按列分组 + COUNT(*)，如何获得每个组合的平均计数？

发布于 2024-11-28 07:56:39 字数 1108 浏览 2 评论 0原文

我有以下（简化的）查询：

SELECT     ResolvedBy, COUNT(*) AS Count, fiCategory, fiSubCategory, fiSymptom
FROM         tContact
WHERE     (ResolvedBy IS NOT NULL)
GROUP BY ResolvedBy, fiCategory, fiSubCategory, fiSymptom
ORDER BY Count DESC

现在我需要 fiCategory、fiSubCategory、fiSymptom 的每个组合的平均计数作为列。怎么做呢？

例如：

ResolvedBy    Count    fiCategory    fiSubCategory    fiSymptom    Average
    1           50         1              2             3            40
    2           30         1              2             3            40
    3           40         1              2             3            40
    1           20         2              3             4            30
    2           40         2              3             4            30

示例中是 fiCategory、fiSubCategory 和 fiSymptom 的两种组合：1,2,3 和 2,3,4。因此，计算出两个平均值：

50+30+40 / 3 = 40
20+40 / 2 = 30。

所以我想对每个组合的计数求和并除以出现的次数。

编辑：该示例是提取所需的查询结果。计数是每个 ResolvedBy 出现的所有该组合的总和。

先感谢您。

原文

I have following (simplified) query:

SELECT     ResolvedBy, COUNT(*) AS Count, fiCategory, fiSubCategory, fiSymptom
FROM         tContact
WHERE     (ResolvedBy IS NOT NULL)
GROUP BY ResolvedBy, fiCategory, fiSubCategory, fiSymptom
ORDER BY Count DESC

Now i need the average count for every combination of fiCategory, fiSubCategory, fiSymptom as column. How to do that?

For example:

ResolvedBy    Count    fiCategory    fiSubCategory    fiSymptom    Average
    1           50         1              2             3            40
    2           30         1              2             3            40
    3           40         1              2             3            40
    1           20         2              3             4            30
    2           40         2              3             4            30

In the example are two combinations of fiCategory,fiSubCategory and fiSymptom: 1,2,3 and 2,3,4. Hence there are two averages that are calculated:

50+30+40 / 3 = 40
20+40 / 2 = 30.

So i want to sum the count of every combination and divide through the number of occurences.

Edit: The example is an extraction of the desired result of the query. The count is the sum of all occurences of this combination for every ResolvedBy.

Thank you in advance.

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白云不回头 2024-12-05 07:56:39

Select ResolvedBy, [Count], fiCategory, fiSubCategory, fiSymptom
    , Avg(Z.Count) Over( Partition By fiCategory, fiSubCategory, fiSymptom ) As AvgByGrp
From    (
        Select ResolvedBy, Count(*) As [Count], fiCategory, fiSubCategory, fiSymptom
        From tContact 
        Group By ResolvedBy, fiCategory, fiSubCategory, fiSymptom
        ) As Z

Order By Z.Count Desc

Select ResolvedBy, [Count], fiCategory, fiSubCategory, fiSymptom
    , Avg(Z.Count) Over( Partition By fiCategory, fiSubCategory, fiSymptom ) As AvgByGrp
From    (
        Select ResolvedBy, Count(*) As [Count], fiCategory, fiSubCategory, fiSymptom
        From tContact 
        Group By ResolvedBy, fiCategory, fiSubCategory, fiSymptom
        ) As Z

Order By Z.Count Desc

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萌辣 2024-12-05 07:56:39

试试这个：

SELECT main.ResolvedBy, COUNT(*) AS Count, 
    main.fiCategory, main.fiSubCategory, main.fiSymptom, average
FROM tContact main
JOIN (SELECT COUNT(*)/count(distinct ResolvedBy) as average,
      fiCategory, fiSubCategory, fiSymptom group by 2,3,4) x
        on x.fiCategory = main.fiCategory
        and x.fiSubCategory = main.fiSubCategory
        and x.fiSymptom = main.fiSymptom
WHERE main.ResolvedBy IS NOT NULL
GROUP BY 1, 3, 4, 5
ORDER BY 2 DESC

Try this:

SELECT main.ResolvedBy, COUNT(*) AS Count, 
    main.fiCategory, main.fiSubCategory, main.fiSymptom, average
FROM tContact main
JOIN (SELECT COUNT(*)/count(distinct ResolvedBy) as average,
      fiCategory, fiSubCategory, fiSymptom group by 2,3,4) x
        on x.fiCategory = main.fiCategory
        and x.fiSubCategory = main.fiSubCategory
        and x.fiSymptom = main.fiSymptom
WHERE main.ResolvedBy IS NOT NULL
GROUP BY 1, 3, 4, 5
ORDER BY 2 DESC

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