Question

采用https://webmasters.stackexchange.com/a/87523

上所述的内容

除了我自己的理解，我想出了我认为会被认为是“回归用户”的内容

1.首先查询显示在两年时间内第一次“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.然后单独查询以选择特定日期范围内的某些字段：

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.将这两个查询连在一起找“返回用户”

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

结果BQ

按用户分组的未抽样新用户/回访者报告GA中的同一时段

问题/问题：

鉴于newVisits（默认字段）和newUsers（我的计算）给出了与GA报告新访问者用户一致的相同结果。为什么GAs返回访客用户与我在BQ中returningUsers的计算不匹配？这两个甚至可以比较，我错过了什么？
我的方法是最有效，更简洁的方法吗？
有没有更好的方法来获取数据，我缺少的东西？

解

根据Martin的回答，我设法在我运行的查询的上下文中创建“返回用户”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

Answer 1

Google Analytics（分析）使用用户近似值（fullvisitorid） - 即使它基于100％＆＃34;表示＆＃34;使用非抽样报告时，您可以获得更好的用户数。

另外需要提及的是：即使totals.visits != 1，也会考虑使用fullvisitorids，而会话仅计入totals.visits = 1

如果用户在新的位置再返回，则会对用户进行重复计算。这意味着，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

如果你想避免重复计算，你可以使用这个，即使她回来，用户也算作新的：

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，这些数字仅与总数相匹配。

在BigQuery

1 个答案: