hive 求两个集合的差集 业务场景是这样的,这里由两个hive表格A和B A的形式大概是这样的:uid B的形式大概是这样的:uid 我想要得到存在A中但是不存在B中的uid 具体代码如下 select a.uid from (select uid from tmp_zidali_500wan_fullinfo_new)a left outer join (select uid from temp_zidali_uid_num_maxvalue_rate)b on a.uid=b.uid wh
Hive在0.11.0版本开始加入了row_number.rank.dense_rank分析函数,可以查询分组排序后的top值 说明: row_number() over ([partition col1] [order by col2]) rank() over ([partition col1] [order by col2]) dense_rank() over ([partition col1] [order by col2]) 它们都是根据col1字段分组,然后对col2字段进行排
阿里交叉面试问到了这个题,当时感觉没有答好,主要是对Hive这块还是不熟悉,其实可以采用row_number()函数. 1.ROW_NUMBER,RANK(),DENSE_RANK() 语法格式:row_number() OVER (partition by COL1 order by COL2 desc ) rank partition by:类似hive的建表,分区的意思: order by :排序,默认是升序,加desc降序: rank:表示别名 表示根据COL1分组,在分组内部根据 CO
select * from bdcdj.lqentry1 a where 顺序号 in (select max(顺序号) from bdcdj.lqentry1 b WHERE b.archival_code IS NOT NULL group by archival_code): 通过archival_code分组 ,取顺序号的最大值.
Single Number III Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. For example: Given nums = [1, 2, 1, 3, 2, 5], return [3, 5]