2

经过多年阅读答案，终于到了我自己提出问题的时候了。

我有一个购买的产品列表和唯一的客户 ID：

+---------+--------+
| Product | Buyer  |
+---------+--------+
| Apples  | Rod    |
| Apples  | Jane   |
| Apples  | Freddy |
| Bananas | Rod    |
| Bananas | Jane   |
| Bananas | Freddy |
| Bananas | Zippy  |
| Pears   | Rod    |
| Pears   | Zippy  |
+---------+--------+

我想在 Netezza SQL 中生成以下输出：

+-----------+-------------+------------------------+---------------------+
| Product A | Buyers of A | A Buyers Also Bought B | No of A Buyers of B |
+-----------+-------------+------------------------+---------------------+
| Apples    |           3 | Bananas                |                   3 |
| Apples    |           3 | Pears                  |                   1 |
| Bananas   |           4 | Apples                 |                   3 |
| Bananas   |           4 | Pears                  |                   2 |
| Pears     |           2 | Apples                 |                   1 |
| Pears     |           2 | Bananas                |                   2 |
+-----------+-------------+------------------------+---------------------+

..这样我就可以看到每个产品的总购买者。至关重要的是，我还想查看对于每种产品，在这些购买者中，有多少人购买了同一列表中的其他产品。 编辑：重要的是要重申，如果他们没有购买产品 A ，我不应该让任何买家出现在 B 的列中。

请问最有效的方法是什么？

（然后我会计算出 B 购买 A 的百分比，但这部分很容易）。

谢谢！

4

2 回答 2

0

常见购买的基本数据是自加入和group by：

select p1.product, p2.product, count(*) as in_common
from purchases p1 join
     purchases p2
     on p1.buyer = p2.buyer
group by p1.product, p2.product;

要获得一个（或另一个）的计数，则为join：

select p1.product, p2.product, pp.cnt, count(*) as in_common
from purchases p1 join
     purchases p2
     on p1.buyer = p2.buyer join
     (select p1.product, count(*) as cnt
      from purchases
      group by p1.product
     ) pp
     on pp.product = p1.product
group by p1.product, p2.product, pp.cnt;

或者，您可以使用窗口函数：

select p1.product, p1.cnt, p2.product, count(*) as in_common
from (select p1.*,
             count(*) over (partition by p1.product) as cnt
      from purchases p1
     ) p1 join
     purchases p2
     on p1.buyer = p2.buyer
group by p1.product, p2.product, p1.cnt;

这是一个显示它工作的rextester。

于 2018-09-02T05:24:22.003 回答
0

您可以创建计数摘要，然后与自身交叉连接，不包括相同的匹配项。

像这样：

SELECT 
    A.Product,
    A.Buyers,
    B.Product,
    B.Buyers
FROM (
    SELECT
        Product
        count(*) AS Buyers
    FROM
        ProductBuyers
    GROUP BY
) AS A
CROSS JOIN (
    SELECT
        Product
        count(*) AS Buyers
    FROM
        ProductBuyers
    GROUP BY
) AS B
WHERE 
    A.Product != B.Product
于 2018-09-02T00:20:02.683 回答