在 BigQuery 中创建 Google Analytics“回访用户”指标时出现问题

Posted 2023-03-24

【中文标题】在 BigQuery 中创建 Google Analytics“回访用户”指标时出现问题

【英文标题】：Issue creating a Google Analytics "Returning Users" metric in BigQuery

【发布时间】：2018-02-14 14:25:06

【问题描述】：

采取https://webmasters.stackexchange.com/a/87523上描述的内容

除了我自己的理解之外，我还提出了我认为会被视为“回访用户”的内容

1.首先查询显示在两年时间段内第一次“最近访问”的用户：

SELECT
  parsedDate,
  CASE
  # return fullVisitorId when the first latest visit is between 2 years and today
    WHEN parsedDate BETWEEN DATE_SUB(CURRENT_DATE(), INTERVAL 2 YEAR) AND CURRENT_DATE() THEN fullVisitorId
  END fullVisitorId
FROM (
  SELECT
    # convert the date field from string to date and get the latest date
    PARSE_DATE('%Y%m%d',
      MAX(date)) parsedDate,
    fullVisitorId
  FROM
    `project.dataset.ga_sessions_*`
  WHERE
    # only show fullVisitorId if first visit
    totals.newVisits = 1
  GROUP BY
    fullVisitorId)

2.然后单独查询选择特定日期范围内的一些字段：

SELECT
  PARSE_DATE('%Y%m%d',
    date) parsedDate,
  fullVisitorId,
  visitId,
  totals.newVisits,
  totals.visits,
  totals.bounces,
  device.deviceCategory
FROM
  `project.dataset.ga_sessions_*`
WHERE
  _TABLE_SUFFIX = "20180118"

3.将这两个查询连接在一起找到“Returning Users”

SELECT
q1.parsedDate date,
COUNT(DISTINCT q1.fullVisitorId) users,
# Default way to determine New Users
SUM(q1.newVisits) newVisits,
# Number of "New Users" based on my queries (matches with default way above)
COUNT(DISTINCT IF(q2.parsedDate < q1.parsedDate, NULL, q2.fullVisitorId)) newUsers,
# Number of "Returning Users" based on my queries
COUNT(DISTINCT IF(q2.parsedDate < q1.parsedDate, q2.fullVisitorId, NULL)) returningUsers
FROM (
(SELECT
  PARSE_DATE('%Y%m%d',
    date) parsedDate,
  fullVisitorId,
  visitId,
  totals.newVisits,
  totals.visits,
  totals.bounces,
  device.deviceCategory
FROM
  `project.dataset.ga_sessions_*`
WHERE
  _TABLE_SUFFIX = "20180118") q1
LEFT JOIN (
SELECT
  parsedDate,
  CASE
  # return fullVisitorId when the first latest visit is between 2 years and today
    WHEN parsedDate BETWEEN DATE_SUB(CURRENT_DATE(), INTERVAL 2 YEAR) AND CURRENT_DATE() THEN fullVisitorId
  END fullVisitorId
FROM (
  SELECT
    # convert the date field from string to date and get the latest date
    PARSE_DATE('%Y%m%d',
      MAX(date)) parsedDate,
    fullVisitorId
  FROM
    `project.dataset.ga_sessions_*`
  WHERE
    # only show fullVisitorId if first visit
    totals.newVisits = 1
  GROUP BY
    fullVisitorId)) q2
ON q1.fullVisitorId = q2.fullVisitorId)
GROUP BY
date

BQ 中的结果

GA 同期按用户报告划分的未抽样新访问者/回访者报告

问题/问题：

鉴于newVisits（默认字段）和newUsers（我的计算）给出的结果相同，这与 GA 报告新访客用户一致。为什么 GA 回访用户和我在 BQ 中计算的 returningUsers 不匹配？这两个甚至可以比较，我错过了什么？

我的方法是最有效、最简洁的方法吗？

有没有更好的方法来获取我缺少的数据？

解决方案

根据 Martin 在下面的回答，我设法在我正在运行的查询的上下文中创建了“回访用户”指标/字段：

SELECT
  date,
  deviceCategory,
  # newUsers - SUM result if it's a new user
  SUM(IF(userType="New Visitor", 1, 0)) newUsers,
  # returningUsers - COUNT DISTINCT fullvisitorId if it's a returning user
  COUNT(DISTINCT IF(userType="Returning Visitor", fullvisitorid, NULL)) returningUsers,
  COUNT(DISTINCT fullvisitorid) users,
  SUM(visits) sessions
FROM (
  SELECT
    date,
    fullVisitorId,
    visitId,
    totals.visits,
    device.deviceCategory,
    IF(totals.newVisits IS NOT NULL, "New Visitor", "Returning Visitor") userType
  FROM
    `project.dataset.ga_sessions_20180118` )
GROUP BY
  deviceCategory,
  date

【参考方案1】：

Google Analytics（分析）使用用户的近似值 (fullvisitorid) - 即使它说“基于 100%”。使用非抽样报告时，您可以获得更好的用户数量。

另外要提一下：即使totals.visits != 1 也会考虑全访问者，而会话仅在totals.visits = 1 的情况下计算

如果用户是新用户然后又被退回，他们也会被重复计算。意思是，这应该给你正确的数字：

SELECT
  totals.newVisits IS NOT NULL AS isNew,
  COUNT(DISTINCT fullvisitorid) AS visitors,
  SUM(totals.visits) AS sessions
FROM
  `project.dataset.ga_sessions_20180214`
GROUP BY
  1

如果您想避免重复计算，可以使用此方法，即使用户返回，也将其计为新用户：

WITH
  visitors AS (
  SELECT
    fullvisitorid,
    -- check if any visit of this visitor was new - will be used for grouping later
    MAX(totals.newVisits ) isNew, 
    SUM(totals.visits) as sessions
  FROM
    `project.dataset.ga_sessions_20180214`
  GROUP BY 1
  )

SELECT
  isNew IS NOT NULL AS isNew,
  COUNT(1) AS visitors,
  sum(sessions) as sessions
FROM
  visitors
GROUP BY 1

当然，这些数字仅在总数上与 GA 匹配。

【讨论】：

谢谢马丁，详细的解释。这些计算会很方便！我希望能够有一个返回用户数量的字段。话虽如此，根据您的解释和计算，我设法做到了。也会用这个更新我的帖子。