GROUP by Largest String for all the substrings - sql

Question

I have a table like this where some rows have the same grp but different names. I want to group them by name such that all the substrings after removing nonalphanumeric characters are aggregated together and grouped by the largest string. The null value is considered the substring of all the strings.
grp
name
value
1
ab&c
10
1
abc d e
56
1
ab
21
1
a
23
1
xy
34
1
[null]
1
2
fgh
87
Desired result
grp
name
value
1
abcde
111
1
xy
34
2
fgh
87
My query-
Select grp,
regexp_replace(name,'[^a-zA-Z0-9]+', '', 'g') name, sum(value) value
from table
group by grp,
regexp_replace(name,'[^a-zA-Z0-9]+', '', 'g');
Result
grp
name
value
1
abc
10
1
abcde
56
1
ab
21
1
a
23
1
xy
34
1
[null]
1
2
fgh
87
What changes should I make in my query?

Answer 1

To solve this problem, I did the following (all of the code below is available on the fiddle here).
CREATE TABLE test
(
grp SMALLINT NOT NULL,
name TEXT NULL,
value SMALLINT NOT NULL
);
and populate it using your data + extra for testing:
INSERT INTO test VALUES
(1, 'ab&c', 10),
(1, 'abc d e', 56),
(1, 'ab', 21),
(1, 'a', 23),
(1, NULL, 1000000),
(1, 'r*&%$s', 100), -- added for testing.
(1, 'rs__t', 101),
(1, 'rs__tu', 101),
(1, 'xy', 1111),
(1, NULL, 1000000),
(2, 'fgh', 87),
(2, 'fgh', 13), -- For Charlieface
(2, NULL, 1000000),
(2, 'x', 50),
(2, 'x', 150),
(2, 'x----y', 100);
Then, you can use this query:
WITH t1 AS
(
SELECT
grp, n_str,
LAG(n_str) OVER (PARTITION BY grp ORDER BY grp, n_str),
CASE
WHEN
LAG(n_str) OVER (PARTITION BY grp ORDER BY grp, n_str) IS NULL
OR
POSITION
(
LAG(n_str) OVER (PARTITION BY grp ORDER BY grp, n_str)
IN
n_str
) = 0
THEN 1
ELSE 0
END AS change,
value
FROM
test t1
CROSS JOIN LATERAL
(
VALUES
(
REGEXP_REPLACE(name,'[^a-zA-Z0-9]+', '', 'g')
)
) AS v(n_str)
WHERE n_str IS NOT NULL
), t2 AS
(
SELECT
ROW_NUMBER() OVER (PARTITION BY grp, s_change ORDER BY grp, n_str DESC) AS rn,
grp, n_str,
SUM(value) OVER (PARTITION BY grp, s_change) AS s_val,
MAX(LENGTH(n_str)) OVER (PARTITION BY grp) AS max_nom
FROM
(
SELECT
grp, n_str, change,
SUM(change) OVER (ORDER BY grp, n_str) AS s_change,
value
FROM
t1
ORDER BY grp, n_str DESC
) AS sub1
), t3 AS
(
SELECT
grp, SUM(value) AS null_sum
FROM
test
WHERE name IS NULL
GROUP BY grp
)
SELECT x.grp, x.n_str, x.s_val + y.null_sum
FROM t2 x
JOIN t3 y
ON x.max_nom = LENGTH(x.n_str) AND x.grp = y.grp
UNION
SELECT grp, n_str, s_val
FROM
t2 WHERE max_nom != LENGTH(n_str) AND rn = 1
ORDER BY grp, n_str;
Result:
grp n_str ?column?
1 abcde 2000110
1 rstu 302
1 xy 1111
2 fgh 1000100
2 xy 300
A few points to note:
Please always provide a fiddle when you ask questions such as this one with tables and data - it provides a single source of truth for the question and eliminates duplication of effort on the part of those trying to help you!
You haven't been very clear about what, exactly, should happen with NULLs - do the values count towards the SUM()? You can vary the CASE statement as required.
What happens when there's a tie in the number of characters in the string? I've included an example in the fiddle, where you get the draws - but you may wish to sort alphabetically (or some other method)?
There appears to be an error in your provided sums for the values (even taking account of counting or not values for NULL for the name field).
Finally, you don't want to GROUP BY the largest string - you want to GROUP BY the grp fields + the SUM() of the values in the the given grp records and then pick out the longest alphanumeric string in that grouping. It would be interesting to know why you want to do this?