SQL group by similar address but different longitude and latitude - sql

Question

I Have a large Postgres dataset table,
the table ('tbl') has 4 columns,
and a data similar to this:
ID
address
x,y
1
22 E 4th Ave, Cordele, GA, 11015
x1,y1
2
22 E 4th Ave, Cordele, GA 11015
x2,y2
3
408 E 5th Ave, Cordele, CA 11215
x2,y2
4
408 E 5th Ave, Cordele, CA, 11215
x2,y2
5
408 E 5th Ave, vic, VA, 11215
x2,y2
6
408 E 5th Ave, vic, VA, 11215
x3,y3
My question is , how to find all the addresses that have similar address (similar address means ignoring the comma between the state and zip, that's the only part that should be ignored), But having different 'x,y' value
In the above example , id 1 and 2 should be returned because they have the same address ( with a diff in the comma) But different 'x,y' values.
Id 3 and 4 should not be returned because their 'x,y' values are identical.
Id 5 and 6 should not be returned because their address values are identical.
*I can count on the address format to always have a state and a zip

Answer 1

It might be overkill, but can you just remove all commas and compare?
select array_agg(distinct address)
from t
group by replace(address, ',', '')
having min(x_y) <> max(x_y);
To specifically remove that comma, you could instead use:
select array_agg(distinct address)
from t
group by (case when address like '%, _____'
then left(address, -7) || right(address, 6)
else address
end)
having min(x_y) <> max(x_y);

Answer 2

I am not sure how many variations you have in your data, but I was able to get what you want on the sample data provided. I inserted the data in a table named locdat, you could change columns and table as per your need.
SELECT
id, address, xy
FROM
(
SELECT
l.*,
COUNT(l.address)
OVER(PARTITION BY replace(l.address, ',', '')) AS addr_count,
COUNT(l.xy)
OVER(PARTITION BY replace(l.address, ',', ''), l.xy) AS xy_count
FROM
locdat l
)
WHERE
( addr_count >= 1
AND xy_count < 2 );