count combination of columns in postgresql matrix - sql

Question

I have a table in postgres like below
I want an sql in postgres that count a combination of 2 columns that has YY
Expecting an output like
Combination Count
AB 2
AC 1
AD 2
AZ 1
BC 1
BD 3
BZ 2
CD 2
CZ 0
DZ 1
Can anyone help me?

Answer 1

WITH stacked AS (
SELECT id
, unnest(array['A', 'B', 'C', 'D', 'Z']) AS col_name
, unnest(array[a, b, c, d, z]) AS col_value
FROM test t
)
SELECT combo, sum(cnt) AS count
FROM (
SELECT t1.id, t1.col_name || t2.col_name AS combo
, (CASE WHEN t1.col_value = 'Y' AND t2.col_value = 'Y' THEN 1 ELSE 0 END) AS cnt
FROM stacked t1
INNER JOIN stacked t2
ON t1.id = t2.id
AND t1.col_name < t2.col_name) t3
GROUP BY combo
ORDER BY combo
yields
| combo | count |
|-------+-------|
| AB | 2 |
| AC | 1 |
| AD | 2 |
| AZ | 2 |
| BC | 1 |
| BD | 3 |
| BZ | 2 |
| CD | 2 |
| CZ | 0 |
| DZ | 1 |
The unnesting recipe for unpivoting the table comes from Stew's post, here.
To count occurrances of YYY among 3 columns you could use:
WITH stacked AS (
SELECT id
, unnest(array['A', 'B', 'C', 'D', 'Z']) AS col_name
, unnest(array[a, b, c, d, z]) AS col_value
FROM test t
)
SELECT combo, sum(cnt) AS count
FROM (
SELECT t1.id, t1.col_name || t2.col_name || t3.col_name AS combo
, (CASE WHEN t1.col_value = 'Y'
AND t2.col_value = 'Y'
AND t3.col_value = 'Y' THEN 1 ELSE 0 END) AS cnt
FROM stacked t1
INNER JOIN stacked t2
ON t1.id = t2.id
INNER JOIN stacked t3
ON t1.id = t3.id
AND t1.col_name < t2.col_name
And t2.col_name < t3.col_name
) t3
GROUP BY combo
ORDER BY combo
;
which yields
| combo | count |
|-------+-------|
| ABC | 0 |
| ABD | 1 |
| ABZ | 2 |
| ACD | 1 |
| ACZ | 0 |
| ADZ | 1 |
| BCD | 1 |
| BCZ | 0 |
| BDZ | 1 |
| CDZ | 0 |
Or, to handle combinations of N columns, you could use WITH RECURSIVE:
For example, for N = 3,
WITH RECURSIVE result AS (
WITH stacked AS (
SELECT id
, unnest(array['A', 'B', 'C', 'D', 'Z']) AS col_name
, unnest(array[a, b, c, d, z]) AS col_value
FROM test t)
SELECT id, array[col_name] AS path, array[col_value] AS path_val, col_name AS last_name
FROM stacked
UNION
SELECT r.id, path || s.col_name, path_val || s.col_value, s.col_name
FROM result r
INNER JOIN stacked s
ON r.id = s.id
AND s.col_name > r.last_name
WHERE array_length(r.path, 1) < 3) -- Change 3 to your value for N
SELECT combo, sum(cnt)
FROM (
SELECT id, array_to_string(path, '') AS combo, (CASE WHEN 'Y' = all(path_val) THEN 1 ELSE 0 END) AS cnt
FROM result
WHERE array_length(path, 1) = 3) t -- Change 3 to your value for N
GROUP BY combo
ORDER BY combo
Note that N = 3 is used in 2 places in the SQL above.

Answer 2

I would do this using a lateral join:
with vals as (
select v.*
from t cross join lateral
(values ('A', A), ('B', B), ('C', C), ('D', D), ('Z', Z)
) v(which, val)
)
select (v1.which || v2.which) as combo,
sum( (val = 'Y')::int ) as count
from vals v1 join
vals v2
on v1.which < v2.which
group by combo
order by combo;
I consider lateral joins to be a more direct way to unpivot the values. There is no need to convert the values to an array an unnest, much less unnest two arrays and align the values.