在 BigQuery 上重新创建 GA 漏斗

Posted 2023-03-24

【中文标题】在 BigQuery 上重新创建 GA 漏斗

【英文标题】：Recreate GA Funnel on BigQuery

【发布时间】：2017-07-26 16:42:26

【问题描述】：

我正在尝试使用 BigQuery 重新创建 GA 漏斗（Google360 上的自定义报告）。 GA 上的漏斗使用每个页面上发生的事件的唯一计数。我在网上发现这个查询大部分都有效：

SELECT
  COUNT( s0.firstHit) AS Landing_Page,
  COUNT( s1.firstHit) AS Model_Selection
from(
SELECT
      s0.fullvisitorID,
      s0.firstHit,
      s1.firstHit,
    FROM (
            # Begin Subquery #1 aka s0
            SELECT
                    fullvisitorID,
                    MIN(hits.hitNumber) AS firstHit
            FROm [64269470.ga_sessions_20170720]
            WHERE
                  hits.eventInfo.eventAction  in ('landing_page') 
                    AND totals.visits = 1
            GROUP BY
                  fullvisitorID
                  ) s0
    # End Subquery #1 aka s0

    left join (

    # Begin Subquery #2 aka s1
          SELECT
              fullvisitorID,
              MIN(hits.hitNumber) AS firstHit
          FROM [64269470.ga_sessions_20170720]
          WHERE
            hits.eventInfo.eventAction  in ('model_selection_page')
            AND totals.visits = 1
          GROUP BY
                fullvisitorID,
                ) s1

      ON
    s0.fullvisitorID = s1.fullvisitorID

    )

查询工作正常，着陆页的值与我在 GA 上获得的值相同，但 Model_Selection 大约高出 10%。这种差异也会随着漏斗而增加（为了清楚起见，我只发布了 2 个步骤）。 知道我在这里想念什么吗？

【参考方案1】：

此查询可以满足您的需要，但在 Standard SQL 版本中：

#standardSQL
SELECT 
  SUM((SELECT COUNTIF(eventInfo.eventAction = 'landing_page') FROM UNNEST(hits))) Landing_Page,
  SUM((SELECT COUNTIF(eventInfo.eventAction = 'model_selection_page') FROM UNNEST(hits) WHERE EXISTS(SELECT 1 FROM UNNEST(hits) WHERE eventInfo.eventAction = 'landing_page'))) Model_Selection  
FROM `64269470.ga_sessions_20170720`

就是这样。 4 条线路，更快更便宜。

您还可以使用模拟数据，例如：

#standardSQL
WITH data AS(
  SELECT '1' AS fullvisitorid, ARRAY<STRUCT<eventInfo STRUCT<eventAction STRING > >> [STRUCT(STRUCT('landing_page' AS eventAction) AS eventInfo)] AS hits UNION ALL
  SELECT '1' AS fullvisitorid, ARRAY<STRUCT<eventInfo STRUCT<eventAction STRING > >> [STRUCT(STRUCT('landing_page' AS eventAction) AS eventInfo), STRUCT(STRUCT('landing_page' AS eventAction) AS eventInfo)] AS hits UNION ALL
  SELECT '1' AS fullvisitorid, ARRAY<STRUCT<eventInfo STRUCT<eventAction STRING > >> [STRUCT(STRUCT('landing_page' AS eventAction) AS eventInfo), STRUCT(STRUCT('model_selection_page' AS eventAction) AS eventInfo)] AS hits UNION ALL
  SELECT '1' AS fullvisitorid, ARRAY<STRUCT<eventInfo STRUCT<eventAction STRING > >> [STRUCT(STRUCT('model_selection_page' AS eventAction) AS eventInfo), STRUCT(STRUCT('model_selection_page' AS eventAction) AS eventInfo)] AS hits
)

SELECT 
  SUM((SELECT COUNTIF(eventInfo.eventAction = 'landing_page') FROM UNNEST(hits))) Landing_Page,
  SUM((SELECT COUNTIF(eventInfo.eventAction = 'model_selection_page') FROM UNNEST(hits) WHERE EXISTS(SELECT 1 FROM UNNEST(hits) WHERE eventInfo.eventAction = 'landing_page'))) Model_Selection  
FROM data

请注意，在 GA 中构建这种类型的报告可能会有点困难，因为您需要选择至少触发过事件“landing_page”然后触发事件“model_selection_page”的访问者。确保您在 GA 中也正确构建了此报告（一种方法可能是首先构建一个自定义报告，其中仅包含已触发“landing_page”的客户，然后应用第二个过滤器查找“model_selection_page”）。

[编辑]：

您在评论中询问有关在会话和用户级别上进行计数的问题。为了计算每个会话，您可以将每个子查询评估的结果限制为 1，如下所示：

SELECT 
  SUM((SELECT 1 FROM UNNEST(hits) WHERE eventInfo.eventAction = 'landing_page' LIMIT 1)) Landing_Page,
  SUM((SELECT 1 FROM UNNEST(hits) WHERE EXISTS(SELECT 1 FROM UNNEST(hits) WHERE eventInfo.eventAction = 'landing_page') AND eventInfo.eventAction = 'model_selection_page' LIMIT 1)) Model_Selection  
FROM data

对于不同的用户计数，想法是相同的，但您必须应用 COUNT(DISTINCT) 操作，如下所示：

SELECT 
  COUNT(DISTINCT(SELECT fullvisitorid FROM UNNEST(hits) WHERE eventInfo.eventAction = 'landing_page' LIMIT 1)) Landing_Page,
  COUNT(DISTINCT(SELECT fullvisitorid FROM UNNEST(hits) WHERE EXISTS(SELECT 1 FROM UNNEST(hits) WHERE eventInfo.eventAction = 'landing_page') AND eventInfo.eventAction = 'model_selection_page' LIMIT 1)) Model_Selection  
FROM data

【讨论】：

您好，威廉，谢谢您的回答。这是您一直在使用的一种有趣的方法。快速的问题。我希望您使用这种结构来区分用户和会话。看起来，在这种情况下，被计数的是总事件。谢谢！

@Jacob 感谢引用此问题的另一个问题我找到了您的评论，很抱歉花了这么长时间才回复。我编辑了我的答案，希望这就是你要找的。让我知道它是否有效:)