如何按 id 对数据进行分组并使用 SQL 获取中值？答案

【问题标题】：How can I group data by id and get the median value using SQL?如何按 id 对数据进行分组并使用 SQL 获取中值？
【发布时间】：2021-09-15 13:41:26
【问题描述】：

我有一张表格，上面有给定商店在几天内开业的时间，如下所示（OPENING_HOUR 设置为 24 小时时间格式，因此表格上的所有时间都是上午）。

>>> BUSINESS_HOURS
    DATE       | STORE_ID | OPENING_HOUR
________________________________________
0   2021-06-01 |   222    |  11
1   2021-06-02 |   222    |  11
2   2021-06-03 |   222    |  11
3   2021-06-04 |   222    |  11
4   2021-06-05 |   222    |  11
5   2021-06-06 |   222    |  11
6   2021-06-07 |   222    |  12
7   2021-06-08 |   222    |  11
8   2021-06-09 |   222    |  11
9   2021-06-10 |   222    |  12

现在我需要按 id 对数据进行分组，并确定哪个 opening_hour 最频繁。在下面的案例中，80% 的案例都在上午 11 点，所以我需要这样的东西：

>>> DATA_GROUPED
    STORE_ID   | OPENING_HOUR | FREQUENCY
________________________________________
0   222        |   11         |  0.8

这可能只使用 SQL 吗？谢谢你们的帮助，伙计们！

【问题讨论】：

标签： sql database google-bigquery databricks

【解决方案1】：

你可以使用窗口函数：

select store_id, opening_hour, count(*) as cnt,
       count(*) * 1.0 / sum(count(*)) over () as ratio
from t
where store_id = 1
group by store_id, opening_hour
order by cnt desc
limit 1;

如果您希望所有商店都使用此功能，可以使用窗口函数：

select t.* except (seqnum)
from (select store_id, opening_hour, count(*) as cnt,
            count(*) * 1.0 / sum(count(*)) over () as ratio,
            row_number() over (partition by store_id order by count(*) desc) as seqnum
     from t
     group by store_id, opening_hour
    ) t
where seqnum = 1;

【讨论】：

【解决方案2】：

我找到了一种使用窗口函数和 CTE 的方法。

WITH Q1 AS (
  SELECT
    DISTINCT STORE_ID,
    OPENING_HOUR,
    COUNT(OPENING_HOUR) AMOUNT,
    ROW_NUMBER() OVER(PARTITION BY STORE_IDORDER BY COUNT(OPENING_HOUR) DESC) as RANK
  FROM T1
  GROUP BY 1, 2
)

SELECT
  STORE_ID,
  OPENING_HOUR,
  ROUND((AMOUNT/SUM(AMOUNT) OVER(PARTITION BY STORE_ID)),2) AS SHARE
FROM Q1-- WHERE RANK = 1

不是最短的答案，但效果很好！

【讨论】：

【解决方案3】：

带开窗功能，这一种解决方案：

WITH business_hours as (
SELECT DATE("2021-06-01") as date, 222 as store_id, 11 as opening_hour
UNION ALL
SELECT "2021-06-02", 222, 11
UNION ALL
SELECT "2021-06-03", 222, 11
UNION ALL
SELECT "2021-06-04", 222, 11
UNION ALL
SELECT "2021-06-05", 222, 11
UNION ALL
SELECT "2021-06-06", 222, 11
UNION ALL
SELECT "2021-06-07", 222, 12
UNION ALL
SELECT "2021-06-08", 222, 11
UNION ALL
SELECT "2021-06-09", 222, 11
UNION ALL
SELECT "2021-06-10", 222, 12)

, agg as (SELECT DISTINCT store_id, opening_hour,
COUNT(store_id) OVER (partition by opening_hour, EXTRACT(MONTH FROM date)) as total_open_per_hour,
COUNT(store_id) OVER (partition by EXTRACT(MONTH FROM date)) as total_open,
from business_hours)

SELECT store_id, opening_hour, safe_divide(total_open_per_hour, total_open) frequency FROM agg

结果：

【讨论】：

【解决方案4】：

考虑以下方法

select * from (
  select distinct store_id, opening_hour, 
    count(1) over(partition by opening_hour) / count(1) over() frequency
  from business_hours
)
where true 
qualify row_number() over(partition by store_id order by frequency desc) = 1

为您提供每家商店最频繁的营业时间

如果应用于您问题中的样本数据 - 输出是

【讨论】：