在 Hive 中每天获取前 N 行 - rank()

Question

我有这个 table，每行捐赠一个销售：

 sale_date  salesman  sale_item_id
 20170102   JohnSmith       309
 20170102   JohnSmith       292
 20170103   AlexHam          93

我试图每天获得前 20 名销售员，我想到了这个：

SELECT sale_date, salesman, sale_count, row_num
FROM (
  SELECT sale_date, salesman,
         count(*) as sale_count,
         rank() over (partition by sale_date order by sale_count desc) as row_num
  from salesforce.sales_data
) T
WHERE sale_date between  '20170101' and '20170110'
 and row_num <= 20

但我得到：

FAILED: SemanticException Failed to breakup Windowing invocations into Groups. At least 1 group must only depend on input columns. Also check for circular dependencies.
Underlying error: org.apache.hadoop.hive.ql.parse.SemanticException: Line 5:35 Expression not in GROUP BY key 'sale_date'

不过我不确定分组会在什么时候生效。有人可以帮忙吗？发送！

Answer 1

您在子查询中缺少 group by:

SELECT sale_date, salesman, sale_count, row_num
FROM (SELECT sale_date, salesman,
             count(*) as sale_count,
             rank() over (partition by sale_date order by count(*) desc) as row_num
      FROM salesforce.sales_data
      GROUP BY sale_date, salesman
     ) T
WHERE sale_date between '20170101' and '20170110' and row_num <= 20;

我认为 Hive 会接受 order by、order by sale_count desc 中的列别名。

另请注意，如果存在并列，您可以得到多于或少于 20 行。如果您恰好需要 20 行，您可能需要 row_number()。

Answer 2

试试这个

SELECT sale_date, salesman, sale_count, row_num from (
SELECT sale_date, salesman, sale_count,
 rank() over (partition by sale_date order by sale_count desc) as         row_num
from 
(
SELECT sale_date, salesman,
   count(*) over (partition by salesman) as sale_count
from  employee
) t1
) t2  where sale_date between  '20170101' and '20170110'
and row_num <= 20;
WHERE sale_date between  '20170101' and '20170110'
and row_num <= 20

编辑和测试。你的问题本质上是你在为你的 over 子句计算它之前尝试使用计数，如果你在推销员的子查询分区中计算你的计数，它将解决问题。您不能在销售查询中进行分组，如果这样做，您将无法访问 sale_date。

在 Hive 中每天获取前 N 行 - rank()

Get top N rows per day in Hive - rank()

sql

hive

rank