功能与 clickhouse 中的 lag partition by 相同

Question

我需要知道每个用户的订购频率。我的意思是每个用户的 2 个订购时间之间的差异。在 SQL 中，我使用了“Lag Partition by”，但我不知道如何在 click house 中计算它。我需要这些数据：

首先我应该用 user_id 和 created_at 对数据进行排序然后我需要为行中的每个用户 ID 设置下一个订单时间。我不能使用邻居功能，因为它不能按 user_id.

进行分区

Answer 1

我不明白为什么 neighbor 不能用于你的情况，但它应该运行良好：

SELECT 
    user_id,
    created,
    if(neighbor(user_id, 1, NULL) != user_id, NULL, neighbor(created, 1, NULL)) AS next_created
FROM 
(
    SELECT 
        number % 3 AS user_id,
        now() + (number * 360) AS created
    FROM numbers(11)
    ORDER BY 
        user_id ASC,
        created ASC
)

/*
┌─user_id─┬─────────────created─┬────────next_created─┐
│       0 │ 2020-10-21 16:00:21 │ 2020-10-21 16:18:21 │
│       0 │ 2020-10-21 16:18:21 │ 2020-10-21 16:36:21 │
│       0 │ 2020-10-21 16:36:21 │ 2020-10-21 16:54:21 │
│       0 │ 2020-10-21 16:54:21 │                ᴺᵁᴸᴸ │
│       1 │ 2020-10-21 16:06:21 │ 2020-10-21 16:24:21 │
│       1 │ 2020-10-21 16:24:21 │ 2020-10-21 16:42:21 │
│       1 │ 2020-10-21 16:42:21 │ 2020-10-21 17:00:21 │
│       1 │ 2020-10-21 17:00:21 │                ᴺᵁᴸᴸ │
│       2 │ 2020-10-21 16:12:21 │ 2020-10-21 16:30:21 │
│       2 │ 2020-10-21 16:30:21 │ 2020-10-21 16:48:21 │
│       2 │ 2020-10-21 16:48:21 │                ᴺᵁᴸᴸ │
└─────────┴─────────────────────┴─────────────────────┘
*/

Answer 2

groupArray 允许转换这个

select 1 uid, 555 time union all select 1, 666 union all select 1, 777
┌─uid─┬─time─┐
│   1 │  555 │
│   1 │  666 │
│   1 │  777 │
└─────┴──────┘

至此

select uid, groupArray(time) dtime from 
   (select * from (select 1 uid, 555 time union all select 1, 666 union all select 1, 777) order by uid, time) 
  group by uid

┌─uid─┬─dtime─────────┐
│   1 │ [555,666,777] │
└─────┴───────────────┘

和

select uid, arrayJoin(arrayPushBack(arrayPopFront(groupArray(time)), null)) dtime from 
   (select * from (select 1 uid, 555 time union all select 1, 666 union all select 1, 777) order by uid, time) 
  group by uid
┌─uid─┬─dtime─┐
│   1 │   666 │
│   1 │   777 │
│   1 │  ᴺᵁᴸᴸ │
└─────┴───────┘


select uid, time, atime from (
  select uid, groupArray(time) as stime, arrayPushBack(arrayPopFront(stime), null) dtime from 
   (select * from (select 1 uid, 555 time union all select 1, 666 union all select 1, 777) order by uid, time) 
  group by uid )
   array join stime as time, dtime as atime
┌─uid─┬─time─┬─atime─┐
│   1 │  555 │   666 │
│   1 │  666 │   777 │
│   1 │  777 │  ᴺᵁᴸᴸ │
└─────┴──────┴───────┘

功能与 clickhouse 中的 lag partition by 相同

function same as lag partition by in clickouse

lag

partition

clickhouse