以上历史记录表捕获了当前存在标签的时间和区域。 我想找出标签何时进入或退出区域。 尝试过
more_data as (
select tag_id,
update_time,
zone_ids as current_zones,
lag(zone_ids, 1) over (partition by tag_id order by update_time asc) as prev_zones
from tag_hist
order by update_time asc
)
select *
from more_data
order by tag_id, update_time asc;
现在我想比较当前区域和上一个区域,以识别进入/退出的区域
with entered zones = current_zones - prev_zones
with exited zones = prev_zones - current_zones
答案 0 :(得分:1)
如果能解决您的需求,您可以尝试以下
with more_data as (
select tag_id,
update_time,
zone_ids as current_zones,
lag(zone_ids, 1) over (partition by tag_id order by update_time asc) as prev_zones
from tag_hist
order by update_time asc
)
select t.*,
(select array(select unnest(current_zones::int[]) except select unnest(prev_zones::int[]))) as "entered zones",
(select array(select unnest(prev_zones::int[]) except select unnest(current_zones::int[]))) as "exited zones"
from more_data t
order by tag_id, update_time asc;
答案 1 :(得分:1)
似乎对未嵌套的数组元素进行操作比对整个数组更容易:
with tag_hist_row_numbers as (
-- add row numbers
select *, row_number() over (order by tag_id, update_time) as rn
from tag_hist
),
tag_hist_enter_exit as (
-- unnest zone_ids and find enters/exits
select
tag_id, zone_id, update_time, rn,
rn - 1 is distinct from lag(rn) over w as enter,
rn + 1 is distinct from lead(rn) over w as exit
from tag_hist_row_numbers
cross join unnest(zone_ids) as zone_id
window w as (partition by tag_id, zone_id order by update_time)
),
tag_hist_times as (
-- assign enter/exit times
select
tag_id,
zone_id,
enter,
update_time as enter_time,
case when exit then update_time else lead(update_time) over w end as exit_time
from tag_hist_enter_exit
where enter or exit
window w as (order by tag_id, zone_id, update_time)
)
select
-- remove redundant rows
-- rows with exit are useless now
tag_id,
zone_id,
enter_time,
exit_time
from tag_hist_times
where enter
order by tag_id, zone_id
如果您希望最终结果采用这种格式,则可以将每个zone_id
的时间汇总到数组中。