我想在使用配置单元时删除数组中的连续重复项。
collect_list()
保留所有重复项,而collect_set()
仅保留不同的条目。我有点需要中间立场。
例如,从下表:
id | number
==============
fk 4
fk 4
fk 2
4f 1
4f 8
4f 8
h9 7
h9 4
h9 7
我想得到这样的东西:
id | aggregate
===========================
fk Array<int>(4,2)
4f Array<int>(1,8)
h9 Array<int>(7,4,7)
答案 0 :(得分:1)
使用lag()
分析函数获取先前的数字,并与当前数字进行比较以检查连续的数字。
演示:
with your_table as (--replace this subquery with your table
select stack(11, --the number of tuples
'fk',4,'2019-01-01 10:10:10.123',
'fk',4,'2019-01-01 10:10:10.124',
'fk',2,'2019-01-01 10:10:10.125',
'4f',1,'2019-01-01 10:10:10.126',
'4f',8,'2019-01-01 10:10:10.127',
'4f',8,'2019-01-01 10:10:10.128',
'h9',7,'2019-01-01 10:10:10.129',
'h9',4,'2019-01-01 10:10:10.130',
'h9',7,'2019-01-01 10:10:10.131',
'h9',7,'2019-01-01 10:10:10.132',
'h9',7,'2019-01-01 10:10:10.133'
) as (id, number, order_ts)
) --replace this subquery with your table
select id, collect_list(case when number = lag_number then null else number end) as aggregate
from
(select id, number, order_ts,
lag(number) over (partition by id order by order_ts) lag_number
from your_table
distribute by id sort by order_ts
)s
group by id;
结果(在此处检查:http://demo.gethue.com/hue/editor?editor=318918):
id aggregate
4f [1,8]
fk [4,2]
h9 [7,4,7]