Question

我想将表传递给UDF或存储过程，然后让它处理数据并从我传递给它的表中返回灵敏度，特异性和95％上/下置信区间（对于每个）。

基本上，我需要从表中计算并返回六个值。

我必须多次这样做，所以自动化会很棒但我还没有创建UDF或SP。我已经阅读了它们（stackoverflow和其他地方，但我仍然坚持如何继续。

我创建了SQL部分来计算感兴趣的参数，但我真的很困惑如何将表传递给它并获得一个表。

DECLARE @R_MODS TABLE(
  SUBJECTID varchar(max),
  ResultCall varchar(max)
)

INSERT INTO @R_MODS VALUES ('11-0001','TP');
INSERT INTO @R_MODS VALUES ('11-0002','TP');
INSERT INTO @R_MODS VALUES ('11-0003','TP');
INSERT INTO @R_MODS VALUES ('11-0004','TP');
INSERT INTO @R_MODS VALUES ('11-0005','TP');
INSERT INTO @R_MODS VALUES ('11-0006','I');
INSERT INTO @R_MODS VALUES ('11-0007','TP');
INSERT INTO @R_MODS VALUES ('11-0008','TP');
INSERT INTO @R_MODS VALUES ('11-0009','I');
INSERT INTO @R_MODS VALUES ('11-0010','TP');
INSERT INTO @R_MODS VALUES ('11-0011','TP');
INSERT INTO @R_MODS VALUES ('11-0012','TN');
INSERT INTO @R_MODS VALUES ('11-0013','TP');
INSERT INTO @R_MODS VALUES ('11-0014','I');
INSERT INTO @R_MODS VALUES ('11-0015','TP');
INSERT INTO @R_MODS VALUES ('11-0016','TP');
INSERT INTO @R_MODS VALUES ('11-0017','TN');
INSERT INTO @R_MODS VALUES ('11-0018','TP');
INSERT INTO @R_MODS VALUES ('11-0019','FP');
INSERT INTO @R_MODS VALUES ('11-0020','FP');

DECLARE @TP float, @TN float, @FP float, @FN float, @SEN float, @SPE float, @M1 float, 
        @M2 float, @Sen95 float, @SpeL float , @SpeU float, @SenU float, @SenL float

SET @TP = (SELECT COUNT(SUBJECTID) FROM @R_MODS WHERE ResultCall='TP')
SET @TN = (SELECT COUNT(SUBJECTID) FROM @R_MODS WHERE ResultCall='TN')
SET @FP = (SELECT COUNT(SUBJECTID) FROM @R_MODS WHERE ResultCall='FP')
SET @FN = (SELECT COUNT(SUBJECTID) FROM @R_MODS WHERE ResultCall='FN')

SET @SEN = @TP/(@TP + @FN)
SET @M1 = @TP + @FN 

SET @SPE = @TN/(@TN + @FP)
SET @M2 =  @FP + @TN 

SET @SenL = ( 2*@M1*@SEN + POWER(1.96,2) - 1 - 1.96 * SQRT(POWER(1.96,2) 
    - 2 -(1/@M1)+ 4*@SEN *(@M1*(1-@SEN) + 1)))/(2*(@M1+POWER(1.96,2)))
SET @SenU = ( 2*@M1*@SEN + POWER(1.96,2) + 1 + 1.96 * SQRT(POWER(1.96,2) 
    + 2 -(1/@M1)+ 4*@SEN *(@M1*(1-@SEN) - 1)))/(2*(@M1+POWER(1.96,2)))

SET @SpeL = ( 2*@M2*@SPE + POWER(1.96,2) - 1 - 1.96 * SQRT(POWER(1.96,2) 
    - 2 -(1/@M2)+ 4*@SPE *(@M2*(1-@SPE) + 1)))/(2*(@M2+POWER(1.96,2)))
SET @SpeU = ( 2*@M2*@SPE + POWER(1.96,2) + 1 + 1.96 * SQRT(POWER(1.96,2) 
    + 2 -(1/@M2)+ 4*@SPE *(@M2*(1-@SPE) - 1)))/(2*(@M2+POWER(1.96,2)))

SELECT @SEN, @SenL, @SenU, 1-@SPE, 1-@SPEL, 1-@SpeU

Answer 1

由于您的代码只是对数据进行计算（即没有表格的副作用或更新）并且没有与表格耦合，您可以做的是创建table valued function来进行这些计算 - 该函数可以将输入表（R_MODS）作为table type，并返回输出表（Sen等）。

这里是SqlFiddle example。

详细信息：

您需要为输入创建表格类型，例如

CREATE TYPE R_MODS_TYPE AS TABLE(
  SUBJECTID varchar(max),
  ResultCall varchar(max)
);

并定义函数：

CREATE FUNCTION dbo.DoCalcs(@TheRMods R_MODS_TYPE READONLY)
RETURNS @Result TABLE
(
    [SEN] DECIMAL(10,4), 
    [SenL] DECIMAL(10,4), 
    [SenU] DECIMAL(10,4),  
    [1-SPE] DECIMAL(10,4), 
    [1-SPEL] DECIMAL(10,4), 
    [1-SpeU] DECIMAL(10,4)
)
AS
BEGIN
    DECLARE @TP float, @TN float, @FP float, @FN float, @SEN float, @SPE float, 
            @M1 float, @M2 float, @Sen95 float, @SpeL float , @SpeU float, 
            @SenU float, @SenL float;

    SET @TP = (SELECT COUNT(SUBJECTID) FROM @TheRMods WHERE ResultCall='TP')
    SET @TN = (SELECT COUNT(SUBJECTID) FROM @TheRMods WHERE ResultCall='TN')
    SET @FP = (SELECT COUNT(SUBJECTID) FROM @TheRMods WHERE ResultCall='FP')
    SET @FN = (SELECT COUNT(SUBJECTID) FROM @TheRMods WHERE ResultCall='FN')

    SET @SEN = @TP/(@TP + @FN)
    SET @M1 = @TP + @FN 

    SET @SPE = @TN/(@TN + @FP)
    SET @M2 =  @FP + @TN 

    SET @SenL = ( 2*@M1*@SEN + POWER(1.96,2) - 1 - 1.96 * SQRT(POWER(1.96,2) 
          - 2 -(1/@M1)+ 4*@SEN *(@M1*(1-@SEN) + 1)))/(2*(@M1+POWER(1.96,2)))
    SET @SenU = ( 2*@M1*@SEN + POWER(1.96,2) + 1 + 1.96 * SQRT(POWER(1.96,2) 
          + 2 -(1/@M1)+ 4*@SEN *(@M1*(1-@SEN) - 1)))/(2*(@M1+POWER(1.96,2)))

    SET @SpeL = ( 2*@M2*@SPE + POWER(1.96,2) - 1 - 1.96 * SQRT(POWER(1.96,2) 
      - 2 -(1/@M2)+ 4*@SPE *(@M2*(1-@SPE) + 1)))/(2*(@M2+POWER(1.96,2)))
    SET @SpeU = ( 2*@M2*@SPE + POWER(1.96,2) + 1 + 1.96 * SQRT(POWER(1.96,2) 
      + 2 -(1/@M2)+ 4*@SPE *(@M2*(1-@SPE) - 1)))/(2*(@M2+POWER(1.96,2)))

    INSERT INTO @Result ([SEN], [SenL], [SenU], [1-SPE], [1-SPEL], [1-SpeU])
        SELECT @SEN, @SenL, @SenU, 1-@SPE, 1-@SPEL, 1-@SpeU;
    RETURN;
END

然后通过声明表类型的实例，填充它并将其传递给函数来调用表函数：

DECLARE @TestData R_MODS_TYPE;

INSERT INTO @TestData VALUES ('11-0001','TP'),
('11-0002','TP'),
('11-0003','TP'),
('11-0004','TP'),
... etc.

SELECT * FROM dbo.DoCalcs(@TestData);

结果：

SEN       SenL     SenU    1-SPE   1-SPEL  1-SpeU
--------- -------- ------- ------- ------- -------
1.0000    0.7166   0.9929  0.5000  0.9081  0.0919

Answer 2

以下是使用Stored Procedure

的另一种方法

您的输入参数需要用户定义的表数据类型。

CREATE TYPE R_MODS_TBL AS TABLE(
    SUBJECTID VARCHAR(MAX),
    ResultCall VARCHAR(MAX)
)

存储过程：

注意变量@TP，@TN，@FP和@FN的分配更改，以使用单个SELECT语句而不是四个独立的。

CREATE PROCEDURE  dbo.YourStoredProcedure(
    @R_MODS R_MODS_TBL READONLY
)
AS

DECLARE 
    @TP FLOAT, @TN FLOAT, @FP FLOAT, @FN FLOAT,
    @SEN FLOAT, @SPE FLOAT, @M1 FLOAT, @M2 FLOAT, @Sen95 FLOAT,
    @SpeL FLOAT, @SpeU FLOAT, @SenU FLOAT, @SenL FLOAT

SELECT
    @TP = COUNT(CASE WHEN ResultCall='TP' THEN SUBJECTID END),
    @TN = COUNT(CASE WHEN ResultCall='TN' THEN SUBJECTID END),
    @FP = COUNT(CASE WHEN ResultCall='FP' THEN SUBJECTID END),
    @FN = COUNT(CASE WHEN ResultCall='FN' THEN SUBJECTID END)
FROM @R_MODS

SET @SEN = @TP/(@TP + @FN)
SET @M1 = @TP + @FN 

SET @SPE = @TN/(@TN + @FP)
SET @M2 =  @FP + @TN 

SET @SenL = (2*@M1*@SEN + POWER(1.96,2) - 1 - 1.96 * SQRT(POWER(1.96,2) - 2 -(1/@M1)+ 4*@SEN *(@M1*(1-@SEN) + 1)))/(2*(@M1+POWER(1.96,2)))
SET @SenU = (2*@M1*@SEN + POWER(1.96,2) + 1 + 1.96 * SQRT(POWER(1.96,2) + 2 -(1/@M1)+ 4*@SEN *(@M1*(1-@SEN) - 1)))/(2*(@M1+POWER(1.96,2)))

SET @SpeL = (2*@M2*@SPE + POWER(1.96,2) - 1 - 1.96 * SQRT(POWER(1.96,2) - 2 -(1/@M2)+ 4*@SPE *(@M2*(1-@SPE) + 1)))/(2*(@M2+POWER(1.96,2)))
SET @SpeU = (2*@M2*@SPE + POWER(1.96,2) + 1 + 1.96 * SQRT(POWER(1.96,2) + 2 -(1/@M2)+ 4*@SPE *(@M2*(1-@SPE) - 1)))/(2*(@M2+POWER(1.96,2)))

SELECT @SEN, @SenL, @SenU, 1-@SPE, 1-@SPEL, 1-@SpeU

要执行存储过程，您需要填充生成的用户定义表数据类型的实例：

DECLARE @R_MODS R_MODS_TBL

INSERT INTO @R_MODS VALUES ('11-0001','TP');
INSERT INTO @R_MODS VALUES ('11-0002','TP');
INSERT INTO @R_MODS VALUES ('11-0003','TP');
INSERT INTO @R_MODS VALUES ('11-0004','TP');
INSERT INTO @R_MODS VALUES ('11-0005','TP');
INSERT INTO @R_MODS VALUES ('11-0006','I');
INSERT INTO @R_MODS VALUES ('11-0007','TP');
INSERT INTO @R_MODS VALUES ('11-0008','TP');
INSERT INTO @R_MODS VALUES ('11-0009','I');
INSERT INTO @R_MODS VALUES ('11-0010','TP');
INSERT INTO @R_MODS VALUES ('11-0011','TP');
INSERT INTO @R_MODS VALUES ('11-0012','TN');
INSERT INTO @R_MODS VALUES ('11-0013','TP');
INSERT INTO @R_MODS VALUES ('11-0014','I');
INSERT INTO @R_MODS VALUES ('11-0015','TP');
INSERT INTO @R_MODS VALUES ('11-0016','TP');
INSERT INTO @R_MODS VALUES ('11-0017','TN');
INSERT INTO @R_MODS VALUES ('11-0018','TP');
INSERT INTO @R_MODS VALUES ('11-0019','FP');
INSERT INTO @R_MODS VALUES ('11-0020','FP');

EXEC dbo.YourStoredProcedure @R_MODS

SQL Fiddle

注意：

在存储过程中使用表值作为输入参数时，需要将参数声明为READONLY。请阅读Mikael Eriksson的answer以获取更多信息。

传递一个表并将表返回给存储过程或函数？

2 个答案: