具有下表和Kusto查询,如何获得具有“购买”列的结果?
let ProductsTable = datatable(Supplier: string, Fruit: string, Price: int, Purchase: datetime)
[
'Contoso', 'Grapes', 220, datetime(2018-10-01 01:00),
'Fabrikam', 'Lemons', 31, datetime(2018-10-01 02:00),
'Contoso', 'Lemons', 29, datetime(2018-10-02 03:00),
'Contoso', 'Grapes', 210, datetime(2018-10-02 04:00),
'Fabrikam', 'Lemons', 30, datetime(2018-10-03 05:00),
'Contoso', 'Bananas', 12, datetime(2018-10-03 06:00),
'Contoso', 'Bananas', 12, datetime(2018-10-04 07:00),
'Contoso', 'Lemons', 29, datetime(2018-10-04 08:00),
'Contoso', 'Grapes', 200, datetime(2018-10-05 09:00),
];
ProductsTable
| summarize Price = min(Price) by Supplier, Fruit
| order by Supplier asc, Fruit asc, Price asc
结果
Contoso Bananas 12
Contoso Grapes 200
Contoso Lemons 29
Fabrikam Lemons 30
期望的结果
Contoso Bananas 12 2018-10-03 06:00
Contoso Grapes 200 2018-10-05 09:00
Contoso Lemons 29 2018-10-02 03:00
Fabrikam Lemons 30 2018-10-03 05:00
我知道可能会有多个结果,例如对于 Contoso-Bananas-12 ,我们可以具有以下任意一种结果
答案 0 :(得分:3)
尝试使用arg_min()
:https://docs.microsoft.com/en-us/azure/kusto/query/arg-min-aggfunction
let ProductsTable = datatable(Supplier: string, Fruit: string, Price: int, Purchase: datetime)
[
'Contoso', 'Grapes', 220, datetime(2018-10-01 01:00),
'Fabrikam', 'Lemons', 31, datetime(2018-10-01 02:00),
'Contoso', 'Lemons', 29, datetime(2018-10-02 03:00),
'Contoso', 'Grapes', 210, datetime(2018-10-02 04:00),
'Fabrikam', 'Lemons', 30, datetime(2018-10-03 05:00),
'Contoso', 'Bananas', 12, datetime(2018-10-03 06:00),
'Contoso', 'Bananas', 12, datetime(2018-10-04 07:00),
'Contoso', 'Lemons', 29, datetime(2018-10-04 08:00),
'Contoso', 'Grapes', 200, datetime(2018-10-05 09:00),
];
ProductsTable
| summarize Price = arg_min(Price, *) by Supplier, Fruit
| order by Supplier asc, Fruit asc, Price asc
答案 1 :(得分:1)
我是Kusto的新手,但我发现下面可以通过在Kusto中使用 partition by 和 row_number()返回相同的输出。
KQL:
let ProductsTable = datatable(Supplier: string, Fruit: string, Price: int, Purchase: datetime)
[
'Contoso', 'Grapes', 220, datetime(2018-10-01 01:00),
'Fabrikam', 'Lemons', 31, datetime(2018-10-01 02:00),
'Contoso', 'Lemons', 29, datetime(2018-10-02 03:00),
'Contoso', 'Grapes', 210, datetime(2018-10-02 04:00),
'Fabrikam', 'Lemons', 30, datetime(2018-10-03 05:00),
'Contoso', 'Bananas', 12, datetime(2018-10-03 06:00),
'Contoso', 'Bananas', 12, datetime(2018-10-04 07:00),
'Contoso', 'Lemons', 29, datetime(2018-10-04 08:00),
'Contoso', 'Grapes', 200, datetime(2018-10-05 09:00),
];
ProductsTable
| partition by Fruit
( sort by Price asc
| extend rn = row_number(1, prev(Supplier) != Supplier)
| where rn == 1
)
| project Supplier, Fruit, Price, Purchase
| sort by Supplier asc, Fruit asc;
结果:
Supplier Fruit Price Purchase
Contoso Bananas 12 2018-10-03 06:00:00.0000000
Contoso Grapes 200 2018-10-05 09:00:00.0000000
Contoso Lemons 29 2018-10-02 03:00:00.0000000
Fabrikam Lemons 30 2018-10-03 05:00:00.0000000
但是,这没有Yoni L的解决方案有效。我曾经使用Redshift / PostgreSQL编写类似的查询,所以我想知道是否可以通过row_num()完成此操作,这是一个非常常见的窗口函数。