标准 SQL - 如何计算数组中值的频率

Question

我得到下表和下面的查询：

SELECT 
  fullVisitorId,
  COUNT(fullVisitorId) as id_count,
  ARRAY_AGG(trafficSource.medium) AS trafic_medium
FROM 
  `bigquery-public-data.google_analytics_sample.ga_sessions_20170101`
GROUP BY
  fullVisitorId
ORDER BY
  id_count DESC

对于列中的每个值trafic_medium（例如：cpc、推荐、有机等），我试图计算出每个值在数组中出现的频率，因此最好添加一个新列“计数”来显示该值的出现频率发生了？

+-----------+---------+------+
| array_agg | medium  | count|
+-----------+---------+------+
| 123       | cpc     |   2  |
+-----------+---------+------+
|           | organic |   1  |
+-----------+---------+------+
|           | cpc     |   2  |
+-----------+---------+------+
| 456       | organic |   2  |
+-----------+---------+------+
|           | organic |   2  |
+-----------+---------+------+
|           | cpc     |   1  |
+-----------+---------+------+

我是 SQL 新手，所以我很困惑。

到目前为止我尝试过：

WITH medium AS
(
    SELECT 
        fullVisitorId,
        COUNT(fullVisitorId) as id_count,
        ARRAY_AGG(trafficSource.medium) AS trafic_medium
    FROM 
        `bigquery-public-data.google_analytics_sample.ga_sessions_20170101`
    GROUP BY
        fullVisitorId
    ORDER BY
        id_count DESC
) 
SELECT
    fullVisitorId,
    trafic_medium,
    (SELECT AS STRUCT Any_Value(trafic_medium) AS name, COUNT(*) AS count
FROM 
    UNNEST(trafic_medium) AS trafic_medium) AS trafic_medium_2,
FROM 
    medium

基于此线程： How to count of elements in a bigquery array field

然而，这仅显示“Any_Value”的数量，而不是所有不同的。

我将不胜感激一些帮助！

ps 我正在 BigQuery 中的“bigquery-public-dataset.google_analytics_sample”上执行此操作

Answer 1

以下是 BigQuery 标准 SQL 的内容，可帮助您入门

#standardSQL
SELECT id, trafic_medium,
  ARRAY(
    SELECT AS STRUCT medium, COUNT(1) `count`
    FROM t.trafic_medium medium
    GROUP BY medium
  ) stats
FROM `project.dataset.table` t

是否适用于您问题中的样本/虚拟数据，如下例所示

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 123 id, ['cpc', 'organic', 'cpc'] trafic_medium UNION ALL
  SELECT 456, ['organic', 'organic', 'cpc']
)
SELECT id, trafic_medium,
  ARRAY(
    SELECT AS STRUCT medium, COUNT(1) `count`
    FROM t.trafic_medium medium
    GROUP BY medium
  ) stats
FROM `project.dataset.table` t
-- ORDER BY id   

结果将是

作为一个选项 - 您可以使用以下版本

#standardSQL
SELECT id, 
  ARRAY(
    SELECT AS STRUCT medium, `count`
    FROM t.trafic_medium medium
    LEFT JOIN (
      SELECT AS STRUCT medium, COUNT(1) `count`
      FROM t.trafic_medium medium
      GROUP BY medium
    ) stats
    USING(medium) 
  ) trafic_medium  
FROM `project.dataset.table` t
-- ORDER BY id   

（如果适用于相同的虚拟数据）将输出如下

该版本看起来更符合您的预期结果