计算postgresql矩阵中的列组合答案

【问题标题】：count combination of columns in postgresql matrix计算postgresql矩阵中的列组合
【发布时间】：2019-07-13 21:01:04
【问题描述】：

我在 postgres 中有一张表，如下所示

我想要一个 postgres 中的 sql，它计算 2 个具有 YY 的列的组合

期待像

这样的输出

组合计数

AB 2
AC 1
AD 2
AZ 1
BC 1
BD 3
BZ 2
CD 2
CZ 0
DZ 1

谁能帮帮我？

【问题讨论】：

标签： sql postgresql matrix combinations

【解决方案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

产量

| combo | count |
|-------+-------|
| AB    |     2 |
| AC    |     1 |
| AD    |     2 |
| AZ    |     2 |
| BC    |     1 |
| BD    |     3 |
| BZ    |     2 |
| CD    |     2 |
| CZ    |     0 |
| DZ    |     1 |

取消透视表的unnesting 方法来自Stew's post, here。

要在您可以使用的 3 列中计算 YYY 的出现次数：

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
;

产生

| combo | count |
|-------+-------|
| ABC   |     0 |
| ABD   |     1 |
| ABZ   |     2 |
| ACD   |     1 |
| ACZ   |     0 |
| ADZ   |     1 |
| BCD   |     1 |
| BCZ   |     0 |
| BDZ   |     1 |
| CDZ   |     0 |

或者，要处理 N 列的组合，您可以使用 WITH RECURSIVE：例如，对于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

请注意，N = 3 在上述 SQL 中的两个位置使用。

【讨论】：

谢谢你，Ubuntu。这正是我想要的。精彩的。知道如何组合 3 列吗？像YYY。只是好奇。提前谢谢！
太棒了！！ N 的情况适用于所有情况。谢谢！

【解决方案2】：

我会使用横向连接来做到这一点：

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;

我认为横向连接是一种更直接的取消透视值的方法。没有必要将值转换为一个未嵌套的数组，更不用说将两个数组取消嵌套并对齐值。

【讨论】：