MDX - TopCount 加上 'Other' 或 'The Rest' 按组（在一组成员上）
MDX - TopCount plus 'Other' or 'The Rest' by group (over a set of members)
我需要按客户组显示前 5 位客户的销售额，但该组内其他客户的销售额汇总为 'Others'。类似于 this question，但针对每个客户组单独计算。
根据 MSDN，要对一组成员执行 TopCount，您必须使用 Generate 函数。
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]
)
但是 'Others' 部分有问题。
我认为我能够按组与其他客户构建集合（数据看起来不错）作为：
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]
)
)
)
但是我不知道如何通过分组来聚合它。
按照 this question
中的方式进行操作
member [Klient].[Klient].[tmp] as
aggregate([OtherCustomersByGroup])
产生一个值，这是合乎逻辑的。
我想我需要每个组中有 'other' 个客户的集合列表，而不是单个 [OtherCustomersByGroup] 集合，但不知道如何构造它们。
大家有什么想法或建议吗？
更新：
对我的需求有些误解。我需要每个客户组中的前 n 个客户按销售额与该组中其他客户的销售额汇总到一个位置（假设称为其他）。
例如这个简化的输入：
| 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 个）：
| 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，我们会在客户端处理。
是的，您已经通过使用 SET for Others 了解了主要思想，但需要进行一些小的补充才能完成任务。
我将使用我的测试数据库，但这可以很容易地转换为您的。
[Report Date] - 日期维度（[Klient] 模拟）[REPORT DATE Y] - 年层次结构 ([Grupa Klientow])[REPORT DATE YM] - 月份层次结构 ([Klient].[Klient])[Measures].[Count] - TopCount ([Measures].[Przychody ze sprzedazy rzeczywiste wartosc])
我也用了top 3只是为了在这里显示结果图像。
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：
..所有成员直到最后一个（即 201606）都在 Other 组中。
希望这对您有所帮助，bardzo dziękuję！
更新： 代码通过在 Report Date Other Groups 计算中删除一个乘法进行了优化。
Update-2：（尚未解决，但正在进行中）
(每个组下使用'Other'个成员)
重要！我们需要额外的层次结构：Group->Client（[Report Date].[REPORT DATE] 和 Year->Month 是我的情况）能够确定每个低级别成员的父级。
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：
我需要把这个结果移到这里，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，因为两个级别是独立的。也许我们可以添加一个 属性 像 'Group_Name' 并使用未链接的关卡，但同样 - 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]
Update-4（在year/month示例中有一个解决方案）
@whytheq 的惊人解决方案帮助完成了我想要的事情：
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];
和图片：
任务已解决，但请标记不是这个答案，而是@whytheq 的！
以下内容反对 AdvWrks，并使用了我在 Chris Webb 的博客上看到的技术，他在此处概述了该技术：
https://cwebbbi.wordpress.com/2007/06/25/advanced-ranking-and-dynamically-generated-named-sets-in-mdx/
我发现创建集合 MyMonthsWithEmployeesSets 的脚本部分很难理解 - 也许@AlexPeshik 可以更清楚地说明以下脚本中发生的事情。
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];
我需要按客户组显示前 5 位客户的销售额，但该组内其他客户的销售额汇总为 'Others'。类似于 this question，但针对每个客户组单独计算。
根据 MSDN，要对一组成员执行 TopCount，您必须使用 Generate 函数。
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]
)
但是 'Others' 部分有问题。
我认为我能够按组与其他客户构建集合（数据看起来不错）作为：
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]
)
)
)
但是我不知道如何通过分组来聚合它。
按照 this question
中的方式进行操作member [Klient].[Klient].[tmp] as
aggregate([OtherCustomersByGroup])
产生一个值，这是合乎逻辑的。
我想我需要每个组中有 'other' 个客户的集合列表，而不是单个 [OtherCustomersByGroup] 集合，但不知道如何构造它们。
大家有什么想法或建议吗？
更新：
对我的需求有些误解。我需要每个客户组中的前 n 个客户按销售额与该组中其他客户的销售额汇总到一个位置（假设称为其他）。
例如这个简化的输入：
| 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 个）：
| 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，我们会在客户端处理。
是的，您已经通过使用 SET for Others 了解了主要思想，但需要进行一些小的补充才能完成任务。
我将使用我的测试数据库，但这可以很容易地转换为您的。
[Report Date]- 日期维度（[Klient]模拟）[REPORT DATE Y]- 年层次结构 ([Grupa Klientow])[REPORT DATE YM]- 月份层次结构 ([Klient].[Klient])[Measures].[Count]- TopCount ([Measures].[Przychody ze sprzedazy rzeczywiste wartosc]) 的度量
我也用了top 3只是为了在这里显示结果图像。
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：
..所有成员直到最后一个（即 201606）都在 Other 组中。
希望这对您有所帮助，bardzo dziękuję！
更新： 代码通过在 Report Date Other Groups 计算中删除一个乘法进行了优化。
Update-2：（尚未解决，但正在进行中）
(每个组下使用'Other'个成员)
重要！我们需要额外的层次结构：Group->Client（[Report Date].[REPORT DATE] 和 Year->Month 是我的情况）能够确定每个低级别成员的父级。
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：
我需要把这个结果移到这里，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，因为两个级别是独立的。也许我们可以添加一个 属性 像 'Group_Name' 并使用未链接的关卡，但同样 - 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]
Update-4（在year/month示例中有一个解决方案）
@whytheq 的惊人解决方案帮助完成了我想要的事情：
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];
和图片：
任务已解决，但请标记不是这个答案，而是@whytheq 的！
以下内容反对 AdvWrks，并使用了我在 Chris Webb 的博客上看到的技术，他在此处概述了该技术：
https://cwebbbi.wordpress.com/2007/06/25/advanced-ranking-and-dynamically-generated-named-sets-in-mdx/
我发现创建集合 MyMonthsWithEmployeesSets 的脚本部分很难理解 - 也许@AlexPeshik 可以更清楚地说明以下脚本中发生的事情。
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];