यह LATERAL . के साथ एक तेज़ संस्करण होना चाहिए उपप्रश्न परीक्षण नहीं किया गया।
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;
यह अभी भी बहुत तेज़ नहीं होने वाला है, क्योंकि प्रत्येक पंक्ति को कई अलग-अलग एकत्रीकरण की आवश्यकता होती है। यहां अड़चन सीपीयू होगी।
वैकल्पिक दृष्टिकोण के साथ अनुवर्ती कार्रवाई भी देखें (जेनरेट किए गए कैलेंडर + विंडो फ़ंक्शंस में शामिल हों):
