2

I have a table like this:

Date Group name value
2022-01-01 A 1.0
2022-01-15 A 0.5
2022-01-31 A 0.2

but this contains only 3 days. I need a table with full-calendar dates, a row for every day of the year where the value is the last available.
For example, all records from 1 Jan 2022 to 14 Jan have value 1.0 (scoped to group_name 'A').

I have tried using LAST_VALUE() but it's not working.

WITH dates AS (
    
    SELECT 
        date::date
    FROM generate_series ( '2022-01-01'::timestamp, '2022-01-31'::timestamp, '1 day'::interval) date

), incomplete_table AS (
     SELECT * FROM (VALUES 
        ('2022-01-01'::date, 'a', 1),
        ('2022-01-15'::date, 'a', 0.5), 
        ('2022-01-31'::date, 'a', 0.2), 
        ('2022-01-02'::date, 'b', 0.1),
        ('2022-01-10'::date, 'b', 0.15),
        ('2022-01-20'::date, 'b', 0.15)
    ) AS t (date,group_name, value)
)
SELECT
    dates.date,
    group_name, 
    value,
    LAST_VALUE(value) OVER (ORDER BY dates.date DESC) as last_value_window
FROM dates
LEFT JOIN incomplete_table ON incomplete_table.date = dates.date
ORDER BY dates.date DESC;
2
  • Can you provide an example of the output you are looking for the query to return? Additionally, considering creating a Fiddle. Feb 21, 2022 at 15:14
  • scoped to group_name A means you want one row per day and group name? Or fill in date and group_namefrom the last nonnull value? Independently? Also, please always declare your version of Postgres. Also, an exact table definition (CREATE TABLE statement) is much better than a CTE, telling us exact data types and constraints. Feb 21, 2022 at 17:14

1 Answer 1

4
+50

Assuming you want one row per day and group name:

WITH incomplete_table(date, group_name, value) AS (
   VALUES 
     ('2022-01-01'::date, 'a', 1)
   , ('2022-01-15'::date, 'a', 0.5)
   , ('2022-01-31'::date, 'a', 0.2)
   , ('2022-01-02'::date, 'b', 0.1)
   , ('2022-01-10'::date, 'b', 0.15)
   , ('2022-01-20'::date, 'b', 0.15)
   )
SELECT d.date, g.group_name, i.value
FROM  (
   SELECT date::date
   FROM   generate_series (timestamp '2022-01-01'
                         , timestamp '2022-01-31'
                         , interval '1 day') date
   ) d
CROSS  JOIN (SELECT DISTINCT group_name FROM incomplete_table) g  -- ①
LEFT   JOIN LATERAL (
   SELECT i.group_name, i.value
   FROM   incomplete_table i
   WHERE  i.group_name = g.group_name
   AND    i.date <= d.date
-- AND    i.date >= timestamp '2022-01-01'  -- ? ②
   ORDER  BY i.date DESC
   LIMIT  1
   ) i ON true
ORDER  BY g.group_name, d.date DESC;

db<>fiddle here

① If available, use a table "groups" providing distinct group names. Faster. Else, if the table is big, consider an emulated index-skip scan. See:

② The search for the latest value is not limited to the given time range unless you spell that out in the LATERAL subquery. About LATERAL:

You still get value IS NULL where no earlier value is found.

If incomplete_table is big, an index on (group_name, date) will help performance (a lot). Possibly even a "covering" index, adding column value. See:

Very similar case with more explanation:

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.