Comparing different columns in SQL for each row - sql

Question

after some transformation I have a result from a cross join (from table a and b) where I want to do some analysis on. The table for this looks like this:
+-----+------+------+------+------+-----+------+------+------+------+
| id | 10_1 | 10_2 | 11_1 | 11_2 | id | 10_1 | 10_2 | 11_1 | 11_2 |
+-----+------+------+------+------+-----+------+------+------+------+
| 111 | 1 | 0 | 1 | 0 | 222 | 1 | 0 | 1 | 0 |
| 111 | 1 | 0 | 1 | 0 | 333 | 0 | 0 | 0 | 0 |
| 111 | 1 | 0 | 1 | 0 | 444 | 1 | 0 | 1 | 1 |
| 112 | 0 | 1 | 1 | 0 | 222 | 1 | 0 | 1 | 0 |
+-----+------+------+------+------+-----+------+------+------+------+
The ids in the first column are different from the ids in the sixth column.
In a row are always two different IDs that are matched with each other. The other columns always have either 0 or 1 as a value.
I am now trying to find out how many values(meaning both have "1" in 10_1, 10_2 etc) two IDs have on average in common, but I don't really know how to do so.
I was trying something like this as a start:
SELECT SUM(CASE WHEN a.10_1 = 1 AND b.10_1 = 1 then 1 end)
But this would obviously only count how often two ids have 10_1 in common. I could make something like this for example for different columns:
SELECT SUM(CASE WHEN (a.10_1 = 1 AND b.10_1 = 1)
OR (a.10_2 = 1 AND b.10_1 = 1) OR [...] then 1 end)
To count in general how often two IDs have one thing in common, but this would of course also count if they have two or more things in common. Plus, I would also like to know how often two IDS have two things, three things etc in common.
One "problem" in my case is also that I have like ~30 columns I want to look at, so I can hardly write down for each case every possible combination.
Does anyone know how I can approach my problem in a better way?
Thanks in advance.
Edit:
A possible result could look like this:
+-----------+---------+
| in_common | count |
+-----------+---------+
| 0 | 100 |
| 1 | 500 |
| 2 | 1500 |
| 3 | 5000 |
| 4 | 3000 |
+-----------+---------+

Answer 1

With the codes as column names, you're going to have to write some code that explicitly references each column name. To keep that to a minimum, you could write those references in a single union statement that normalizes the data, such as:
select id, '10_1' where "10_1" = 1
union
select id, '10_2' where "10_2" = 1
union
select id, '11_1' where "11_1" = 1
union
select id, '11_2' where "11_2" = 1;
This needs to be modified to include whatever additional columns you need to link up different IDs. For the purpose of this illustration, I assume the following data model
create table p (
id integer not null primary key,
sex character(1) not null,
age integer not null
);
create table t1 (
id integer not null,
code character varying(4) not null,
constraint pk_t1 primary key (id, code)
);
Though your data evidently does not currently resemble this structure, normalizing your data into a form like this would allow you to apply the following solution to summarize your data in the desired form.
select
in_common,
count(*) as count
from (
select
count(*) as in_common
from (
select
a.id as a_id, a.code,
b.id as b_id, b.code
from
(select p.*, t1.code
from p left join t1 on p.id=t1.id
) as a
inner join (select p.*, t1.code
from p left join t1 on p.id=t1.id
) as b on b.sex <> a.sex and b.age between a.age-10 and a.age+10
where
a.id < b.id
and a.code = b.code
) as c
group by
a_id, b_id
) as summ
group by
in_common;

Answer 2

The proposed solution requires first to take one step back from the cross-join table, as the identical column names are super annoying. Instead, we take the ids from the two tables and put them in a temporary table. The following query gets the result wanted in the question. It assumes table_a and table_b from the question are the same and called tbl, but this assumption is not needed and tbl can be replaced by table_a and table_b in the two sub-SELECT queries. It looks complicated and uses the JSON trick to flatten the columns, but it works here:
WITH idtable AS (
SELECT a.id as id_1, b.id as id_2 FROM
-- put cross join of table a and table b here
)
SELECT in_common,
count(*)
FROM
(SELECT idtable.*,
sum(CASE
WHEN meltedR.value::text=meltedL.value::text THEN 1
ELSE 0
END) AS in_common
FROM idtable
JOIN
(SELECT tbl.id,
b.*
FROM tbl, -- change here to table_a
json_each(row_to_json(tbl)) b -- and here too
WHERE KEY<>'id' ) meltedL ON (idtable.id_1 = meltedL.id)
JOIN
(SELECT tbl.id,
b.*
FROM tbl, -- change here to table_b
json_each(row_to_json(tbl)) b -- and here too
WHERE KEY<>'id' ) meltedR ON (idtable.id_2 = meltedR.id
AND meltedL.key = meltedR.key)
GROUP BY idtable.id_1,
idtable.id_2) tt
GROUP BY in_common ORDER BY in_common;
The output here looks like this:
in_common | count
-----------+-------
2 | 2
3 | 1
4 | 1
(3 rows)