2.name;

处理三重匹配并非易事；进一步扩展它肯定是相当困难的。

Answer 2

如果您预先知道所有产品是什么，您可以通过像这样旋转数据来做到这一点。

如果您事先不知道产品是什么，您可以在存储过程中动态构建此查询。如果产品数量很大，这两种方法的实用性都会崩溃，但我认为无论如何实现这一要求，这都可能是正确的。

select
    product_combination, 
    case product_combination
        when 'AAA, BBB' then aaa_bbb
        when 'AAA, CCC' then aaa_ccc
        when 'BBB, CCC' then bbb_ccc
        when 'AAA, BBB, CCC' then aaa_bbb_ccc
    end as number_of_transactions
from
(
    select 'AAA, BBB' as product_combination union all
    select 'AAA, CCC' union all
    select 'BBB, CCC' union all
    select 'AAA, BBB, CCC'
) as combination_list
cross join
(
    select
        sum(case when aaa = 1 and bbb = 1 then 1 else 0 end) as aaa_bbb,
        sum(case when aaa = 1 and ccc = 1 then 1 else 0 end) as aaa_ccc,
        sum(case when bbb = 1 and ccc = 1 then 1 else 0 end) as bbb_ccc,
        sum(case when aaa = 1 and bbb = 1 and ccc = 1 then 1 else 0 end) as aaa_bbb_ccc
    from
    (
        select
            count(case when a.product_name = 'AAA' then 1 else null end) as aaa,
            count(case when a.product_name = 'BBB' then 1 else null end) as bbb,
            count(case when a.product_name = 'CCC' then 1 else null end) as ccc,
            b.transaction_id
        from
            products a
        inner join
            transaction_has_products b
        on
            a.id = b.product_id
        group by
            b.transaction_id
    ) as product_matrix
) as combination_counts

结果：

product_combination  number_of_transactions
AAA, BBB             2
AAA, CCC             1
BBB, CCC             1
AAA, BBB, CCC        1

Answer 3

根据您对查询的控制程度（对于 postgresql，TSQL 可能需要更改）

SELECT COUNT(*) FROM transactions t WHERE
(
     SELECT COUNT(DISTINCT tp.product) 
     FROM transaction_has_products tp 
     WHERE tp.[transaction_id] = t.id and tp.product IN (1, 2, 3)
) = 3

，其中 (1,2,3) 是您要检查的 ID 列表， = 3 等于列表中的条目数量。

Answer 4

1. 生成所有可能的组合。我用这个来支持自己：https://stackoverflow.com/a/9135162/2244766（这有点棘手，我不完全理解逻辑...但它有效！）
2. 创建一个子查询，将 products_in_transactions 聚合到每个 transaction_id 的产品数组中
3. 使用数组包含运算符将它们连接起来

经过上述步骤后，您可能会得到类似的结果：

with all_combis as (
    with RECURSIVE y1 as (
            with x1 as (
                --select id from products
                select distinct product_id as a from transaction_has_products 
            )
            select array[a] as b ,a as c ,1 as d 
            from x1
            union all
            select b||a,a,d+1
            from x1
            join y1 on (a < c)
    )
    select *
    from y1
)
, grouped_transactions as (
  SELECT 
    array_agg(product_id) as products
  FROM transaction_has_products
  GROUP BY transaction_id
)
SELECT all_combis.b, count(*)
from all_combis
left JOIN grouped_transactions ON grouped_transactions.products @> all_combis.b 
--WHERE array_upper(b, 1) > 1 -- or whatever
GROUP BY all_combis.b
order by array_upper(b, 1) desc, count(*) desc

您可以加入您的products 表来用产品名称替换产品 id - 但我想你会从这里得到它。
这是小提琴（sqlfiddle 今天过得很糟糕 - 所以请在您的db，以防它抛出一些奇怪的错误，例如超时或类似的错误）

GL，HF：D