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Question

我有一个MS SQL Server 2008数据库，我存储了供应食物的地方（咖啡馆，餐馆，食客等）。在连接到此数据库的网站上，人们可以按1到3的等级对地点进行评级。

在网站上有一个页面，人们可以在这个页面中查看某个城市中排名前25位（评分最高）的名单。数据库结构看起来像这样（表中存储了更多信息，但这里是相关信息）： Database structure: Cities->Places->Votes

一个地方位于一个城市，投票放在一个地方。

到目前为止，我刚刚计算了每个地方的平均投票得分，我将某个地方的所有选票总和除以该地点的投票数，这样的话（伪代码）：

vote_count = total number of votes for the place
vote_sum = total sum of all the votes for the place

vote_score = vote_sum/vote_count

如果一个地方没有选票，我还必须处理除零。所有这些都在存储过程中完成，该存储过程获取我想在顶部列表中显示的其他数据。以下是当前存储过程，它获取投票得分最高的前25个位置：

ALTER PROCEDURE [dbo].[GetTopListByCity]
    (
    @city_id Int
    )
AS
    SELECT TOP 25 dbo.Places.place_id, 
           dbo.Places.city_id,
           dbo.Places.place_name,
           dbo.Places.place_alias,
           dbo.Places.place_street_address,
           dbo.Places.place_street_number,
           dbo.Places.place_zip_code,
           dbo.Cities.city_name,
           dbo.Cities.city_alias,
           dbo.Places.place_phone,
           dbo.Places.place_lat,
           dbo.Places.place_lng,
           ISNULL(SUM(dbo.Votes.vote_score),0) AS vote_sum,
           (SELECT COUNT(*) FROM dbo.Votes WHERE dbo.Votes.place_id = dbo.Places.place_id) AS vote_count,
           COALESCE((CONVERT(FLOAT,SUM(dbo.Votes.vote_score))/(CONVERT(FLOAT,(SELECT COUNT(*) FROM dbo.Votes WHERE dbo.Votes.place_id = dbo.Places.place_id)))),0) AS vote_score

    FROM dbo.Places INNER JOIN dbo.Cities ON dbo.Places.city_id = dbo.Cities.city_id
    LEFT OUTER JOIN dbo.Votes ON dbo.Places.place_id = dbo.Votes.place_id
    WHERE dbo.Places.city_id = @city_id
    AND dbo.Places.hidden = 0
    GROUP BY dbo.Places.place_id,
             dbo.Places.city_id,
             dbo.Places.place_name,
             dbo.Places.place_alias,
             dbo.Places.place_street_address,
             dbo.Places.place_street_number,
             dbo.Places.place_zip_code,
             dbo.Cities.city_name,
             dbo.Cities.city_alias,
             dbo.Places.place_phone,
             dbo.Places.place_lat,
             dbo.Places.place_lng
    ORDER BY vote_score DESC, vote_count DESC, place_name ASC

    RETURN

正如你所看到的，它不仅仅取得了投票得分 - 我需要有关这个地方，它所在的城市等的数据。这样做很好，但有一个大问题：投票得分太简单了，因为它没有考虑到投票数。使用简单的计算方法，一个投票得分为3的地方将在列表中高于一个有十四票，得分为3，一票得分为2的地方：

3/1 = 3
(14*3 + 1*2) = 44/15 = 2.933333333333

为了解决这个问题，我一直在研究使用某种形式的加权平均值/加权指数。我发现了一个真正的贝叶斯估计的例子看起来很有希望。它看起来像这样：

weighted rating (WR) = (v ÷ (v+m)) × R + (m ÷ (v+m)) × C

where:

R = average for the place (mean) = (Rating)
v = number of votes for the place = (votes)
m = minimum number of votes required to be listed in the Top 25 (unsure how many, but somewhere between 2-5 seems realistic)
C = the mean vote across the whole database

当我尝试在存储过程中实现此加权评级时，问题就开始了 - 它很快变得复杂，我纠结于括号和松散跟踪存储过程的作用。

现在我需要一些帮助来解决两个问题：

这是计算网站加权指数的合适方法吗？

在存储过程中实现时，这个（或其他合适的计算方法）会是什么样子？

Answer 1

我看不出你的计算有什么问题。但我可以看到你做了很多次同样的事情。我的建议将帮助您在一个地方进行聚合，然后选择非常简单。

;WITH CTE
(
    SELECT
        SUM(dbo.Votes.vote_score) AS SumOfVoteScore,
        COUNT(*) AS CountOfVotes,
        Votes.place_id
    FROM
        Votes
    GROUP BY
        Votes.place_id
)
 SELECT TOP 25 
    dbo.Places.place_id, 
    dbo.Places.city_id,
    dbo.Places.place_name,
    dbo.Places.place_alias,
    dbo.Places.place_street_address,
    dbo.Places.place_street_number,
    dbo.Places.place_zip_code,
    dbo.Cities.city_name,
    dbo.Cities.city_alias,
    dbo.Places.place_phone,
    dbo.Places.place_lat,
    dbo.Places.place_lng,
    ISNULL(CTE.SumOfVoteScore,0) AS vote_sum,
    CTE.CountOfVotes AS vote_count,
    COALESCE((CONVERT(FLOAT,CTE.SumOfVoteScore)/
    (CONVERT(FLOAT,CTE.CountOfVotes))),0) AS vote_score

FROM dbo.Places INNER JOIN dbo.Cities ON dbo.Places.city_id = dbo.Cities.city_id
LEFT JOIN CTE ON dbo.Places.place_id=CTE.place_id
WHERE dbo.Places.city_id = @city_id
AND dbo.Places.hidden = 0
GROUP BY dbo.Places.place_id,
         dbo.Places.city_id,
         dbo.Places.place_name,
         dbo.Places.place_alias,
         dbo.Places.place_street_address,
         dbo.Places.place_street_number,
         dbo.Places.place_zip_code,
         dbo.Cities.city_name,
         dbo.Cities.city_alias,
         dbo.Places.place_phone,
         dbo.Places.place_lat,
         dbo.Places.place_lng
ORDER BY vote_score DESC, vote_count DESC, place_name ASC

CTE功能帮助我们重复使用计算。这样我们就不必使用SUM(vote_score)和SELECT COUNT(*) FROM Votes WHERE...倍数。那么当你选择计算时很容易理解。

我希望这会有所帮助

修改

您不必在CTE中定义表列。此CTE (SumOfVoteScore, CountOfVotes, place_id) AS的效果与CTE AS一样好。如果使用递归cte，则需要定义列。因为另一部分你是union。

参考here和here，您会找到有关CTE功能的一些信息

Answer 2

感谢Arion！

我一直在寻找CTE的东西，但我只是不知道那是我在寻找的东西！学习新东西总是很好，我知道我会在其他项目中使用CTE。当我在我的存储过程中实现您的CTE时，我得到了这段代码：

ALTER PROCEDURE dbo.GetTopListByCityCTE
    (
    @city_id Int
    )
AS

;WITH CTE (SumOfVoteScore, CountOfVotes, place_id) AS
(
    SELECT
        SUM(dbo.Votes.vote_score) AS SumOfVoteScore,
        COUNT(*) AS CountOfVotes,
        Votes.place_id
    FROM
        Votes
    GROUP BY
        Votes.place_id

)

 SELECT TOP 25 
    dbo.Places.place_id, 
    dbo.Places.city_id,
    dbo.Places.place_name,
    dbo.Places.place_alias,
    dbo.Places.place_street_address,
    dbo.Places.place_street_number,
    dbo.Places.place_zip_code,
    dbo.Cities.city_name,
    dbo.Cities.city_alias,
    dbo.Places.place_phone,
    dbo.Places.place_lat,
    dbo.Places.place_lng,
    ISNULL(CTE.SumOfVoteScore,0) AS vote_sum,
    CTE.CountOfVotes AS vote_count,
    COALESCE((CONVERT(FLOAT,CTE.SumOfVoteScore)/
    (CONVERT(FLOAT,CTE.CountOfVotes))),0) AS vote_score

FROM dbo.Places INNER JOIN dbo.Cities ON dbo.Places.city_id = dbo.Cities.city_id
LEFT JOIN CTE ON dbo.Places.place_id = CTE.place_id
WHERE dbo.Places.city_id = @city_id
AND dbo.Places.hidden = 0
GROUP BY dbo.Places.place_id,
         dbo.Places.city_id,
         dbo.Places.place_name,
         dbo.Places.place_alias,
         dbo.Places.place_street_address,
         dbo.Places.place_street_number,
         dbo.Places.place_zip_code,
         dbo.Cities.city_name,
         dbo.Cities.city_alias,
         dbo.Places.place_phone,
         dbo.Places.place_lat,
         dbo.Places.place_lng,
         CTE.SumOfVoteScore,
         CTE.CountOfVotes
ORDER BY vote_score DESC, vote_count DESC, place_name ASC

快速检查显示它返回与前一代码相同的结果，但它更容易阅读和遵循，希望更有效。

现在我将不得不做一些尝试，用一个考虑了投票数的新的（简单的）评级计算方法来替换它。

Answer 3

好的 - 所以这是我提出的存储过程：

ALTER PROCEDURE dbo.GetTopListByCityCTE
    (
    @city_id Int
    )
AS

DECLARE @MinimumNumber float;
DECLARE @TotalNumberOfVotes int;
DECLARE @AverageRating float;
DECLARE @AverageNumberOfVotes float;

/* MINIMUM NUMBER */
SET @MinimumNumber = 1;

/* TOTAL NUMBER OF VOTES -- ALL PLACES */
SET @TotalNumberOfVotes = (
    SELECT COUNT(*) FROM Votes
);

/* AVERAGE RATING -- ALL PLACES */
SET @AverageRating = (
    SELECT
        CONVERT(FLOAT,(SUM(dbo.Votes.vote_score))) / CONVERT(FLOAT,COUNT(*)) AS AverageRating
    FROM 
        Votes);

/* AVERAGE NUMBER OF VOTES -- ALL PLACES */
/* CURRENTLY NOT USED IN INDEX - KEPT FOR REFERENCE */
SET @AverageNumberOfVotes = (
    SELECT AVG(CONVERT(FLOAT,NumberOfVotes)) FROM (SELECT COUNT(*) AS NumberOfVotes FROM Votes GROUP BY place_id) AS AverageNumberOfVotes

);
/* SUM OF ALL VOTE SCORES AND COUNT OF ALL VOTES -- INDIVIDUAL PLACES */
WITH CTE AS (
    SELECT
        CONVERT(FLOAT, SUM(dbo.Votes.vote_score)) AS SumVotesForPlace,
        CONVERT(FLOAT, COUNT(*)) AS CountVotesForPlace,
        Votes.place_id
    FROM
        Votes
    GROUP BY
        Votes.place_id
)

 SELECT 
    dbo.Places.place_id, 
    dbo.Places.city_id,
    dbo.Places.place_name,
    dbo.Places.place_alias,
    dbo.Places.place_street_address,
    dbo.Places.place_street_number,
    dbo.Places.place_zip_code,
    dbo.Cities.city_name,
    dbo.Cities.city_alias,
    dbo.Places.place_phone,
    dbo.Places.place_lat,
    dbo.Places.place_lng,
    ISNULL(CTE.SumVotesForPlace,0) AS vote_sum,
    ISNULL(CTE.CountVotesForPlace,0) AS vote_count,
    COALESCE((CTE.SumVotesForPlace/
    CTE.CountVotesForPlace),0) AS vote_score,
    ISNULL((CTE.CountVotesForPlace / (CTE.CountVotesForPlace + @MinimumNumber)) * (COALESCE((CTE.SumVotesForPlace / CTE.CountVotesForPlace),0)) + (@MinimumNumber / (CTE.CountVotesForPlace + @MinimumNumber)) * @AverageRating,0) AS WeightedIndex

FROM dbo.Places INNER JOIN dbo.Cities ON dbo.Places.city_id = dbo.Cities.city_id
LEFT JOIN CTE ON dbo.Places.place_id = CTE.place_id
WHERE dbo.Places.city_id = @city_id
AND dbo.Places.hidden = 0
GROUP BY dbo.Places.place_id,
         dbo.Places.city_id,
         dbo.Places.place_name,
         dbo.Places.place_alias,
         dbo.Places.place_street_address,
         dbo.Places.place_street_number,
         dbo.Places.place_zip_code,
         dbo.Cities.city_name,
         dbo.Cities.city_alias,
         dbo.Places.place_phone,
         dbo.Places.place_lat,
         dbo.Places.place_lng,
         CTE.SumVotesForPlace,
         CTE.CountVotesForPlace
ORDER BY WeightedIndex DESC, vote_count DESC, place_name ASC

有一个名为@AverageNumberOfVotes的变量，它没有在计算中使用，但我把它保存在那里以供参考，以备不时之需。

针对我拥有的数据运行此操作，我得到的结果与之前略有不同，但这不是革命，也不是我需要的。以下是执行上述SP时返回的前10行：

vote_sum        vote_count  vote_score          WeightedIndex
1110            409         2,71393643031785    2,7140960047496
807             310         2,60322580645161    2,60449697749787
38              15          2,53333333333333    2,56708633093525
25              10          2,5                 2,55442722744881
2               1           2                   2,55188848920863
2               1           2                   2,55188848920863
2               1           2                   2,55188848920863
2               1           2                   2,55188848920863
2               1           2                   2,55188848920863
2               1           2                   2,55188848920863

这里的问题似乎是，只有一票，得分为2，加权指数变为2,55188848920863？

计算此索引的公式取自IMDB（http://www.imdb.com/chart/top），我认为我做错了或者我的数据库中的数据与数据无法比较（数字） IMDB的投票或投票比例？

修改

有没有办法可以调整这个功能，这对我来说效果更好？是否有更好的功能/方法？我仍然需要在存储过程中进行计算。

计算存储过程中的加权（贝叶斯）平均分数/指数？

3 个答案:

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