Postgresql 用更有效的方法来计算单独分组的列，替换几十个联接

有一个表（在PostgreSQL中）有多个列（从n1到n10），每个列的每一行都包含单独的数字（为了简单起见，在下面的示例中，数字是1、2和3）。表中有数千行。表中数据摘录：

+----+----+----+------+-----+
| n1 | n2 | n3 | n... | n10 |
+----+----+----+------+-----+
|  3 |  2 |  1 |    1 |   1 |
|  2 |  1 |  2 |    2 |   3 |
|  1 |  1 |  2 |    3 |   1 |
|  2 |  3 |  1 |    1 |   2 |
|  3 |  2 |  1 |    1 |   2 |
|  1 |  3 |  1 |    3 |   3 |
|  2 |  3 |  1 |    3 |   3 |
|  1 |  1 |  3 |    3 |   1 |
|  3 |  2 |  3 |    1 |   2 |
|  2 |  1 |  2 |    2 |   1 |
+----+----+----+------+-----+

我要做的是计算每一列中的各个数字，这样结果表如下所示：

+--------+----------+----------+----------+------------+-----------+
| number | n1_count | n2_count | n3_count | n..._count | n10_count |
+--------+----------+----------+----------+------------+-----------+
|      1 |        3 |        4 |        5 |          4 |         4 |
|      2 |        4 |        3 |        3 |          2 |         3 |
|      3 |        3 |        3 |        2 |          4 |         3 |
+--------+----------+----------+----------+------------+-----------+

我通过使用多个左连接成功地实现了这一点：

SELECT number, n1_count, n2_count, n3_count, n..._count, n10_count FROM
  (VALUES ('1'), ('2'), ('3') AS t (number)
LEFT JOIN
  (SELECT n1, COUNT(n1) AS n1_count FROM table GROUP BY n1) AS n1_ ON n1 = number
LEFT JOIN
  (SELECT n2, COUNT(d2) AS n2_count FROM table GROUP BY n2) AS n2_ ON n2 = number
LEFT JOIN
  (SELECT n3, COUNT(d3) AS n3_count FROM table GROUP BY n3) AS n3_ ON n3 = number
LEFT JOIN
  (SELECT n..., COUNT(d4) AS n..._count FROM table GROUP BY n...) AS n..._ ON n... = number
LEFT JOIN
  (SELECT n10, COUNT(d5) AS n10_count FROM table GROUP BY n10) AS n10_ ON n10 = number;

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但是最终的查询（10个左连接）看起来非常庞大和复杂，所以我想知道是否可以通过更优雅和高效的方式实现相同的结果？请给我指出我的选项。

您可以使用过滤聚合和使用数组的单个连接：

select t.number, 
       count(*) filter (where tt.n1 = t.number) as n1_count,
       count(*) filter (where tt.n2 = t.number) as n2_count,
       count(*) filter (where tt.n3 = t.number) as n3_count,
       count(*) filter (where tt.n4 = t.number) as n4_count
from the_table tt
  join (values (1),(2),(3) ) as t(number) on t.number = any(array[tt.n1,tt.n2,tt.n3,tt.n4])
group by t.number;  

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谢谢你的解决方案，它正在工作，我不知道这种方式。但不幸的是，查询需要42毫秒才能完成，而我的初始查询需要14毫秒。尽管您的解决方案看起来更简洁，但它并没有那么有效。。。