使用 postgresql 连接查询计算百分比

Question

我试图通过连接 3 个表数据来计算百分比，以获得每个用户推文的 positive_count、negative_count、neutral_count 的百分比。我已经成功地获得了正面、负面和中性的计数，但未能获得相同的百分比而不是计数。这是获取计数的查询：

SELECT 
   t1.u_id,count() as total_tweets_count , 
   (
     SELECT count() from t1,t2,t3 c 
      WHERE 
            t1.u_id='18839785' AND 
            t1.u_id=t2.u_id AND 
            t2.ts_id=t3.ts_id AND 
            t3.sentiment='Positive'
   ) as pos_count , 
   (
    SELECT count() from t1,t2,t3
     WHERE
           t1.u_id='18839785' AND 
           t1.u_id=t2.u_id AND 
           t2.ts_id=t3.ts_id AND
           t3.sentiment='Negative'
   ) as neg_count , 
   (
    SELECT count() from t1,t2,t3
     WHERE
           t1.u_id='18839785' AND
           t1.u_id=t2.u_id AND
           t2.ts_id=t3.ts_id AND
           t3.sentiment='Neutral'
   ) as neu_count
 FROM t1,t2,t3 
WHERE
      t1.u_id='18839785' AND
      t1.u_id=t2.u_id AND
      t2.ts_id=t3.ts_id

分组依据t1.u_id；

**OUTPUT:**  

u_id | total_tweets_count | pos_count | neg_count | neu_count 
-----------------+--------------------+-----------+-----------+------- 
18839785|         88 |        38 |        25 |        25
(1 row)

现在我想要相同的百分比而不是计数。我用下面的方式写了查询但是失败了。

SELECT 
       total_tweets_count,pos_count,
       round((pos_count * 100.0) / total_tweets_count, 2) AS pos_per,neg_count,
       round((neg_count * 100.0) / total_tweets_count, 2) AS neg_per,
       neu_count, round((neu_count * 100.0) / total_tweets_count, 2) AS neu_per 
 FROM (
       SELECT
             count(*) as total_tweets_count,
             count(
                   a.u_id='18839785' AND
                   a.u_id=b.u_id AND
                   b.ts_id=c.ts_id AND
                   c.sentiment='Positive'
                  ) AS pos_count,
             count(
                   a.u_id='18839785' AND
                   a.u_id=b.u_id AND
                   b.ts_id=c.ts_id AND 
                   c.sentiment='Negative'
                  ) AS neg_count,
             count(
                   a.u_id='18839785' AND
                   a.u_id=b.u_id AND
                   b.ts_id=c.ts_id AND
                   c.sentiment='Neutral') AS neu_count
         FROM t1,t2, t3
        WHERE 
              a.u_id='18839785' AND 
              a.u_id=b.u_id AND
              b.ts_id=c.ts_id 
     GROUP BY a.u_id
    ) sub;

谁能帮我实现如下每个用户数据的百分比？

u_id | total_tweets_count | pos_count | neg_count | neu_count 
------------------+--------------------+-----------+-----------+-----
18839785|      88 |        43.18 |        28.4 |        28.4
(1 row)

Answer 1

我不太确定你在找什么。

对于初学者，您可以使用条件聚合而不是三个标量子查询来简化查询（顺便说一下，不需要在 a.u_id 上重复 where 条件）

您声明要 "count for all users"，因此您需要删除主查询中的 WHERE 子句。简化也摆脱了重复的 WHERE 条件。

select u_id, 
       total_tweets_count, 
       pos_count,
       round((pos_count * 100.0) / total_tweets_count, 2) AS pos_per,
       neg_count,
       round((neg_count * 100.0) / total_tweets_count, 2) AS neg_per,
       neu_cont,
       round((neu_count * 100.0) / total_tweets_count, 2) AS neu_per
from (
  SELECT 
     t1.u_id,
     count(*) as total_tweets_count, 
     count(case when t3.sentiment='Positive' then 1 end) as pos_count, 
     count(case when t3.sentiment='Negative' then 1 end) as neg_count, 
     count(case when t3.sentiment='Neutral' then 1 end) as neu_count 
  FROM t1
    JOIN t2 ON t1.u_id=t2.u_id 
    JOIN t3 t2.ts_id=t3.ts_id
  -- no WHERE condition on the u_id here
  GROUP BY t1.u_id
) t

请注意，我将 WHERE 子句中过时、古老且脆弱的隐式连接替换为 "modern" 显式 JOIN 运算符

使用更新的 Postgres 版本，表达式 count(case when t3.sentiment='Positive' then 1 end) as pos_count 也可以重写为：

count(*) filter (where t3.sentiment='Positive') as pos_count

这更具可读性（我认为也更容易理解）。

在您的查询中，您可以通过使用共同相关的子查询来实现对 u_id 的全局 WHERE 条件的重复，例如：

(
  SELECT count(*) 
  FROM t1 inner_t1 --<< use different aliases than in the outer query
    JOIN t2 inner_t2 ON inner_t2.u_id = inner_t1.u_id
    JOIN t3 inner_t3 ON inner_t3.ts_id = inner_t2.ts_id
  -- referencing the outer t1 removes the need to repeat the hardcoded ID
  WHERE innter_t1.u_id = t1.u_id 
) as pos_count

tablet1的重复也没有必要，所以上面可以改写为：

(
  SELECT count(*) 
  FROM t2 inner_t2 
    JOIN t3 inner_t3 ON inner_t3.ts_id = inner_t2.ts_id
  WHERE inner_t2.u_id = t1.u_id --<< this references the outer t1 table
) as pos_count

但是使用条件聚合的版本仍然比使用三个标量子查询快很多（即使你删除了 t1 table).

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