问题
I have a sample data as below and wanting to get a desired o/p, please help me with some idea.
I want the o/p of prev_diff_value of the 3rd,4th row to be 2015-01-01 00:00:00 instead of 2015-01-02 00:00:00.
with dat as (
select 1 as id,'20150101 02:02:50'::timestamp as dt union all
select 1,'20150101 03:02:50'::timestamp union all
select 1,'20150101 04:02:50'::timestamp union all
select 1,'20150102 02:02:50'::timestamp union all
select 1,'20150102 02:02:50'::timestamp union all
select 1,'20150102 02:02:51'::timestamp union all
select 1,'20150103 02:02:50'::timestamp union all
select 2,'20150101 02:02:50'::timestamp union all
select 2,'20150101 03:02:50'::timestamp union all
select 2,'20150101 04:02:50'::timestamp union all
select 2,'20150102 02:02:50'::timestamp union all
select 1,'20150104 02:02:50'::timestamp
)-- select * from dat
select id , dt , lag(trunc(dt)) over(partition by id order by dt asc) prev_diff_value
from dat
order by id,dt desc
O/P :
id dt prev_diff_value
1 2015-01-04 02:02:50 2015-01-03 00:00:00
1 2015-01-03 02:02:50 2015-01-02 00:00:00
1 2015-01-02 02:02:51 2015-01-02 00:00:00
1 2015-01-02 02:02:50 2015-01-02 00:00:00
1 2015-01-02 02:02:50 2015-01-01 00:00:00
回答1:
As I understand you want to get the previous different date for each timestamp within id partition. I would then apply lag against the unique combination of id and date and join back to the original dataset like this:
with dat as (
select 1 as id,'20150101 02:02:50'::timestamp as dt union all
select 1,'20150101 03:02:50'::timestamp union all
select 1,'20150101 04:02:50'::timestamp union all
select 1,'20150102 02:02:50'::timestamp union all
select 1,'20150102 02:02:50'::timestamp union all
select 1,'20150102 02:02:51'::timestamp union all
select 1,'20150103 02:02:50'::timestamp union all
select 2,'20150101 02:02:50'::timestamp union all
select 2,'20150101 03:02:50'::timestamp union all
select 2,'20150101 04:02:50'::timestamp union all
select 2,'20150102 02:02:50'::timestamp union all
select 1,'20150104 02:02:50'::timestamp
)
,dat_unique_lag as (
select *, lag(date) over(partition by id order by date asc) prev_diff_value
from (
select distinct id,trunc(dt) as date
from dat
)
)
select *
from dat
join dat_unique_lag
using (id)
where trunc(dat.dt)=dat_unique_lag.date
order by id,dt desc;
However, this is not super performant. If nature of your data is that you have a limited number of timestamps for the same day you might just extend your lag with a conditional statement like this:
with dat as (
select 1 as id,'20150101 02:02:50'::timestamp as dt union all
select 1,'20150101 03:02:50'::timestamp union all
select 1,'20150101 04:02:50'::timestamp union all
select 1,'20150102 02:02:50'::timestamp union all
select 1,'20150102 02:02:50'::timestamp union all
select 1,'20150102 02:02:51'::timestamp union all
select 1,'20150103 02:02:50'::timestamp union all
select 2,'20150101 02:02:50'::timestamp union all
select 2,'20150101 03:02:50'::timestamp union all
select 2,'20150101 04:02:50'::timestamp union all
select 2,'20150102 02:02:50'::timestamp union all
select 1,'20150104 02:02:50'::timestamp
)
select id, dt,
case
when lag(trunc(dt)) over(partition by id order by dt asc)=trunc(dt)
then case
when lag(trunc(dt),2) over(partition by id order by dt asc)=trunc(dt)
then case
when lag(trunc(dt),3) over(partition by id order by dt asc)=trunc(dt)
then lag(trunc(dt),4) over(partition by id order by dt asc)
else lag(trunc(dt),3) over(partition by id order by dt asc)
end
else lag(trunc(dt),2) over(partition by id order by dt asc)
end
else lag(trunc(dt)) over(partition by id order by dt asc)
end as prev_diff_value
from dat
order by id,dt desc;
Basically, you look at the previous record and if it doesn't suit you then you look back to the record before that one and so on until you find the correct record or run out of your statement depth. Here it looks until the 4th record back.
来源：https://stackoverflow.com/questions/44645751/lag-function-to-get-the-last-different-valueredshift