SQL（BigQuery） - 在整个时间查找不寻常的值序列

Question

我有一个星期几和一天中的小时数的美元价值列表（这是从时间戳，所以我只做dayOfWeek和hourOfDay为1周）

Id | dayOfWeek | hourOfDay | dollars 
1  1   1   0 
1  1   2   0 
1  1   3   0 
1  1   4   0 
1  1   5   6 
1  1   6   5 
1  1   7   7 
1  1   8   18 
1  1   9   13 
1  1   10   19 
1  1   11   18 
1  1   12   13 
1  1   13   19 
1  1   14   10 
1  1   15   16 
1  1   16   15 
1  1   17   17 
1  1   18   18 
1  1   19   13 
1  1   20   0 
1  1   21   0 
1  1   22   0 
1  1   23   0 
1  2   1   0 
1  2   2   0 
1  2   3   0 
1  2   4   0 
1  2   5   16 
1  2   6   15 
1  2   7   27 
1  2   8   11 
1  2   9   13 
1  2   10   11 
1  2   11   18 
1  2   12   14 
1  2   13   14 
1  2   14   10 
1  2   15   16 
1  2   16   15 
1  2   17   17 
1  2   18   18 
1  2   19   13 
1  2   20   10 
1  2   21   22 
1  2   22   0 
1  2   23   0 

我想找到高于平均连续0的日期结束时的Ids。我正在考虑使用诸如percent_rank（）之类的方法来查找“高于平均水平”的情况，但我无法将每个Id的0个情况的连续实例组合在一起。

任何帮助将非常感激，但请让我知道，如果我没有正确思考方式，或者如果我应该考虑一个不同的方向。非常感谢。

Answer 1

下面是BigQuery的标准SQL

#standardSQL 
WITH outages AS (
    SELECT 
    id, 
    MIN(dayOfWeek) AS dayOfWeek, 
    MIN(hourOfDay) AS hourOfDay, 
    COUNT(1) AS len 
    FROM (
    SELECT 
     id, seq, 
     FIRST_VALUE(dayOfWeek) OVER(win) AS dayOfWeek, 
     FIRST_VALUE(hourOfDay) OVER(win) AS hourOfDay 
    FROM (
     SELECT 
     id, dayOfWeek, hourOfDay, dollars, 
     COUNTIF(dollars <> 0) OVER(PARTITION BY id ORDER BY dayOfWeek, hourOfDay) AS seq 
     FROM `yourTable` 
    ) 
    WHERE dollars = 0 
    WINDOW win AS (PARTITION BY id, seq ORDER BY dayOfWeek, hourOfDay) 
) 
    GROUP BY id, seq 
), 
averages AS (
    SELECT id, AVG(len) AS len 
    FROM outages 
    GROUP BY id 
) 
SELECT o.* 
FROM outages AS o JOIN averages AS a 
ON o.id = a.id AND o.len > a.len 

您可以测试/使用其虚拟数据从你的问题如下

#standardSQL 
WITH yourTable AS (
    SELECT * FROM UNNEST([STRUCT<id INT64, dayOfWeek INT64, hourOfDay INT64, dollars INT64>(1, 1, 1, 0),(1, 1, 2, 0),(1, 1, 3, 0),(1, 1, 4, 0),(1, 1, 5, 6),(1, 1, 6, 5),(1, 1, 7, 7),(1, 1, 8, 18),(1, 1, 9, 13),(1, 1, 10, 19),(1, 1, 11, 18),(1, 1, 12, 13),(1, 1, 13, 19),(1, 1, 14, 10),(1, 1, 15, 16),(1, 1, 16, 15),(1, 1, 17, 17),(1, 1, 18, 18),(1, 1, 19, 13),(1, 1, 20, 0),(1, 1, 21, 0),(1, 1, 22, 0),(1, 1, 23, 0),(1, 2, 0, 0),(1, 2, 1, 0),(1, 2, 2, 0),(1, 2, 3, 0),(1, 2, 4, 0),(1, 2, 5, 16),(1, 2, 6, 15),(1, 2, 7, 27),(1, 2, 8, 11),(1, 2, 9, 13),(1, 2, 10, 11),(1, 2, 11, 18),(1, 2, 12, 14),(1, 2, 13, 14),(1, 2, 14, 10),(1, 2, 15, 16),(1, 2, 16, 15),(1, 2, 17, 17),(1, 2, 18, 18),(1, 2, 19, 13),(1, 2, 20, 10),(1, 2, 21, 22),(1, 2, 22, 0),(1, 2, 23, 0)]) 
), 
outages AS (
    SELECT 
    id, 
    MIN(dayOfWeek) AS dayOfWeek, 
    MIN(hourOfDay) AS hourOfDay, 
    COUNT(1) AS len 
    FROM (
    SELECT 
     id, seq, 
     FIRST_VALUE(dayOfWeek) OVER(win) AS dayOfWeek, 
     FIRST_VALUE(hourOfDay) OVER(win) AS hourOfDay 
    FROM (
     SELECT 
     id, dayOfWeek, hourOfDay, dollars, 
     COUNTIF(dollars <> 0) OVER(PARTITION BY id ORDER BY dayOfWeek, hourOfDay) AS seq 
     FROM `yourTable` 
    ) 
    WHERE dollars = 0 
    WINDOW win AS (PARTITION BY id, seq ORDER BY dayOfWeek, hourOfDay) 
) 
    GROUP BY id, seq 
), 
averages AS (
    SELECT id, AVG(len) AS len 
    FROM outages 
    GROUP BY id 
) 
SELECT o.* 
FROM outages AS o JOIN averages AS a 
ON o.id = a.id AND o.len > a.len 

正如你可以看到这里玩 - outages子选择计算所有具有序列长度和该序列开始的零序列，并输出如下

id dayOfWeek hourOfDay len 
1 1   1   4  
1 1   20   9  
1 2   22   2  

个

最终选择输出从中断只有行，其中各长度大于平均长度（从averages子选择），选择那些ID

id dayOfWeek hourOfDay len 
1 1   20   9