当查询具有GROUP BY时，如何获得总计的百分比? [英] How to get a percentage of total when the query has a GROUP BY?

问题描述

说我有一个非标准化的表，其中包含电影演员姓名及其所去过的电影.例如

Say I have a non-normalized table with movie actor names and the movies they've been in. eg.

CREATE TABLE movies_actors (
  movies_actors_id INT,
  movie VARCHAR(255),
  actor VARCHAR(255),
  PRIMARY KEY (movies_actors_id)
);

我执行SELECT actor, COUNT(1) FROM movies_actors GROUP BY actor来查找演员参加过多少部电影.但是我也想知道演员参加过多少部电影.

I do a SELECT actor, COUNT(1) FROM movies_actors GROUP BY actor to find out how many movies the actor has been in. But I also want to find out what percentage of movies that actor has been in.

我想我可以做到:

SELECT
  actor,
  COUNT(1) AS total,
  COUNT(1) / (SELECT COUNT(1) FROM movies_actors) * 100 AS avg
FROM movies_actors
GROUP BY actor;

但是这似乎... idk ...令人讨厌.

But that just seems... idk... yucky.

有什么想法吗?

推荐答案

对于大型集合，JOIN的性能可能优于子查询.

For large sets, a JOIN may perform better than the subquery.

SELECT ma.actor
     , COUNT(1) AS total
     , COUNT(1) / t.cnt * 100 AS `percentage`
  FROM movies_actors ma
 CROSS
  JOIN (SELECT COUNT(1) AS cnt FROM movies_actors) t
 GROUP
    BY ma.actor
     , t.cnt  

对于大型集合，当返回很大比例的行时，JOIN操作通常可以胜过子查询.在您的情况下，它不是相关的子查询，因此MySQL不必多次执行该子查询，因此它可能没有任何区别.

For large sets, and when a large percentage of the rows are being returned, the JOIN operation can usually outperform a subquery. In your case, it's not a correlated subquery, so MySQL shouldn't have to execute that multiple times, so it may not make any difference.

注意COUNT(1)的非粉丝...我们可以用COUNT(*)或IFNULL(SUM(1),0)替换所有出现的COUNT(1)，以实现相同的结果.

Note to non-fans of COUNT(1)... we could replace any and all occurrences of COUNT(1) with COUNT(*) or IFNULL(SUM(1),0) to achieve equivalent result.

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