count combination of columns in postgresql matrix

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.