postgresql：表中给定id y的id x的最后一次出现值

Question

Hi I'm new to postgresql and need a little help.

i have the following table, where rows belong to class 1 or 2 based on the value of column pf. The rows are ordered by charttime.

table

i want a new column called prev_val, and the prev_val corresponding row should be equal to the last occurring value of the the class diff from the corresponding rows class. ie ex- if pf=2 for a given row then the prev_val of this row must be equal to the previous value of a row with pf=1.

ie

for a given pf=2, prev_val=valuenum of pf=1 with charrttime less than current row

for a given pf=1, prev_val=valuenum of pf=2 with charrttime less than current row

Answer 1

You are looking for the function lag() . If I understand the question correctly:

select t.*,
       lag(valuenum) over (partition by pf order by charttime) as prev_val
from t;

EDIT:

One way to express the logic would be:

select t.*,
       (case when pf = 1
             then lag(case when pf = 2 then valuenum end ignore nulls) over (partition by pf order by charttime) 
             then lag(case when pf = 1 then valuenum end ignore nulls) over (partition by pf order by charttime) 
        end) as prev_other_val
from t;

Unfortunately, Postgres does not yet support ignore nulls , so this doesn't work. The following might work:

select t.*,
       (case when pf = 1
             then lag(valuenum) filter (where pf = 2) over (partition by pf order by charttime) 
             then lag(valuenum) filter (where pf = 1) over (partition by pf order by charttime) 
        end) as prev_other_val
from t;

This uses the filter clause for the same (and more efficient) effect.

An alternative that should definitely work (but not on large tables) is a lateral join or subquery:

select t.*,
       (select t2.valuenum
        from t t2
        where t2.pf <> t.pf and t2.chartime < t.chartime
        order by t2.chartime desc
        fetch first 1 row only
       ) as prev_other_val
from t;