Question

EmpNumber City                                                        Total Sales
----------------------------------------------------------------------------------
     1811 Boston                                                      $14557260.03
     1803 Chichago                                                    $18266965.58

我的桌子看起来像这样。我能知道如何根据他们的销售额从特定城市中找到最佳员工吗？

所需的输出：

select employeenumber, city, max(TotalSales) 
from(
select employeenumber, a.city, sum(quantityordered*priceeach)  as TotalSales
from offices a, employees b, customers c, orders d, orderdetails e
where a.officeCode = b.officeCode
and   b.employeenumber = c.salesrepemployeenumber
and   c.customernumber = d.customernumber
and   d.ordernumber = e.ordernumber
group by employeenumber, a.city
order by a.city)
group by employeenumber, city;

我尝试过

bind_rows()

但是我仍然从波士顿获得3名员工，从Chichago获得3名员工。我想要的只是每个城市的一名员工。谢谢

Answer 1

只需使用row_number()分析函数：

0).
nodeID id_0 |
nodeName Test 3 |
parentID id_0
********************************************
1).
nodeID id_72864 |
nodeName Element A |
parentID id_0 
********************************************
2).
nodeID id_72865 |
nodeName Element B |
parentID id_0
********************************************
3).
nodeID id_72866 |
nodeName Element C |
parentID id_0
********************************************
4).
nodeID id_72867 |
nodeName Element D |
parentID id_0
********************************************
5).
nodeID id_72868 |
nodeName Element E |
parentID id_0
********************************************
6).
nodeID id_72869 |
nodeName Element 1 |
parentID id_72864
********************************************
7).
nodeID id_72870 |
nodeName Element 2 |
parentID id_72865
********************************************
8).
nodeID id_108185 |
nodeName Element |
parentID id_72865
********************************************
9).
nodeID id_72871 |
nodeName Element 3 |
parentID id_72866
********************************************
10).
nodeID id_72872 |
nodeName Element 4 |
parentID id_72867
********************************************
11).
nodeID id_73527 |
nodeName Element 5 |
parentID id_72868
*********************************************

如果对于TotalSales的最高值发生平局（等于TotalSales），则应将它们包括在结果中，然后将select employeenumber, city, TotalSales from ( select employeenumber, a.city, nvl(quantityordered,0)*nvl(priceeach,0) as TotalSales row_number() over ( partition by o.city order by nvl(quantityordered,0)*nvl(priceeach,0) desc ) as rn from offices off join employees e on off.officeCode = e.officeCode join customers c on e.employeenumber = c.salesrepemployeenumber join orders ord on c.customernumber = ord.customernumber join orderdetails odd on ord.ordernumber = odd.ordernumber ) where rn = 1替换为dense_rank()，这是另一个分析函数。

Answer 2

从上面共享的第一个数据集中创建名为tbl的临时表后，这将为您提供所需的答案。

  select EmpNumber, City, Max_Sales as `Max Sales` from
    (select City, max(`Total Sales`) as `Max_Sales`
      from tbl group by City) a
        left join
    (select `Total Sales` as drop_later, EmpNumber from tbl) b
       on a.Max_Sales = b.drop_later

这是Spark SQL的输出：

   EmpNumber      City    Max Sales
0       1811    Boston  14557260.03
1       1803  Chichago  18266965.58

选择取决于其他列的每个组的最大值

2 个答案: