Sql BigQuery中Top N查询的推广

这是一个后续问题，用于概括案例。现在让我们获取以下数据：

 year   genre         studio            title       revenue
2014    fantasy       fox               avatar      10
2015    fantasy       fox               avatar      12
2016    fantasy       fox               avatar      12
2015    action        sony              spider-man  10
2015    romance       paramount         love letter 15
2015    action        sony              spider-man  10
2015    action        sony              spider-man  10
2015    action        disney            toy story   10
2015    action        sony              edgar       4
2015    action        sony              thomas      1
2015    fantasy       fox               avatar      2

我希望得到以下结果来构建树结构：

Past 2 years, Top 2 genres (Alphabetically), Top 2 studios (by Count), Top 2 titles by SUM Revenue DESC

因此，我们会得到如下结果：

从概念上讲，我希望实现的查询是这样的：

SELECT year, genre, studio, title, SUM(revenue)
FROM titles
GROUP BY year, genre, studio, title

// in pseudocode
ORDER BY
    (year DESC) LIMIT 2,
    (genre ASC) LIMIT 10,
    (COUNT(studio) DESC) LIMIT 2,
    (SUM(revenue) DESC) LIMIT 2

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做上述工作的最佳方法是什么，这更像是在BQ中构建树结构的一般化。

我无法在您的数据集中找到“avatar2”，但结果中有它。因此，我无法验证边缘的答案。这是我提出的SQL Server查询。我希望不会有太多的变化需要

 WITH A as 
    (SELECT 
    year, 
    genre, 
    studio,
    COUNT(*) OVER (PARTITION BY year, genre, studio) AS studio_movie_count,
    title,
    revenue,
    SUM(revenue) OVER (PARTITION BY year, genre, studio,title) AS revenue_sum FROM movies),

    B as

    (SELECT 
    year,
    DENSE_RANK() OVER (ORDER BY year DESC) AS year_num, 
    genre,
    DENSE_RANK() OVER (PARTITION BY year ORDER BY genre ASC) AS genre_num,
    studio,
    DENSE_RANK() OVER (PARTITION BY year, genre ORDER BY studio_movie_count DESC) AS studio_num,
    title,
    DENSE_RANK() OVER (PARTITION BY year, genre, studio ORDER BY revenue_sum DESC) AS title_num,
    revenue

    FROM A)

    SELECT year, genre, studio, title, revenue
    FROM B
    WHERE year_num < 3 AND
    genre_num < 3 AND
    studio_num < 3 AND
    title_num < 3;

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在子查询中筛选前2年的行，同时按工作室查找电影计数，按标题查找收入总和

然后按流派、工作室、收入和过滤器排名前2名

select year, genre, studio, title, revenue 
from (
    select year, genre, studio, title, revenue,
        dense_rank() over (partition by year order by genre) as genre_rank,
        dense_rank() over (partition by year, genre order by count_by_studio desc) as studio_rank,
        dense_rank() over (partition by year, genre, studio order by revenue_by_title desc) as title_rank
    from (
        select year,
            genre,
            studio,
            title,
            revenue,
            dense_rank() over (order by year desc) as year_rank,
            count(*) over (partition by year, genre, studio) as count_by_studio,
            sum(revenue) over (partition by year, genre, studio, title) as revenue_by_title
        from titles
    ) where year_rank <= 2
) where genre_rank <= 2
and studio_rank <= 2
and title_rank <= 2;

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你的第一个问题已经没有什么答案了！你试过自己概括一下吗？你应该表现出一些努力，否则看起来你只是在外包你的工作tasks@MikhailBerlyant我在这里加了一笔赏金。我一直在使用array_agg使用您的方法来实现这一点，但我认为另一种方法可能更通用于处理各种类型/级别。