Question

在BigQuery上考虑这个表模式：

Table User
{
user_id: STRING (REQUIRED)
user_name: STRING (REQUIRED)
actions: RECORD (REPEATED) 
    {
        action_id: STRING (REQUIRED)
        action_type: INTEGER (REQUIRED)
        action_date: TIMESTAMP (REQUIRED)
    }
}

我想找到多次创建某种类型操作的所有用户（user_id和user_name），这些操作之间的最短时间少于X天。

未定义每个用户的存储操作数（可以是1,2或n）。这些操作不按任何标准排序（但我认为可以使用ORDER BY解决这个问题。）

例如，与用户：

{
    user_id: "u1", 
    user_name: "User 1", 
    actions: 
    {action_id: "a1", action_type: 1, action_date: "2016-02-22"},
    {action_id: "a2", action_type: 1, action_date: "2016-01-22"},
    {action_id: "a3", action_type: 1, action_date: "2015-12-22"}
},
{
    user_id: "u2", 
    user_name: "User 2", 
    actions: 
    {action_id: "a4", action_type: 1, action_date: "2016-02-22"},
    {action_id: "a5", action_type: 2, action_date: "2016-01-22"},
    {action_id: "a6", action_type: 1, action_date: "2015-12-22"}
},
{
    user_id: "u3", 
    user_name: "User 3", 
    actions: 
    {action_id: "a7", action_type: 1, action_date: "2016-02-22"}
},
{
    user_id: "u4", 
    user_name: "User 4", 
    actions: 
    {action_id: "a8", action_type: 1, action_date: "2016-02-22"},
    {action_id: "a9", action_type: 1, action_date: "2015-02-22"},
    {action_id: "a10", action_type: 1, action_date: "2015-01-22"}
},

查询＆＃34;选择多次执行1类型操作的用户，每次执行之间的最短时间小于45天＆＃34;应该返回User 1和User 4。

关于如何在BigQuery上执行此操作的任何想法？

Answer 1

请尝试以下
写在路上，因此没有经过测试，但我觉得它应该工作并做你需要的事情

SELECT 
  user_id, 
  user_name, 
  action_type, 
  MIN(DATEDIFF(action_date_next, action_date)) AS min_distance
FROM (
  SELECT 
    user_id, 
    user_name, 
    action_type, 
    action_date, 
    LAG(action_date) 
        OVER(PARTITION BY user_id, action_type 
        ORDER BY action_date DESC) AS action_date_next
  FROM (
    SELECT 
      user_id, 
      user_name, 
      actions.action_type AS action_type, 
      actions.action_date AS action_date 
    FROM table_users 
  )
)
WHERE action_date_next IS NOT NULL
GROUP BY user_id, user_name, action_type
HAVING action_type = 1 AND min_distance < 45

以下版本更紧凑 - 尝试一下

SELECT 
  user_id, 
  user_name, 
  action_type, 
  MIN(DATEDIFF(action_date_next, action_date)) AS min_distance
FROM (
  SELECT 
    user_id, 
    user_name, 
    actions.action_type AS action_type, 
    actions.action_date AS action_date, 
    LAG(actions.action_date) 
        OVER(PARTITION BY user_id, actions.action_type 
        ORDER BY actions.action_date DESC) AS action_date_next
  FROM table_users
)
WHERE action_date_next IS NOT NULL
GROUP BY user_id, user_name, action_type
HAVING action_type = 1 AND min_distance < 45

BigQuery：选择重复记录中字段之间的最小差异

1 个答案: