如何找到特定列中零的平均值

Question

我的情况是我有如下记录。

+-----------+---------+---+--------+
|     a     |    b    | c |   d    |
+-----------+---------+---+--------+
| 537196605 | HZA-LOC | 0 | 201701 |
| 537196605 | HZA-LOC | 0 | 201702 |
| 537196605 | HZA-LOC | 0 | 201703 |
| 537196605 | HZA-LOC | 0 | 201704 |
| 537196605 | HZA-LOC | 0 | 201705 |
| 537196605 | HZA-LOC | 2 | 201706 |
| 537196605 | HZA-LOC | 0 | 201707 |
| 537196605 | HZA-LOC | 4 | 201708 |
| 537196605 | HZA-LOC | 0 | 201709 |
| 537196605 | HZA-LOC | 0 | 201710 |
| 537196605 | HZA-LOC | 0 | 201711 |
| 537196605 | HZA-LOC | 0 | 201712 |
+-----------+---------+---+--------+

现在我想在这里计算零的平均值。

我的意思是我需要计算为计数(零(/总零对，即 10/3 = 3.33

所有在一起的零都被视为一个值。前 5 个值将计为 1，第 7 个值计为 1,9-12 计为 1，总计为 3。请注意，C 列的值是根据 A 和 B 的分组生成的。期望的结果应如下所示：-

阿 b H537196605 HZA-LOC 3.33

请帮助我为此提供一个通用解决方案，因为我有数百万条这样的记录。

Answer 1

这是使用窗口函数的一种方法

SELECT [Avg] = (Count(*) * 1.0) / Sum(CASE WHEN prev_c = c THEN 0 ELSE 1 END) 
FROM   (SELECT *,
               prev_c = Lag(c)OVER(ORDER BY d)
        FROM   yourtable) a
WHERE  c = 0 

Answer 2

您可以使用

一个ROW_NUMBER()技巧来创建group_id值。

确定组后，计算

每个组的成员数和组数，就完成了。

http://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=e10b07b6fd9f8bc7549a3b98aaff3af6

WITH
  enumerated_example AS
(
  SELECT
    ROW_NUMBER() OVER (PARTITION BY a, b    ORDER BY d)   AS row_id,
    ROW_NUMBER() OVER (PARTITION BY a, b, c ORDER BY d)   AS zero_row_id,
    *
  FROM
    example
),
  zero_groups AS
(
  SELECT
    row_id - zero_row_id   AS group_id,
    COUNT(*)               AS group_size
  FROM
    enumerated_example
  WHERE
    c = 0
  GROUP BY
    row_id - zero_row_id
)
SELECT
  CAST(SUM(group_size) AS DECIMAL(13, 8)) / COUNT(*)
FROM
  zero_groups

Answer 3

一种方法是使用运行总和设置组。然后计算每个a，b组合的组数。此后，您只需要group by.

select a,b,1.0*grp_count/count(distinct grp) as result
from (select a,b,c,d,
      sum(case when c<>0 then 1 else 0 end) over(partition by a,b order by d) as grp,
      count(case when c=0 then 1 end) over(partition by a,b) as grp_count
      from tbl
     ) t
group by a,b,grp_count

Answer 4

一个不同(但相似(的解决方案：

CREATE TABLE #Sample (a int,
                      b varchar(7),
                      c tinyint,
                      d int)
INSERT INTO #Sample
VALUES
    (537196605,'HZA-LOC',0,201701),
    (537196605,'HZA-LOC',0,201702),
    (537196605,'HZA-LOC',0,201703),
    (537196605,'HZA-LOC',0,201704),
    (537196605,'HZA-LOC',0,201705),
    (537196605,'HZA-LOC',2,201706),
    (537196605,'HZA-LOC',0,201707),
    (537196605,'HZA-LOC',4,201708),
    (537196605,'HZA-LOC',0,201709),
    (537196605,'HZA-LOC',0,201710),
    (537196605,'HZA-LOC',0,201711),
    (537196605,'HZA-LOC',0,201712);
GO
WITH Groups AS(
    SELECT *,
           ROW_NUMBER() OVER (PARTITION BY a, b ORDER BY d) -
           ROW_NUMBER() OVER (PARTITION BY a, b, CASE c WHEN 0 THEN 0 ELSE 1 END ORDER BY d) AS Grp
    FROM #Sample)
SELECT a, b,
       CONVERT(decimal(6,2),COUNT(*)) / COUNT(DISTINCT Grp) AS h
FROM Groups
WHERE c = 0
GROUP BY a,b;
GO

DROP TABLE #Sample;
GO