Question

我试图从我们的一个数据库表中提取与“G”类型（G GAR GER GR）匹配的特定类型研究的频率以及该周的研究总数（匹配所有类型）和'G'的相对频率。我希望将输出分组为几周（从星期一开始到星期五结束）。

我使用generate_series在感兴趣的时段内获得一系列星期一

SELECT (current_date- cast(extract(dow from current_date) as int) + 1) - s.a AS dates FROM generate_series(0,182,7) AS s(a);

使用以下命令

FROM (SELECT (current_date- cast(extract(dow from current_date) as int) + 1) - s.a AS dates FROM generate_series(0,182,7) AS s(a)) d
    LEFT JOIN study ON
study_date <= d.dates 
WHERE study_name ~ 'T(C|L)S' and study_type ~'G'                                            
GROUP BY 1
ORDER BY 1;

我得到以下输出

  dates    | count 
------------+-------
 26/06/2017 |     6
 03/07/2017 |    11
 10/07/2017 |    15
 17/07/2017 |    30
 24/07/2017 |    38
 31/07/2017 |    47
 07/08/2017 |    61
 14/08/2017 |    77

这给我一个正确的滚动计数，因为我们对'G'类型进行了更多的研究。但是，我尝试以下操作将其拆分为每周范围

SELECT d.dates, count(*)                                
FROM (SELECT (current_date- cast(extract(dow from current_date) as int) + 1) - s.a AS dates FROM generate_series(0,182,7) AS s(a)) d
    LEFT JOIN study ON
study_date >= d.dates  
WHERE study_name ~ 'T(C|L)S' and study_type ~ 'G' AND study_date <= myo_date + interval '5' day
GROUP BY 1
ORDER BY 1;

给出了我不理解的输出（我期待上面输出中行之间的差异）

  dates    | count 
------------+-------
 13/02/2017 |    79
 20/02/2017 |    79
 27/02/2017 |    79
 06/03/2017 |    79
 13/03/2017 |    79
 20/03/2017 |    79
 27/03/2017 |    79
 03/04/2017 |    79
 10/04/2017 |    79
 17/04/2017 |    79
 24/04/2017 |    79
 01/05/2017 |    79
 08/05/2017 |    79
 15/05/2017 |    79
 22/05/2017 |    79
 29/05/2017 |    79
 05/06/2017 |    79
 12/06/2017 |    79
 19/06/2017 |    79
 26/06/2017 |    74
 03/07/2017 |    70
 10/07/2017 |    68
 17/07/2017 |    53
 24/07/2017 |    44
 31/07/2017 |    34
 07/08/2017 |    22
 14/08/2017 |     3

如何每周获得计数？另外，如何在study_type可以是任何内容的情况下添加额外的列总计数。

预期输出

 dates    | count 
    ------------+-------
     26/06/2017 |    6
     03/07/2017 |    5
     10/07/2017 |    4
     17/07/2017 |    15
     24/07/2017 |    8
     31/07/2017 |    9
     07/08/2017 |    14
     14/08/2017 |    16

Answer 1

您可以尝试在此使用LAG()：

WITH cte AS (
    SELECT
        (current_date- cast(extract(dow from current_date) as int) + 1) - s.a AS dates,
    COUNT(*) AS cnt
    FROM generate_series(0,182,7) AS s(a)) d
    LEFT JOIN study
        ON study_date <= d.dates 
    WHERE study_name ~ 'T(C|L)S' and study_type ~'G'
    GROUP BY 1
)

SELECT
    dates,
    cnt - LAG(cnt, 0) OVER (ORDER BY dates)
FROM cte;

psql提取每周计数和总数的百分比

1 个答案: