BigQuery SQL，获取中位数，按日期分组？

Question

当试图通过 window 函数使用分区获取中位数时，我收到一条错误消息“SELECT list expression references column seller_stock既不分组也不聚合”，这是为什么，我必须如何以不同的方式写这个SQL？我每天有很多记录，我想 return 每天的中位数 ...

SELECT date(snapshot_date) AS period, 
  PERCENTILE_DISC(**seller_stock**, 0.5) OVER (PARTITION BY snapshot_date) AS median_stock
FROM `table.name`  
WHERE snapshot_date >= "2022-04-01" 
GROUP BY snapshot_date

Answer 1

问题是您不能按 AGG 函数分组，因为您已经通过行获得中位数，您将只需要该语句的第一行。

您可以使用中级 table 或辅助

这是一个例子：

with median_data as (
  select 
    date(snapshot_date) AS period,
    PERCENTILE_DISC(seller_stock, 0.5) OVER (PARTITION BY snapshot_date) AS median_stock,
    row_number() over(order by snapshot_date) as r
  from `table.name`
  where snapshot_date >= "2022-04-01" 
)

select period,median_stock from median_data where r =  1

BigQuery SQL，获取中位数，按日期分组？

BigQuery SQL, Obtain Median, Grouped by Date?

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