Dette burde være en hurtigere variant med LATERAL underforespørgsler. Uafprøvet.
SELECT s.record_id, s.security_id, s.date
, s.price / l.pmax AS price_to_peak_earnings
, s.price / l.pmin AS price_to_minimum_earnings
-- , ...
, s.price / l.cape1 AS cape1
, s.price / l.cape2 AS cape2
-- , ...
, s.price / l.cape10 AS cape10
, s.price / l.capb1 AS capb1
, s.price / l.capb2 AS capb2
-- , ...
, s.price / l.capb10 AS capb10
-- , ...
FROM (
SELECT *
, (date - interval '1 y')::date AS date1
, (date - interval '2 y')::date AS date2
-- ...
, (date - interval '10 y')::date AS date10
FROM (
SELECT *, min(date) OVER (PARTITION BY security_id) AS min_date
FROM security_data
) s1
) s
LEFT JOIN LATERAL (
SELECT CASE WHEN s.date10 >= s.min_date THEN NULLIF(max(earnings) , 0) END AS pmax
, CASE WHEN s.date10 >= s.min_date THEN NULLIF(min(earnings) , 0) END AS pmin
-- ...
, NULLIF(avg(earnings) FILTER (WHERE date >= s.date1), 0) AS cape1 -- no case
, CASE WHEN s.date2 >= s.min_date THEN NULLIF(avg(earnings) FILTER (WHERE date >= s.date2), 0) END AS cape2
-- ...
, CASE WHEN s.date10 >= s.min_date THEN NULLIF(avg(earnings) , 0) END AS cape10 -- no filter
, NULLIF(avg(book) FILTER (WHERE date >= s.date1), 0) AS capb1
, CASE WHEN s.date2 >= s.min_date THEN NULLIF(avg(book) FILTER (WHERE date >= s.date2), 0) END AS capb2
-- ...
, CASE WHEN s.date10 >= s.min_date THEN NULLIF(avg(book) , 0) END AS capb10
-- ...
FROM security_data
WHERE security_id = s.security_id
AND date >= s.date10
AND date < s.date
) l ON s.date1 >= s.min_date -- no computations if < 1 year of trailing data
ORDER BY s.security_id, s.date;
Det kommer stadig ikke til at være lynende hurtigt, da hver række har brug for flere separate sammenlægninger. Flaskehalsen her vil være CPU.
Se også opfølgningen med en alternativ tilgang (JOIN til genererede kalender + vinduesfunktioner):