假设你的表2，最后一列将与表2一致。（我的意思是，在这里作用于表2本身，你可以得到pix\u id的结果，coo\u id将在表2中正确匹配。）如果我的假设是错误的请原谅。

hive (sflow)> desc table2;
OK
col_name    data_type   comment
id  string  from deserializer
time_stamp  string  from deserializer
pix_id  string  from deserializer
coo_id  string  from deserializer
Time taken: 0.277 seconds

Hive（sflow）>

SELECT t2.id,t2.time_stamp,t2.pix_id,t2.coo_id
   FROM table2 t2 JOIN
        ( SELECT pix_id,coo_id, Max(UNIX_TIMESTAMP(time_stamp)) as max_epoch 
          FROM table2 
          GROUP BY pix_id,coo_id)  temp   
WHERE t2.pix_id=temp.pix_id AND t2.coo_id=temp.coo_id AND UNIX_TIMESTAMP(t2.time_stamp) = max_epoch ;

ps：复制完整的日志（请注意，我运行的是伪模式hadoop，配置单元0.9，2gb ram）：

hive (sflow)> from table2 t2 join (select pix_id,coo_id, Max(UNIX_TIMESTAMP(time_stamp)) as max_epoch from table2 group by pix_id,coo_id) temp
            > select t2.id,t2.time_stamp,t2.pix_id,t2.coo_id where t2.pix_id=temp.pix_id and t2.coo_id=temp.coo_id and UNIX_TIMESTAMP(t2.time_stamp) = max_epoch ;

Total MapReduce jobs = 2
Launching Job 1 out of 2
Number of reduce tasks not specified. Estimated from input data size: 1
Total MapReduce CPU Time Spent: 24 seconds 0 msec
OK
id  time_stamp  pix_id  coo_id
id4 2014-01-01 22:51:39,487 px1 coo1
id6 2014-01-01 02:09:42,675 px1 coo2
id8 2014-01-01 23:34:51,782 px1 coo3
id11    2014-01-01 06:59:43,999 px2 coo2
id12    2014-01-01 09:21:57,325 px3 coo3
id14    2014-01-01 17:27:05,128 px4 coo4
Time taken: 145.17 seconds

hive (sflow)> 
hive (sflow)> desc table2;
OK
col_name    data_type   comment
id  string  from deserializer
time_stamp  string  from deserializer
pix_id  string  from deserializer
coo_id  string  from deserializer
Time taken: 0.277 seconds
hive (sflow)>

你可以用 collect_max 来自砖厂的自定义项（http://github.com/klout/brickhouse )只使用一个作业步骤生成数据。

select array_index( map_keys( max_map ), 0) as id,
    from_unixtime( array_index( map_values( max_map), 0) as time_stamp,
    pix_id,
    coo_id
from (
   select pix_id, coo_id, 
       collect_max( id, unix_timestamp(time_stamp) ) as max_map
   from table2
   group by pix_id, coo_id ) cm ;

对于小数据集，这并不重要，但是对于非常大的数据集，它允许您只通过一次数据传递来解决问题。