功能与clickouse中的滞后分区相同 [英] function same as lag partition by in clickouse

问题描述

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个订购时间之间存在差异. 在SQL中，我使用了延迟分区依据"但我不知道如何在点击房子中计算出这一点. 我需要以下数据:

I need to know the frequency of order for each user. I mean difference between 2 order time for each user. In SQL I used "Lag Partition by" but I don't know how I can calculate this in click house. I need this data:

首先，我应该使用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设置下一个订购时间.我不能使用邻居功能，因为它无法按user_id进行分区.

at first I should sort data with user_id and created_at then I need to have next order time for each user id in row. I can't use neighbor function because it can't do partition by user_id.

推荐答案

我不明白为什么

I didn't understand why neighbor cannot be used in your case, but it should works well:

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 │
└─────────┴─────────────────────┴─────────────────────┘
*/

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