Question

我有如下数据：

CustomerId   Category
  100            2
  100            2
  100            3
  100            6
  100            4
  200            3
  200            6
  200            7
  300            2

所以我想要的输出是Jaccard Similarity Index：

Jaccard（100,200）= 2（共享项目）/ 5（项目联合）
Jaccard（100,300）= 1（共享项目）/ 4（项目联合）
Jaccard（200,300）= 0（共享项目）/ 4（项目联合）。

我最初尝试的是找到联盟和条款的交集，但我不确定它是否是最有效的方式。另外，我想避免像Jaccard（100,300）和Jaccard（300,100）一起出现的重复。有人可以帮忙吗？

     select t1.customer_id, t2.customer_id,
              sum(case when t1.category_id = t2.category_id then 1 else 0 end) intersection,
sum(case when t1.category = t2.category then 1
         when t1.category <> t2.category then 1 else 0 end)
union
    from t t1 cross join
         t t2
  Where t1.customer_id <> t2.customer_id
    group by t1.customer_id, t2.customer_id

不幸的是，我还检查过我有一位顾客在同一类别购买多件商品。因此，我编辑了表格，以反映客户100在类别2中有两个项目。但是，它不应该更改Jaccard相似性度量值。

Answer 1

您不需要cross join。通过计算一对的不同category_id的总和并从中减去相交的category_id来获得分母。

SELECT t1.customer_id AS id1,
       t2.customer_id AS id2,
       1.0*sum(CASE WHEN t1.category_id = t2.category_id THEN 1 ELSE 0 END)
     / (count(DISTINCT t1.category_id)+count(DISTINCT t2.category_id)-sum(CASE WHEN t1.category_id = t2.category_id THEN 1 ELSE 0 END)) AS jaccard_similarity
FROM t t1
JOIN t t2 ON t1.customer_id<t2.customer_id
GROUP BY t1.customer_id, t2.customer_id

如果join不支持不等式，请使用

SELECT t1.customer_id AS id1,
       t2.customer_id AS id2,
       1.0*sum(CASE WHEN t1.category_id = t2.category_id THEN 1 ELSE 0 END)
       / (count(DISTINCT t1.category_id)+count(DISTINCT t2.category_id)-sum(CASE WHEN t1.category_id = t2.category_id THEN 1 ELSE 0 END)) AS jaccard_similarity
FROM t t1
CROSS JOIN t t2 
WHERE t1.customer_id<t2.customer_id
GROUP BY t1.customer_id, t2.customer_id

如果你只需要成对的交叉点数，下面的查询就足够了。

select t1.customer_id as id1, t2.customer_id as id2
,sum(case when t1.category_id = t2.category_id then 1 else 0 end) as intersection
from t t1 
join t t2 on t1.customer_id<t2.customer_id
group by t1.customer_id, t2.customer_id

编辑：根据OP的评论，客户可以多次拥有相同的类别，但只应计算一次。

SELECT t1.customer_id AS id1,
       t2.customer_id AS id2,
       1.0*COUNT(DISTINCT CASE WHEN t1.category_id = t2.category_id THEN t1.category_id END)
       / (COUNT(DISTINCT t1.category_id)+COUNT(DISTINCT t2.category_id)
         -COUNT(DISTINCT CASE WHEN t1.category_id = t2.category_id THEN t1.category_id END)) AS jaccard_similarity
FROM t t1
CROSS JOIN t t2 
WHERE t1.customer_id<t2.customer_id
GROUP BY t1.customer_id, t2.customer_id

蜂巢中的Jaccard相似度计算

1 个答案: