LOG 和 EXP 函数中的舍入问题 [英] Rounding issue in LOG and EXP functions

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本文介绍了LOG 和 EXP 函数中的舍入问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在尝试执行累积乘法.我正在尝试两种方法来做到这一点

I am trying to perform cumulative multiplication. I am trying two methods to do this

DECLARE @TEST TABLE
  (
     PAR_COLUMN INT,
     PERIOD     INT,
     VALUE      NUMERIC(22, 6)
  ) 
INSERT INTO @TEST VALUES 
(1,601,10 ),
(1,602,20 ),
(1,603,30 ),
(1,604,40 ),
(1,605,50 ),
(1,606,60 ),
(2,601,100),
(2,602,200),
(2,603,300),
(2,604,400),
(2,605,500),
(2,606,600)

注意: value 列中的数据永远不会是整数,并且值会有小数部分.为了显示近似问题,我将示例值保留为整数.

Note: The data in value column will never be integer and values will have decimal part. To show approximation problem I have kept example values as integers.

在此方法中,我使用 EXP + LOG + SUM() Over(Order by) 技术来查找累积乘法.在这种方法中,数值不准确;结果中存在一些舍入和近似问题.

In this method am using EXP + LOG + SUM() Over(Order by) technique to find cumulative multiplication. In this method values are not accurate; there is some rounding and approximation issue in the result.

SELECT *,
       Exp(Sum(Log(Abs(NULLIF(VALUE, 0))))
             OVER(
               PARTITION BY PAR_COLUMN
               ORDER BY PERIOD)) AS CUM_MUL
FROM   @TEST;

结果:

PAR_COLUMN  PERIOD  VALUE       CUM_MUL
----------  ------  ---------   ----------------
1           601     10.000000   10
1           602     20.000000   200             -- 10 * 20 = 200(correct)
1           603     30.000000   6000.00000000001 -- 200 * 30 = 6000.000000000 (not 6000.00000000001) incorrect
1           604     40.000000   240000
1           605     50.000000   12000000
1           606     60.000000   720000000.000001  -- 12000000 * 60 = 720000000.000000 (not 720000000.000001) incorrect
2           601     100.000000  100
2           602     200.000000  20000
2           603     300.000000  5999999.99999999 -- 20000.000000 *300.000000 = 6000000.000000 (not 5999999.99999999) incorrect
2           604     400.000000  2399999999.99999  
2           605     500.000000  1199999999999.99
2           606     600.000000  719999999999998

<小时>

方法 2:传统乘法(递归 CTE)

此方法完美运行,没有任何舍入或近似问题.


Method 2: Tradictional Multiplication (Recursive CTE)

This method works perfectly without any rounding or approximation problem.

;WITH CTE
     AS (SELECT TOP 1 WITH TIES PAR_COLUMN,
                                PERIOD,
                                VALUE,
                                CUM_MUL = VALUE
         FROM   @TEST
         ORDER  BY PERIOD
         UNION ALL
         SELECT T.PAR_COLUMN,
                T.PERIOD,
                T.VALUE,
                Cast(T.VALUE * C.CUM_MUL AS NUMERIC(22, 6))
         FROM   CTE C
                INNER JOIN @TEST T
                        ON C.PAR_COLUMN = T.PAR_COLUMN
                           AND T.PERIOD = C.PERIOD + 1)
SELECT *
FROM   CTE 
ORDER BY PAR_COLUMN,PERIOD

结果

PAR_COLUMN  PERIOD  VALUE       CUM_MUL
----------  ------  ---------   ----------------
1           601     10.000000   10.000000
1           602     20.000000   200.000000
1           603     30.000000   6000.000000
1           604     40.000000   240000.000000
1           605     50.000000   12000000.000000
1           606     60.000000   720000000.000000
2           601     100.000000  100.000000
2           602     200.000000  20000.000000
2           603     300.000000  6000000.000000
2           604     400.000000  2400000000.000000
2           605     500.000000  1200000000000.000000
2           606     600.000000  720000000000000.000000

<小时>

谁能告诉我为什么方法 1 中的值不准确以及如何修复它?我尝试将数据类型更改为 Float 并增加 numeric 中的 scale 但没有用.


Can anyone tell me why in method 1 values are not accurate and how to fix it? I tried by changing the data types to Float and by increasing the scale in numeric but no use.

我真的很想使用方法 1,它比方法 2 快得多.

I really want to use method 1 which is much faster than method 2.

现在我知道近似值的原因了.任何人都可以找到解决此问题的方法吗?

Now I know the reason for approximation. Can anyone find a fix for this problem?

推荐答案

对于您的数据,您可以四舍五入到大倍数:

You can round to big multiple, for your data:

--720000000000000 must be multiple of 600

select
   round( 719999999999998/600,  0 ) * 600

--result: 720000000000000

在 SQLFiddle 上测试

create TABLE T 
  (
     PAR_COLUMN INT,
     PERIOD     INT,
     VALUE      NUMERIC(22, 6)
  ) 
INSERT INTO T VALUES 
(1,601,10.1 ),    --<--- I put decimals just to test!
(1,602,20 ),
(1,603,30 ),
(1,604,40 ),
(1,605,50 ),
(1,606,60 ),
(2,601,100),
(2,602,200),
(2,603,300),
(2,604,400),
(2,605,500),
(2,606,600)

查询 1:

with T1 as (
SELECT *,
       Exp(Sum(Log(Abs(NULLIF(VALUE, 0))))
             OVER(
               PARTITION BY PAR_COLUMN
               ORDER BY PERIOD)) AS CUM_MUL,
       VALUE AS CUM_MAX1,
       LAG( VALUE , 1, 1.) 
             OVER(
               PARTITION BY PAR_COLUMN
               ORDER BY PERIOD ) AS CUM_MAX2,
       LAG( VALUE , 2, 1.) 
             OVER(
               PARTITION BY PAR_COLUMN
               ORDER BY PERIOD ) AS CUM_MAX3
FROM   T )
select PAR_COLUMN,  PERIOD,  VALUE, 
       ( round( ( CUM_MUL  / ( CUM_MAX1 * CUM_MAX2 * CUM_MAX3) ) ,6) 
         * 
         cast( ( 1000000 * CUM_MAX1 * CUM_MAX2 * CUM_MAX3) as bigint )
       ) / 1000000.
       as CUM_MUL
FROM T1

结果:

| PAR_COLUMN | PERIOD | VALUE |         CUM_MUL |
|------------|--------|-------|-----------------|
|          1 |    601 |  10.1 |            10.1 | --ok! because my data
|          1 |    602 |    20 |             202 |
|          1 |    603 |    30 |            6060 |
|          1 |    604 |    40 |          242400 |
|          1 |    605 |    50 |        12120000 |
|          1 |    606 |    60 |       727200000 |
|          2 |    601 |   100 |             100 |
|          2 |    602 |   200 |           20000 |
|          2 |    603 |   300 |         6000000 |
|          2 |    604 |   400 |      2400000000 |
|          2 |    605 |   500 |   1200000000000 |
|          2 |    606 |   600 | 720000000000000 |

注意我 x1000000 不用小数

这篇关于LOG 和 EXP 函数中的舍入问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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