BigQuery 中带有 GROUPBY 的百分位数函数

Question

在我的 CENSUS 表中，我想按州分组，并为每个州获取县人口中位数和县数。

在 psql、redshift 和雪花中，我可以这样做：

psql=> SELECT state, count(county), PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY "population2000") AS median FROM CENSUS GROUP BY state;
        state         | count |  median
----------------------+-------+----------
 Alabama              |    67 |    36583
 Alaska               |    24 |   7296.5
 Arizona              |    15 |   116320
 Arkansas             |    75 |    20229
...

我正在尝试在标准 BigQuery 中找到一种很好的方法来执行此操作。我注意到有未记录的percentile_cont 分析功能可用，但我必须做一些重大的黑客攻击才能让它做我想做的事。

我希望能够用我收集到的正确论点来做同样的事情：

SELECT
  state,
  COUNT(county),
  PERCENTILE_CONT(population2000,
    0.5) OVER () AS `medPop`
FROM
  CENSUS
GROUP BY
  state;

但是这个查询会产生错误

SELECT list expression references column population2000 which is neither grouped nor aggregated at

我可以得到我想要的答案，但如果这是推荐的方法来做我想做的事，我会非常失望：

SELECT
  MAX(nCounties) AS nCounties,
  state,
  MAX(medPop) AS medPop
FROM (
  SELECT
    nCounties,
    T1.state,
    (PERCENTILE_CONT(population2000,
        0.5) OVER (PARTITION BY T1.state)) AS `medPop`
  FROM
    census T1
  LEFT OUTER JOIN (
    SELECT
      COUNT(county) AS `nCounties`,
      state
    FROM
      census
    GROUP BY
      state) T2
  ON
    T1.state = T2.state) T3
GROUP BY
  state

有没有更好的方法来做我想做的事？另外，PERCENTILE_CONT 函数是否会被记录下来？

感谢阅读！

Answer 1

感谢您的关注。 PERCENTILE_CONT 正在开发中，一旦它是 GA，我们将发布文档。我们将首先支持它作为分析函数，我们计划稍后支持它作为聚合函数（允许 GROUP BY）。在这两个版本之间，一个更简单的解决方法是

SELECT
  state,
  ANY_VALUE(nCounties) AS nCounties,
  ANY_VALUE(medPop) AS medPop
FROM (
  SELECT
    state,
    COUNT(county) OVER (PARTITION BY state) AS nCounties,
    PERCENTILE_CONT(population2000,
      0.5) OVER (PARTITION BY state) AS medPop
  FROM
    CENSUS)
GROUP BY
  state