统计postgresql矩阵中的列组合
我在博士后中有一张表,如下所示 我想要一个postgres中的sql,它计算两个具有YY的列的组合 期望像这样的输出 组合计数统计postgresql矩阵中的列组合,sql,postgresql,matrix,combinations,Sql,Postgresql,Matrix,Combinations,我在博士后中有一张表,如下所示 我想要一个postgres中的sql,它计算两个具有YY的列的组合 期望像这样的输出 组合计数 AB 2 AC 1 AD 2 AZ 1 BC 1 BD 3 BZ 2 CD 2 CZ 0 DZ 1 有人能帮我吗 WITH stacked AS ( SELECT id , unnest(array['A', 'B', 'C', 'D', 'Z']) AS col_name , unnest(array[a, b, c, d,
AB 2
AC 1
AD 2
AZ 1
BC 1
BD 3
BZ 2
CD 2
CZ 0
DZ 1
有人能帮我吗
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
取消对桌子的注意的令人不安的方法来自
要在3列中计算YYY的发生率,可以使用:
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
;
或者,要处理N列的组合,您可以使用递归:
例如,对于N=3
请注意,在上面的SQL中,N=3用于两个位置。我将使用横向联接:
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方法来解开这些值。不需要将值转换为一个数组,更不用说两个数组,并对齐值。
谢谢你,Ubuntu。这正是我想要的。精彩的你知道如何把它组合成3列吗?像YYY。只是好奇。提前谢谢你!明亮的N的情况适用于所有情况。非常感谢。| combo | count |
|-------+-------|
| ABC | 0 |
| ABD | 1 |
| ABZ | 2 |
| ACD | 1 |
| ACZ | 0 |
| ADZ | 1 |
| BCD | 1 |
| BCZ | 0 |
| BDZ | 1 |
| CDZ | 0 |
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
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;