Question

我要求按客户群显示前5位客户销售额，但与其他客户在集团内的销售额汇总为“其他”。与this question类似，但为每个客户群分别计算。

根据MSDN执行TopCount，对于一组成员，您必须使用Generate function。

这部分工作正常：

with 

set [Top5CustomerByGroup] AS
GENERATE
( 
    [Klient].[Grupa Klientow].[Grupa Klientow].ALLMEMBERS,
    TOPCOUNT
    (
        [Klient].[Grupa Klientow].CURRENTMEMBER * [Klient].[Klient].[Klient].MEMBERS
        , 5
        , [Measures].[Przychody ze sprzedazy rzeczywiste wartosc]
    )
)

SELECT 
{ [Measures].[Przychody ze sprzedazy rzeczywiste wartosc]} ON COLUMNS,
{
[Klient].[Grupa Klientow].[Grupa Klientow].ALLMEMBERS * [Klient].[Klient].[All], --for drilldown purposes
[Top5CustomerByGroup]
}
ON ROWS
FROM 
(
  SELECT ({[Data].[Rok].&[2013]} ) ON COLUMNS
      FROM [MyCube]
)

然而我对'其他'部分有疑问。

我认为我能够按组建立与其他客户的集合（数据看起来不错）：

set [OtherCustomersByGroup] AS
GENERATE
( 
    [Klient].[Grupa Klientow].[Grupa Klientow].ALLMEMBERS,
    except
    (
        {[Klient].[Grupa Klientow].CURRENTMEMBER * [Klient].[Klient].[Klient].MEMBERS},
        TOPCOUNT
        (
            [Klient].[Grupa Klientow].CURRENTMEMBER * [Klient].[Klient].[Klient].MEMBERS
            , 5
            , [Measures].[Przychody ze sprzedazy rzeczywiste wartosc]
        )
    )
)

但是我不知道如何将其与分组聚合在一起。

在this question

中执行此操作

member [Klient].[Klient].[tmp] as
aggregate([OtherCustomersByGroup])

生成一个值，这是合乎逻辑的。

我认为我需要在每个组中使用“其他”客户的集合列表，而不是单个[OtherCustomersByGroup]集合，但不知道如何构建它们。

有没有人有任何想法或建议？

更新

对我的需求存在一些误解。我需要通过销售来满足每个客户群中的前N个客户，并将该组中其他客户的销售汇总到一个位置（假设称为其他）。

例如，对于此简化输入：

| Group  | Client   | Sales  |
|--------|----------|--------|
| Group1 | Client1  |    300 |
| Group1 | Client2  |      5 |
| Group1 | Client3  |    400 |
| Group1 | Client4  |    150 |
| Group1 | Client5  |    651 |
| Group1 | Client6  | null   |
| Group2 | Client7  |     11 |
| Group2 | Client8  |     52 |
| Group2 | Client9  |     44 |
| Group2 | Client10 |     21 |
| Group2 | Client11 |    201 |
| Group2 | Client12 |    325 |
| Group2 | Client13 |    251 |
| Group3 | Client14 |     15 |

我需要这样的输出（这里是前2名）：

| Group  | Client   | Sales  |
|--------|----------|--------|
| Group1 | Client5  |    651 |
| Group1 | Client3  |    400 |
| Group1 | Others   |    455 |
| Group2 | Client12 |    325 |
| Group2 | Client13 |    251 |
| Group2 | Others   |    329 |
| Group3 | Client14 |     15 |
| Group3 | Others   |  null  | <- optional row

不需要排序，我们将在客户端处理它。

Answer 1

是的，您通过使用SET for Others获得了主要想法，但完成任务需要几个小的补充。

我将使用我的测试数据库，但这可以很容易地转换为你的。

[Report Date] - 日期维度（[Klient] analogue）
[REPORT DATE Y] - 年等级（[Grupa Klientow]）
[REPORT DATE YM] - 月份层次结构（[Klient].[Klient]）
[Measures].[Count] - 衡量TopCount（[Measures].[Przychody ze sprzedazy rzeczywiste wartosc]）

我也使用前三名来显示结果图片。

这是代码：

with

/* first, add empty [Other] member to the group level */
member [Report Date].[REPORT DATE Y].[Other] as null

/* second, copy measure by fixing the lowest level */
member [Measures].[Count with Other Groups] as ([Report Date].[REPORT DATE YM],[Measures].[Count])

/* third, create top 10 by group */
set [Report Date Top 10 Groups] as
Generate([Report Date].[REPORT DATE Y].Children
,TopCount([Report Date].[REPORT DATE Y].CurrentMember
 * [Report Date].[REPORT DATE YM].Children,3,[Measures].[Count with Other Groups]))

/* this is the part for Other group mapping */
set [Report Date Other Groups] as
[Report Date].[REPORT DATE Y].[Other]
 * ([Report Date].[REPORT DATE YM].Children
    - Extract([Report Date Top 10 Groups],[Report Date].[REPORT DATE YM]))

select {[Measures].[Count],[Measures].[Count with Other Groups]} on 0
,
{
[Report Date Top 10 Groups],[Report Date Other Groups]}
on 1
from 
[DATA]

结果如下：

TopCount_withGroups

..直到最后一位（即201606）的所有成员都在Other组。

希望这有帮助，bardzodziękuję！

通过删除Report Date Other Groups计算中的一个乘法来优化

更新代码。

更新-2 :(尚未解决，但正在进行中）

（在每组下使用'其他'成员）

重要！我们需要额外的层次结构：Group->Client（[Report Date].[REPORT DATE]，Year->Month是我的情况），以便能够为每个低级别成员确定父级。

with

/* create top 10 by group */
set [Report Date Top 10 Groups] as
Generate([Report Date].[REPORT DATE Y].Children
,TopCount([Report Date].[REPORT DATE Y].CurrentMember
 * [Report Date].[REPORT DATE].Children,3,[Measures].[Count]))

/* this is the part for Other group the lowest level non-aggregated members */
set [Report Date Other Members] as
[Report Date].[REPORT DATE Y].Children
* ([Report Date].[REPORT DATE].[Month].AllMembers
    - [Report Date].[REPORT DATE].[All])
- [Report Date Top 10 Groups]

/* add empty [Other] member to the group level, HERE IS AN ISSUE */
member [Report Date].[REPORT DATE].[All].[Other] as null

set [Report Date Other Groups] as
[Report Date].[REPORT DATE Y].[All].Children
* [Report Date].[REPORT DATE].[Other]

member [Measures].[Sum of Top] as
IIF([Report Date].[Report Date].CurrentMember is [Report Date].[REPORT DATE].[Other]
,null /* HERE SHOULD BE CALCULATION, but only
 {[Report Date].[Report Date Y].[All].[Other]}
 is shown, because 'Other' is added to the entire hierarchy */
,SUM([Report Date].[REPORT DATE Y].CurrentMember
        * ([Report Date].[Report Date].CurrentMember.Parent.Children
            - Extract([Report Date Other Members],[Report Date].[REPORT DATE]))
    ,[Measures].[Count]))

member [Measures].[Sum of Group] as
([Report Date].[Report Date].CurrentMember.Parent,[Measures].[Count])

select {[Measures].[Count],[Measures].[Sum of Group],[Measures].[Sum of Top]} on 0
,
Order(Hierarchize({[Report Date Top 10 Groups]
,[Report Date Other Groups]}),[Measures].[Count],DESC)

on 1
from 
[DATA]

这是中间结果：

TopN_v2

我需要在此处移动此结果，但不知道该怎么做。

我也尝试过使用每个级别的平面层次结构。 Other成员显示正确，但无法计算SUM，因为两个级别都是独立的。也许我们可以添加像'Group_Name'这样的属性并使用未链接的级别，但是再次 - 它会大大降低性能。所有这些IIF([bla-bla-bla low level member].Properties("Group_Name")=[bla-bla-bla group level].Member_Name都非常慢。

Update-3（上述代码的AdvWorks版本）

with

/* create top 10 by group */
set [Top 10 Groups] as
Generate([Customer].[Country].Children
,TopCount([Customer].[Country].CurrentMember
 * [Customer].[Customer Geography].Children,3,[Measures].[Internet Order Count]))

/* this is the part for Other group the lowest level non-aggregated members */
set [Other Members] as
[Customer].[Country].Children
* ([Customer].[Customer Geography].[State-Province].AllMembers
    - [Customer].[Customer Geography].[All])
- [Top 10 Groups]

/* add empty [Other] member to the group level */
member [Customer].[Customer Geography].[All].[Other] as
([Customer].[Country],[Measures].[Internet Order Count])

set [Other Groups] as
[Customer].[Country].[All].Children
* [Customer].[Customer Geography].[Other]

member [Measures].[Sum of Top] as
IIF([Customer].[Customer Geography].CurrentMember is [Customer].[Customer Geography].[Other]
,null
,SUM([Customer].[Country].CurrentMember
        * ([Customer].[Customer Geography].CurrentMember.Parent.Children
            - Extract([Other Members],[Customer].[Customer Geography]))
    ,[Measures].[Internet Order Count]))

member [Measures].[Sum of Group] as
([Customer].[Customer Geography].CurrentMember.Parent,[Measures].[Internet Order Count])

select {[Measures].[Internet Order Count],[Measures].[Sum of Group],[Measures].[Sum of Top]} on 0
,
Order(Hierarchize({[Top 10 Groups],[Other Groups]}),[Measures].[Internet Order Count],DESC) on 1
from [Adventure Works]

TopCount_withGroups_AdvWorks

更新-4（以年/月为例的解决方案）

@whytheq的惊人解决方案帮助我做了我想要的事情：

WITH 
  SET [All Grupa Klientow]  AS ([Report Date].[Report Date Y].Children) 
  SET [All Klient] AS ([Report Date].[Report Date YM].Children)
  SET [Top N Members] AS 
    Generate
    (
      [All Grupa Klientow]
     ,TopCount
      (
        (EXISTING 
          [All Klient])
       ,3
       ,[Measures].[Count]
      )
    ) 
  MEMBER [Report Date].[Report Date YM].[Other] AS 
    Aggregate({(EXISTING {[All Klient]} - [Top N Members])}) 
SELECT 
  {[Measures].[Count]} ON 0
 ,{
      [All Grupa Klientow]
    * 
      {
        [Top N Members]
       ,[Report Date].[Report Date YM].[Other]
      }
  } ON 1
FROM [DATA];

图片：

TopCount_withGroups_FinalResult

任务已经解决，但请注明不是这个答案，而是@ whytheq's！

Answer 2

以下是针对AdvWrks并使用我在Chris Webb博客上看到的技术，他在此概述： https://cwebbbi.wordpress.com/2007/06/25/advanced-ranking-and-dynamically-generated-named-sets-in-mdx/

创建MyMonthsWithEmployeesSets集的脚本部分我觉得很难理解 - 也许@AlexPeshik可以更多地了解下面脚本中发生的事情。

WITH 
  SET MyMonths AS 
    TopPercent
    (
      [Date].[Calendar].[Month].MEMBERS
     ,20
     ,[Measures].[Reseller Sales Amount]
    ) 
  SET MyEmployees AS 
    [Employee].[Employee].[Employee].MEMBERS 
  SET MyMonthsWithEmployeesSets AS 
    Generate
    (
      MyMonths
     ,Union
      (
        {[Date].[Calendar].CurrentMember}
       ,StrToSet
        ("
             Intersect({}, 
             {TopCount(MyEmployees, 10, ([Measures].[Reseller Sales Amount],[Date].[Calendar].CurrentMember))
             as EmployeeSet"
            + 
              Cstr(MyMonths.CurrentOrdinal)
          + "})"
        )
      )
    ) 
  MEMBER [Employee].[Employee].[RestOfEmployees] AS 
    Aggregate
    (
      Except
      (
        MyEmployees
       ,StrToSet
        (
          "EmployeeSet" + Cstr(Rank([Date].[Calendar].CurrentMember,MyMonths))
        )
      )
    ) 
  MEMBER [Measures].[EmployeeRank] AS 
    Rank
    (
      [Employee].[Employee].CurrentMember
     ,StrToSet
      (
        "EmployeeSet" + Cstr(Rank([Date].[Calendar].CurrentMember,MyMonths))
      )
    ) 
SELECT 
  {
    [Measures].[EmployeeRank]
   ,[Measures].[Reseller Sales Amount]
  } ON 0
 ,Generate
  (
    Hierarchize(MyMonthsWithEmployeesSets)
   ,
      [Date].[Calendar].CurrentMember
    * 
      {
        Order
        (
          Filter
          (
            MyEmployees
           ,
            [Measures].[EmployeeRank] > 0
          )
         ,[Measures].[Reseller Sales Amount]
         ,BDESC
        )
       ,[Employee].[Employee].[RestOfEmployees]
      }
  ) ON 1
FROM [Adventure Works];

编辑 - Alex第三次尝试的解决方案：

WITH 
  SET [AllCountries] AS [Country].[Country].MEMBERS 
  SET [AllStates]    AS [State-Province].[State-Province].MEMBERS 
  SET [Top2States] AS 
    Generate
    (
      [AllCountries]
     ,TopCount
      (
        (EXISTING 
          [AllStates])
       ,3
       ,[Measures].[Internet Order Count]
      )
    ) 
  MEMBER [State-Province].[All].[RestOfCountry] AS 
    Aggregate({(EXISTING {[AllStates]} - [Top2States])}) 
SELECT 
  {[Measures].[Internet Order Count]} ON COLUMNS
 ,{
      [AllCountries]
    * 
      {
        [Top2States]
       ,[State-Province].[All].[RestOfCountry]
       ,[State-Province].[All]
      }
  } ON ROWS
FROM [Adventure Works];

MDX - TopCount加上'其他'或'休息'按组（通过一组成员）

2 个答案: