动态循环并按名称对列进行分组的脚本所以我试图弄清楚这一点，它只是不太有效。
我有一个带有键值对的表，我将其从高到宽拆分。 我最终得到了大约 10k 列，它们都有一个常规的命名结构，比如
“15.02.0 参数 1”、“15.02 参数 2”、“15.02 参数 3”......“15.03 参数 1”、“15.03 参数 2”、“15.03 参数 3”、...
如何查看所有这些列并按起始术语对其进行动态分组？ IE。组中所有 15.02% 的列，所有 15.03% 列，所有 15.04%，...所有 17.06%....并根据 groupby 编号命名组？
我的脚本只是将所有内容分组到一个组中。乌。
//**split Aggregate Data from TALL to WIDE, grouping WIDE columns by StepID
dt = current data table();
Summarize( dt, RecipeStepIndex = by( :RecipeStepId) );//get list of unique RecipeStepID's
//split the table TALL to WIDE
dt << Split(
Split By( :RecipeStepId, :ParameterName, :AggregateFunction ),
Split( :Value ),
Group( :Recipe, :ProcessTime, :Tool, :Chamber, :Lot, :WaferScribe ),
Output Table( "AggWide" ),
Remaining Columns( Drop All ),
Sort by Column Property,
dt << sort( by( :Recipe, :EquipmentId, :ChamberId, :LotId, :SlotNo ))
);
dtAggWide = current data table();
ColNamesList = dtAggWide << get column names(string, continuous); //get the column names of the WIDE table
//Create an empty list for the matching columns
RecipeStepList = {};
// Find matching column names
For( i = 1, i <= N items( RecipeStepIndex ), i++,
For( j = 1, j <= N Items( ColNamesList ), j++,
If(
Contains( ColNamesList[j], RecipeStepIndex[i] ),
Insert Into( RecipeStepList, ColNamesList[j] ),
);
);
);
// If groups of columns are found, create Col groupings
If( N Items( RecipeStepList ) > 0,
dtAggWide << select columns( RecipeStepList );
dtAggWide << group columns();
);脚本回复：脚本以动态循环并按名称对列进行分组有很多方法可以做到这一点。
这是一个示例，它首先将组构建为关联数组，其中键作为组的名称，值是组中的列列表，然后循环遍历该关联数组以对数据表中的列进行分组。 要获取组名，我们对列名使用 Word() 函数。 由于使用 For Each，这需要 JMP16
Names Default To Here(1);
// this solution requires JMP16
dtAggWide = New Table("Untitled",
Add Rows(0),
New Column("15.02.0 param1", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.02.0 param2", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.02.0 param3", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.03 param1", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.03 param2", Numeric, "Continuous", Format("Best", 12), Set Values([]))
);
col_list = dtAggWide << Get Column Names("String");
// first get the column group names and columns<BR />// in this solution case we use associative array
aa_groups = Associative Array();
For Each({col_name}, col_list,
col_group = Word(1, col_name, " ");
If(!Contains(aa_groups, col_group),
aa_groups[col_group] = {};
);
Insert Into(aa_groups[col_group], col_name)
);
Show(aa_groups);
For Each({{key, value}}, aa_groups,
dtAggWide << Group Columns(key, value);
);
回复：脚本以动态循环并按名称对列进行分组谢谢，很快就会试试这个。
但是动态构建分组数组列表呢？ 基本上， :RecipeStepId 的每个唯一值都应该是一个分组名称。 我一开始就尝试这样做，在拆分表之前，将其保存到“RecipeStepIndex”
Summarize( dt, RecipeStepIndex = by( :RecipeStepId) );//get list of unique RecipeStepID's
回复：脚本以动态循环并按名称对列进行分组例如，您可以使用基于 RecipeStepIndex 的列名“预填充”关联数组。 很可能会从拆分列名称的开头找到 RecipeStepIndex，因此这应该仍然有效。
Names Default To Here(1);
// this solution requires JMP16
dtAggWide = New Table("Untitled",
Add Rows(0),
New Column("15.02.0 param1", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.02.0 param2", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.02.0 param3", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.03 param1", Numeric, "Continuous", Format("Best", 12), Set Values([])),
New Column("15.03 param2", Numeric, "Continuous", Format("Best", 12), Set Values([]))
);
col_list = dtAggWide << Get Column Names("String");
// first get the column group names and columns// in this solution case we use associative array
aa_groups = Associative Array();
//Summarize(dt, RecipeStepIndex = by(:RecipeStepId));//get list of unique RecipeStepID's
RecipeStepIndex = {"15.02.0", "15.03"};
aa_groups = Associative Array(RecipeStepIndex, Repeat({{}}, N Items(RecipeStepIndex)));
For Each({col_name}, col_list,
col_group = Word(1, col_name, " ");
Insert Into(aa_groups[col_group], col_name)
);
For Each({{key, value}}, aa_groups,
dtAggWide << Group Columns(key, value);
);
