Question

嗨，我有一个餐桌设施。其中包含每天的分数（每天可以报告多个分数，两者都有效）

我需要按月计算得分的第90个百分点，标准差和平均值。

设施：

Id   Month Date  score
1    Jan     1    5
1    Jan     1    5
1    Jan     2    3
1    Jan     3    4
1    Jan     4    4
1    Jan     5    4
1    Feb     1    5
1    Feb     1    5
1    Feb     2    3
1    Feb     3    4
1    Feb     4    4
1    Feb     5    4

有什么办法吗？

感谢您的帮助。

Answer 1

您可以使用SQL Server 2012中引入的新的分析函数套件：

SELECT DISTINCT
            [Month],
            Mean   = AVG(Score) OVER (PARTITION BY [Month]),
            StdDev = STDEV(Score) OVER (PARTITION BY [Month]),
            P90    = PERCENTILE_CONT(0.9) WITHIN GROUP (ORDER BY Score) OVER (PARTITION BY [Month])
FROM        my_table

有2个百分位函数：PERCENTILE_CONT用于连续分布，PERCENTILE_DISC用于离散分布。选择一个适合您需求的产品。

Answer 2

这是设置......

CREATE TABLE Facility (Id INT NOT NULL, Month nvarchar(3) NOT NULL, Date INT NOT NULL, score INT NOT NULL)

INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Jan', 1, 5)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Jan', 1, 5)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Jan', 2, 3)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Jan', 3, 4)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Jan', 4, 4)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Jan', 5, 4)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Feb', 1, 5)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Feb', 1, 5)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Feb', 2, 3)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Feb', 3, 4)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Feb', 4, 4)
INSERT INTO Facility (Id, Month, Date, score) VALUES (1, 'Feb', 5, 4)

现在，标准偏差和均值足够直接 - 它们内置了聚合函数......

SELECT
    [Month],
    AVG(CONVERT(real, score)) AS [Mean],
    STDEV(score) AS [Standard Deviation]
FROM
    Facility
GROUP BY
    [Month]

对于你的第90个百分点，你需要发明一个函数......

CREATE FUNCTION NintythPercentile(@Month nvarchar(3)) RETURNS INT AS
BEGIN
    DECLARE @ReturnValue INT

    SELECT 
        @ReturnValue = MIN(DerivedTopTenPercent.score) --AS [90th Percentile]
    FROM
        (
        SELECT TOP 10 PERCENT
            score
        FROM
            Facility
        WHERE
            [Month] = @Month
        ORDER BY
            score DESC
        ) DerivedTopTenPercent

    RETURN @ReturnValue
END

使用该功能，您的最终查询将如下所示......

SELECT
    [Month],
    AVG(CONVERT(real, score)) AS [Mean],
    STDEV(score) AS [Standard Deviation],
    dbo.NintythPercentile([Month]) AS [90th Percentile]
FROM
    Facility
GROUP BY
    [Month]

如何计算SQL中数据的90th Percentile，SD，Mean

2 个答案: