如何获取按评论数量分组的用户数量？

Question

我想按他们制作的 cmets 数分组的用户数。

[User]: ID
[Comment]: ID, UserID

所以如果user A has made 1 comment, user B has made 1 comment and user C has made 2 comments，那么输出将是：

0 comments => 0 users
1 comment  => 2 users (A+B)
2 comments => 1 user  (C)

你会如何查询这个？

Answer 1

这取决于您的特定数据库结构，但假设您有一个 users 表和一个 cmets 表：

users table:
id: serial
name: text

comments table:
id: serial
user_id: integer (foreign key to the users table)
comment: text

您可以计算每个用户使用此查询进行的 cmets 数：

  SELECT users.id, users.name, count(comments.id) as comment_cnt
    FROM users LEFT JOIN
         comments ON users.id = comments.user_id
GROUP BY users.id, users.name

然后您可以在嵌套查询中使用此查询的结果来计算每个 cmets 数量的用户数量：

  SELECT comment_cnt, count(*) FROM
  (SELECT users.id, users.name, count(comments.id) as comment_cnt
    FROM users LEFT JOIN
         comments ON users.id = comments.user_id
GROUP BY users.id, users.name) AS comment_cnts
GROUP BY comment_cnt;

我不知道有什么优雅的方法可以填补给定数量的 cmets 用户为零的空白，但是临时表和另一个级别的嵌套可以工作：

CREATE TABLE wholenumbers (num integer);

INSERT INTO wholenumbers VALUES (0), (1), (2), (3), (4), (5), (6);

   SELECT num as comment_cnt, COALESCE(user_cnt,0) as user_cnt
     FROM wholenumbers
LEFT JOIN (SELECT comment_cnt, count(*) AS user_cnt
             FROM (  SELECT users.id, users.name, count(comments.id) AS comment_cnt
                       FROM users LEFT JOIN comments ON users.id = comments.user_id
                   GROUP BY users.id, users.name) AS comment_cnts
         GROUP BY comment_cnt) AS user_cnts ON wholenumbers.num = user_cnts.comment_cnt
ORDER BY num;

Answer 2

基于表格布局@ClaytonC provided:

WITH cte AS (
   SELECT msg_ct, count(*) AS users
   FROM  (
      SELECT count(*) AS msg_ct
      FROM   comments 
      GROUP  BY user_id
      ) sub
   GROUP  BY 1
   )
SELECT msg_ct, COALESCE(users, 0) AS users
FROM   generate_series(0, (SELECT max(msg_ct) FROM cte)) msg_ct
LEFT   JOIN cte USING (msg_ct)
ORDER  BY 1;

要点

• 首先，计算每个用户的 cmets (msg_ct)。只要引用完整性由外键强制执行，您就完全不需要加入users 表来聚合每个用户的cmets。只需计算 comments 中的行数。
  接下来，按消息计数计算用户数 (users)。

• 我在 CTE 中执行此操作，因为我在最终查询中使用了两次派生表。
  首先让generate_series()动态生成从最小值到最大值的所有计数，包括间隙。
  然后for table到LEFT JOIN去，得到最终结果。

• 计数从 0 开始（在我更新之后）。如果你想让它从最小的实际msg_ct 开始，请考虑我在编辑历史中的答案的初稿。

• 解释基础知识的密切相关的答案：

  • Select all integers that are not already in table in postgres

统计没有 cmets 的用户

正如@ClaytonC 评论的那样，上述答案确实不包括没有 cmets 的用户。

要解决这个问题（如果你真的需要它），要么在开始时就 LEFT JOIN 到 users：

WITH cte AS (
   SELECT msg_ct, count(*) AS users
   FROM  (
      SELECT count(c.user_id) AS msg_ct
      FROM   users u
      LEFT   JOIN comments c ON c.user_id = u.id
      GROUP  BY u.id
      ) sub
   GROUP  BY 1
   )
SELECT ...

或者，由于加入只是为了寻找没有 cmets 的用户，我们可能会更便宜：统计 所有个用户并减去有cmets（无论如何我们都处理过）：

WITH cte AS (
   SELECT msg_ct, count(*)::int AS users
   FROM  (
      SELECT count(*)::int AS msg_ct
      FROM   comments 
      GROUP  BY user_id
      ) sub
   GROUP  BY 1
   )
, agg AS (
   SELECT max(msg_ct)   AS max_ct      -- maximum for generate_series
         ,((SELECT count(*) FROM users) - sum(users))::int AS users
                                       -- quiet rest with 0 comments
   FROM cte
   )
SELECT 0 AS msg_ct, users FROM agg     -- users with 0 comments
UNION  ALL
SELECT msg_ct, COALESCE(users, 0)
FROM  (SELECT generate_series(1, max_ct) AS msg_ct FROM agg) g
LEFT   JOIN cte USING (msg_ct)
ORDER  BY 1;

查询变得有点复杂，但对于大表可能更快。没有把握。使用EXPLAIN ANALYZE 进行测试（如果对结果发表评论，我将不胜感激。）