查询Hive QL另一列中与每个值关联的最短字符串值的更有效方法

Question

I have a table in Hive containing store names, order IDs, and User IDs (as well as some other columns including item ID). There is a row in the table for every item purchased (so there can be more than one row per order if the order contains multiple items). Order IDs are unique within a store, but not across stores. A single order can have more than one user ID associated with it.

I'm trying to write a query that will return a list of all stores and order IDs and the shortest user ID associated with each order.

So, for example, if the data looks like this:

 STORE | ORDERID | USERID | ITEMID
 ------+---------+--------+-------
|  a   |    1    |  bill  |  abc  |
|  a   |    1    |  susan |  def  |
|  a   |    2    |  jane  |  abc  |
|  b   |    1    |  scott |  ghi  |
|  b   |    1    |  tony  |  jkl  |

Then the output would look like this:

 STORE | ORDERID | USERID 
 ------+---------+-------
   a   |    1    |  bill 
   a   |    2    |  jane 
   b   |    1    |  tony 

I've written a query that will do this, but I feel like there must be a more efficient way to go about it. Does anybody know a better way to produce these results?

This is what I have so far:

select 
    users.store, users.orderid, users.userid
from 
    (select 
         store, orderid, userid, length(userid) as len 
     from 
         sales) users
join 
    (select distinct 
         store, orderid, 
         min(length(userid)) over (partition by store, orderid) as len 
     from 
         sales) len on users.store = len.store
                    and users.orderid = len.orderid
                    and users.len = len.len

Answer 1

Probably rank() is the best way:

select s.*
from (select s.*, rank() over (partition by store order by length(userid) as seqnum
      from sales s
     ) s
where seqnum = 1;

Answer 2

Check out probably this will work for you, here you can achieve your goal of single "SELECT" clause with no extra overhead on SQL.

select distinct 
    store, orderid, 
    first_value(userid) over(partition by store, orderid order by length(userid) asc) f_val 
from 
    sales;

The result will be:

store   orderid    f_val
a       1          bill
a       2          jane
b       1          tony