Question

我正在尝试计算找到part_id的最后2个批次的每个part_id的平均时间以及所有批次的每个part_id的平均时间。

我已经设法隔离了找到的最后2个批次的part_id的平均值，但我无法将其集成到代码中，以便对每个零件ID进行计算。我收到错误未知栏＆＃39; tst.part_id＆＃39;在＆＃39; where子句＆＃39;中，我需要将该tst.part_id值传递给我的嵌套选择查询。

以下是我的小提琴：

http://sqlfiddle.com/#!9/77bea5/85

我正在使用的查询：

#include<stdio.h>
#include<stdlib.h>
#define speedpersecond 5

int main()

{

char time  [5];

double distance;

printf("Please enter time in seconds");

gets(time);

distance = atoi(time) * speedpersecond;

printf("Distance is %g ", distance);

return 8585;

}

以下是我想要的内容：

SELECT tst.part_id,
  AVG(tst.est_time) AS 'Average Time Overall',

(SELECT AVG(ft.avgLastMax)
FROM 
(SELECT t2.avglst as 'avgLastMax', t2.numval as 'numval'
 FROM (SELECT 
      @num:=CASE WHEN @last != tst3.batch_id
      THEN @num:=(@num + 1)
      ELSE @num:=@num END 'numval',
      @last:=tst3.batch_id,
       @name:=CASE WHEN @num > 2 
        THEN @name:=@name
        ELSE @name:=(tst3.est_time) END 'avglst'
      FROM test AS tst3,
      ( select @last:=0, @avg := 0, @name :=0 , @num :=0) var
       /* GET AVERAGE FOR A SINGLE PART ID
       WHERE tst3.part_id = 1 */ 
        WHERE tst3.part_id = tst.part_id
      ORDER BY tst3.run_id DESC) as t2

  )
 as ft
WHERE ft.numval <3) as 'AVG on last 2 batches'

FROM test AS tst
GROUP BY tst.part_id;

表架构：

part_id   AVG on last 2 batches    Average Time Overall

1         27.25                    25.67

2         22.5                     22.5

3         16.67                    16.67

4         47.5                     47.5

已编辑：通过tst3.run_id DESC更正了表和订单以获取最后一个batch_id。

Answer 1

假设你的第一个结果是错误的......

sqlContext.read.jdbc(...).filter(...)
  .select("ORDINAL_POSITION", "COLUMN_NAME")
  .rdd
  .map { case Row(i: java.math.BigDecimal, c: String) => (i, c) }

或者，更快......带变量......

SELECT a.part_id
     , AVG(a.est_time) 
  FROM test a 
  JOIN 
     ( SELECT x.part_id
            , x.batch_id 
         FROM 
            ( SELECT DISTINCT part_id 
                            , batch_id 
                         FROM test
            ) x 
         JOIN 
            ( SELECT DISTINCT part_id
                            , batch_id 
                         FROM test
            ) y 
           ON y.part_id = x.part_id 
          AND y.batch_id >= x.batch_id 
        GROUP
           BY x.part_id
            , x.batch_id 
       HAVING COUNT(*) <= 2
     ) b 
    ON b.part_id = a.part_id 
   AND b.batch_id = a.batch_id 
 GROUP 
    BY a.part_id;
+---------+-----------------+
| part_id | AVG(a.est_time) |
+---------+-----------------+
|       1 |         27.2500 |
|       2 |         22.5000 |
|       3 |         16.6667 |
|       4 |         47.5000 |
+---------+-----------------+

Answer 2

SELECT W.*,T1.ALLAVG
FROM (
SELECT V.PART_ID, AVG(V.EST_TIME) LAST2 FROM
(
SELECT U.PART_ID,U.MAXRN,X.RUN_ID,X.EST_TIME
FROM
(
/*LAST 2*/
SELECT S.PART_ID,MAX(S.RN) MAXRN
FROM
(
/*DERIVE A ROW NUMBER*/
SELECT T.PART_ID , T.RUN_ID,
         IF (CONCAT(T.PART_ID,T.RUN_ID)  <> @R ,@RN:=@RN+1,@RN:=1) RN,
         @R:=CONCAT(T.PART_ID,T.RUN_ID) R
FROM     (SELECT @RN:=0,@P:=0,@R:=0) RN, TEST T
ORDER   BY T.PART_ID,T.RUN_ID
) S
GROUP   BY S.PART_ID
) U

/*USING THE MAX ABOVE GET THE LAST 2 */
JOIN 

(SELECT T.PART_ID , T.RUN_ID,T.EST_TIME,
         IF (CONCAT(T.PART_ID,T.RUN_ID)  <> @R1 ,@RN1:=@RN1+1,@RN1:=1) RN,
          @R1:=CONCAT(T.PART_ID,T.RUN_ID) R
FROM     (SELECT @RN1:=0,@P1:=0,@R1:=0) RN, TEST T
ORDER   BY T.PART_ID,T.RUN_ID

) X ON X.PART_ID = U.PART_ID AND (X.RN BETWEEN U.MAXRN -1 AND U.MAXRN) #AMEND THIS FOR THE LAST N REQUIRED
) V
GROUP BY V.PART_ID
) W
JOIN
/*ALL*/
(SELECT T.PART_ID,AVG(T.EST_TIME) 'ALLAVG' FROM TEST T GROUP BY T.PART_ID) T1 ON T1.PART_ID = W.PART_ID

RESULT

+---------+---------+---------+
| PART_ID | LAST2   | ALLAVG  |
+---------+---------+---------+
|       1 | 32.5000 | 25.6667 |
|       2 | 22.5000 | 22.5000 |
|       3 | 17.5000 | 16.6667 |
|       4 | 47.5000 | 47.5000 |
+---------+---------+---------+

计算属于不同组集的最后x个记录的平均值

2 个答案: