Time series querying in Postgres

Question

This is a follow on question from @Erwin's answer to Efficient time series querying in Postgres.

In order to keep things simple I'll use the same table structure as that question

id | widget_id | for_date | score |

The original question was to get score for each of the widgets for every date in a range. If there was no entry for a widget on a date then show the score from the previous entry for that widget. The solution using a cross join and a window function worked well if all the data was contained in the range you were querying for. My problem is I want the previous score even if it lies outside the date range we are looking at.

Example data:

INSERT INTO score (id, widget_id, for_date, score) values
(1, 1337, '2012-04-07', 52),
(2, 2222, '2012-05-05', 99),
(3, 1337, '2012-05-07', 112),
(4, 2222, '2012-05-07', 101);

When I query for the range May 5th to May 10th 2012 (ie generate_series('2012-05-05'::date, '2012-05-10'::date, '1d')) I would like to get the following:

DAY          WIDGET_ID  SCORE
May, 05 2012    1337    52
May, 05 2012    2222    99
May, 06 2012    1337    52
May, 06 2012    2222    99
May, 07 2012    1337    112
May, 07 2012    2222    101
May, 08 2012    1337    112
May, 08 2012    2222    101
May, 09 2012    1337    112
May, 09 2012    2222    101
May, 10 2012    1337    112
May, 10 2012    2222    101

The best solution so far (also by @Erwin) is:

SELECT a.day, a.widget_id, s.score
FROM  (
   SELECT d.day, w.widget_id
         ,max(s.for_date) OVER (PARTITION BY w.widget_id ORDER BY d.day) AS effective_date
   FROM  (SELECT generate_series('2012-05-05'::date, '2012-05-10'::date, '1d')::date AS day) d
   CROSS  JOIN (SELECT DISTINCT widget_id FROM score) AS w
   LEFT   JOIN score s ON s.for_date = d.day AND s.widget_id = w.widget_id
   ) a
LEFT JOIN  score s ON s.for_date = a.effective_date AND s.widget_id = a.widget_id
ORDER BY a.day, a.widget_id;

But as you can see in this SQL Fiddle it produces null scores for widget 1337 on the first two days. I would like to see the earlier score of 52 from row 1 in its place.

Is it possible to do this in an efficient way?

Answer 1

As @Roman mentioned, DISTINCT ON can solve this. Details in this related answer:

• Select first row in each GROUP BY group?

Subqueries are generally a bit faster than CTEs, though:

SELECT DISTINCT ON (d.day, w.widget_id)
       d.day, w.widget_id, s.score
FROM   generate_series('2012-05-05'::date, '2012-05-10'::date, '1d') d(day)
CROSS  JOIN (SELECT DISTINCT widget_id FROM score) AS w
LEFT   JOIN score s ON s.widget_id = w.widget_id AND s.for_date <= d.day
ORDER  BY d.day, w.widget_id, s.for_date DESC;

You can use a set returning function like a table in the FROM list.

SQL Fiddle

One multicolumn index should be the key to performance:

CREATE INDEX score_multi_idx ON score (widget_id, for_date, score)

The third column score is only included to make it a covering index in Postgres 9.2 or later. You would not include it in earlier versions.

Of course, if you have many widgets and a wide range of days, the CROSS JOIN produces a lot of rows, which has a price-tag. Only select the widgets and days you actually need.