BigQuery 的所有时间唯一性（按行）

Question

我有一个与上周非常相似的问题：Get all time unique values in BigQuery

我有一个这样的数据库：

ID	Day	Value
1	2021-09-01	a
2	2021-09-01	b
3	2021-09-01	c
4	2021-09-02	d
5	2021-09-02	a
6	2021-09-02	a
7	2021-09-02	e
8	2021-09-03	c
9	2021-09-03	f
10	2021-09-03	a

我想计算我每天和所有时间有多少不同的行，但所有时间的唯一性应该只计算之前的日期（如果用户是新用户，我想计算其背后的业务逻辑） .与之前的问题不同，我想留下行，但我想按行查看唯一性（作为新列）。这与我们作为新用户或老用户在 Google Analytics（分析）上所拥有的几乎相同。因此，如果用户在 2021 年 9 月 2 日访问该站点并在 2021 年 9 月 3 日访问该站点，首先我希望看到新用户，但在 2021 年 9 月 3 日我希望看到返回用户。 所以我想看看这个输出

ID	Day	Value	Type
1	2021-09-01	a	New
2	2021-09-01	b	New
3	2021-09-01	c	New
4	2021-09-02	d	New
5	2021-09-02	a	Returning
6	2021-09-02	a	Returning
7	2021-09-02	e	Returning
8	2021-09-03	c	New
9	2021-09-03	f	New
10	2021-09-03	a	Returning

如果我只检查一天我可以做到，但如果我在整个数据库上检查这些，我就不能这样做，因为检查前的日期。

Answer 1

看来您想使用doc 中详述的分析功能

使用OVER 和PARTITION BY 的分析功能，您可以通过值对数据进行分区，然后使用ORDER BY 按日期对其进行排序。现在检查它是否是该分区中的第一行并相应地分配类型。

这个查询应该得到你想要的；

WITH data as(
    SELECT "2021-09-01" day,"a" value
    UNION ALL ( SELECT "2021-09-01", "b" )
    UNION ALL ( SELECT "2021-09-01", "c" )
    UNION ALL ( SELECT "2021-09-02", "d" )
    UNION ALL ( SELECT "2021-09-02", "a" )
    UNION ALL ( SELECT "2021-09-02", "a" )
    UNION ALL ( SELECT "2021-09-02", "e" )
    UNION ALL ( SELECT "2021-09-03", "c" )
    UNION ALL ( SELECT "2021-09-03", "f" )
    UNION ALL ( SELECT "2021-09-03", "a" )
    )
    
    SELECT day, value,
      IF(ROW_NUMBER() OVER (PARTITION BY value ORDER BY day) = 1, 'New','Returning') as type
    
    FROM data

结果

Row	day	value	type
1	2021-09-01	a	New
2	2021-09-02	a	Returning
3	2021-09-02	a	Returning
4	2021-09-03	a	Returning
5	2021-09-01	b	New
6	2021-09-01	c	New
7	2021-09-03	c	Returning
8	2021-09-02	d	New
9	2021-09-02	e	New
10	2021-09-03	f	New

修改了附加要求

要将具有相同日期的所有分组 values 作为第一个事件提供 New 类型，您可以使用另一个分析函数 FIRST_VALUE 并结合当前日期值。

WITH data as
(SELECT "2021-09-01" day,"a" value
UNION ALL ( SELECT "2021-09-01","b")
UNION ALL ( SELECT "2021-09-01","c")
UNION ALL ( SELECT "2021-09-02","d")
UNION ALL ( SELECT "2021-09-02","a")
UNION ALL ( SELECT "2021-09-01","a")
UNION ALL ( SELECT "2021-09-02","a")
UNION ALL ( SELECT "2021-09-02","e")
UNION ALL ( SELECT "2021-09-03","c")
UNION ALL ( SELECT "2021-09-03","f")
UNION ALL ( SELECT"2021-09-03","a"))

SELECT *,
IF(ROW_NUMBER() OVER (PARTITION BY value ORDER BY day) = 1 OR FIRST_VALUE(day) OVER (PARTITION BY value ORDER BY day) = day, 'New','Returning')  as type
FROM data

结果

Row	day	value	type
1	2021-09-01	a	New
2	2021-09-01	a	New
3	2021-09-02	a	Returning
4	2021-09-02	a	Returning
5	2021-09-03	a	Returning
6	2021-09-01	b	New
7	2021-09-01	c	New
8	2021-09-03	c	Returning
9	2021-09-02	d	New
10	2021-09-02	e	New
11	2021-09-03	f	New

Answer 2

考虑下面的方法

select *, if(0 = count(*) over prev_days, 'New', 'Returning') as type
from your_table
window prev_days as (
  partition by value order by unix_date(date(day)) 
  range between unbounded preceding and 1 preceding 
)