根据 table 名称更改所有列的名称（KQL/Kusto/Azure 数据资源管理器）

Question

我加入了 2 个 table，它们都有数百个名称相似的列。我想更改每个 table 中的所有列名称以包含 table 名称。为了使查询简单，我不想显式地调用每个列名。下面的查询实现了这个目标。但是，以下查询在应用于大型数据集时非常慢。我假设性能低下是由于 replace_regex() 函数在整个数据集上是运行。是否有另一种方法可以在提高更大数据集的性能的同时实现相同的结果？

let T1 = datatable (Key:string , Col2:string , Col3:string )
[
  "1", "b", "c",
  "2", "e", "f",
  "3", "h", "i"
] 
| project PackedRecord = todynamic(replace_regex(tostring(pack_all()), '"([a-zA-Z0-9_]*)":"', @'"T1_":"'))
| evaluate bag_unpack(PackedRecord);
let T2 = datatable (Key:string , Col2:string , Col3:string )
[
  "1", "B", "C",
  "2", "E", "F",
  "4", "H", "I"
] 
| project PackedRecord = todynamic(replace_regex(tostring(pack_all()), '"([a-zA-Z0-9_]*)":"', @'"T2_":"'))
| evaluate bag_unpack(PackedRecord);
let JoinTable = T1 | join kind=inner T2 on $left.T1_Key == $right.T2_Key;
JoinTable

上一题供参考

Answer 1

在执行 bag_unpack 操作时，您可以在不使用 replace_regex() 并依赖 OutputColumnPrefix 的情况下获得相同的结果。修改了原始 kql 代码段中的几行。

根据 kusto 文档，OutputColumnPrefix 参数允许传递一个公共前缀以添加到插件生成的所有列。

let T1 = datatable (Key:string , Col2:string , Col3:string )
[
  "1", "b", "c",
  "2", "e", "f",
  "3", "h", "i"
]
| project Key, PackedRecord = pack_all()
| evaluate bag_unpack(PackedRecord, OutputColumnPrefix = "T1_") | project-away T1_Key; // get rid of additional key;
let T2 = datatable (Key:string , Col2:string , Col3:string )
[
  "1", "B", "C",
  "2", "E", "F",
  "4", "H", "I"
]
| project Key, PackedRecord = pack_all()
| evaluate bag_unpack(PackedRecord, OutputColumnPrefix = "T2_") | project-away T2_Key; // get rid of additional key;
T1 
| join kind=inner T2 on $left.Key == $right.Key | project-away Key1 // get rid of second key after join

根据 table 名称更改所有列的名称（KQL/Kusto/Azure 数据资源管理器）

Change name of all columns based on table names (KQL / Kusto / Azure Data Explorer)

azure-data-explorer

kql

kusto-explorer