How to use sql LAG() properly - sql

Question

I have the following SQL line that has a syntax error. I'm trying to reference prior day close in my SQL query how do i fix my query to not error out?
Thanks!
SELECT *
FROM "daily_data"
WHERE date >'2018-01-01' and (open-LAG(close))/LAG(close)>=1.4 and volume > 1000000 and open > 1
Error:
Query execution failed
Reason: SQL Error [42809]: ERROR: window function lag requires an OVER
clause Position: 63

Answer 1

You need to use a subquery. You cannot use window functions in the where clause. You also need an ORDER BY and potentially a PARTITION BY clause:
SELECT *
FROM (SELECT dd.*,
LAG(close) OVER (ORDER BY date) as prev_close
FROM "daily_data" dd
) dd
WHERE date > '2018-01-01' AND
(open - prev_close) / prev_close >= 1.4 AND
volume > 1000000 AND
open > 1;

Answer 2

lag(close) means "the value of close from the prior record." So the phrase by itself is missing something fundamental, specifically how do you define "prior record" since there is never any implied order in a RDBMS.
As with functions such as rank and row_number, to propery form the lead and lag commands you need to establish the prior (or next) record by defining the order of output. In other words, "if you were to sort the output by x, the prior record's close" would look like this:
lag (close) over (order by x)
To order by something descending:
lag (close) over (order by x desc)
You can optionally chunk the data by a field using partition by which may or may not be useful in your problem. For example, "for each item, if you were to sort the output by x, the prior record's close:"
lag (close) over (partition by item order by x)
To the question here is prior record (lag)... how? By which fields, in which order?
As a final thought, analytic/windowing functions cannot be used in the where clause in PostgreSQL. To accomplish this, wrap them in a subquery:
with daily as (
SELECT
d.*,
LAG (d.close) over (order by d.<something>) as prior_close
FROM "daily_data" d
WHERE
d.date >'2018-01-01' and
d.volume > 1000000 and
d.open > 1
)
select *
from daily
where
(open - prior_close) / prior_close >= 1.4