Question

我正在尝试获取特定日期的独特事件数量，并在90/30/7天后回滚。我已经在有限数量的行上使用查询，但是对于大型数据集，我从聚合字符串中获得内存错误，这变得很大。

我正在寻找一种更有效的方法来实现相同的结果。

表格看起来像这样：

+---+------------+-------------+
|   |     date   |     userid  |
+---+------------+-------------+
| 1 | 2013-05-14 | xxxxx       |
| 2 | 2017-03-14 | xxxxx       |
| 3 | 2018-01-24 | xxxxx       |
| 4 | 2013-03-21 | xxxxx       |
| 5 | 2014-03-19 | xxxxx       |
| 6 | 2015-09-03 | xxxxx       |
| 7 | 2014-02-06 | xxxxx       |
| 8 | 2014-10-30 | xxxxx       |
| ..| ...        | ...         |
+---+------------+-------------+

所需结果的格式：

+---+------------+---------------------------------------------+
|   |     date   | active_users_7_days | active_users_90_days  |
+---+------------+---------------------------------------------+
| 1 | 2013-05-14 | 1240                | 34339                 |
| 2 | 2017-03-14 | 4334                | 54343                 |
| 3 | 2018-01-24 | .....               | .....                 |
| 4 | 2013-03-21 | .....               | .....                 |
| 5 | 2014-03-19 | .....               | .....                 |
| 6 | 2015-09-03 | .....               | .....                 |
| 7 | 2014-02-06 | .....               | .....                 |
| 8 | 2014-10-30 | .....               | .....                 |
| ..| ...        | .....               | .....                 |
+---+------------+---------------------------------------------+

我的查询如下：

#standardSQL
    WITH
      T1 AS(
      SELECT
        date,
        STRING_AGG(DISTINCT userid) AS IDs
      FROM
        `consumer.events`
      GROUP BY
        date ),
      T2 AS(
      SELECT
        date,
        STRING_AGG(IDs) OVER(ORDER BY UNIX_DATE(date) RANGE BETWEEN 90 PRECEDING
          AND CURRENT ROW) AS IDs
      FROM
        T1 )
    SELECT
      date,
      (
      SELECT
        COUNT(DISTINCT (userid))
      FROM
        UNNEST(SPLIT(IDs)) AS userid) AS NinetyDays
    FROM
      T2

Answer 1

计算唯一身份用户需要大量资源，如果您希望在滚动窗口中获得结果，则需要更多资源。对于可扩展的解决方案，请查看HLL ++等近似算法：

https://medium.freecodecamp.org/counting-uniques-faster-in-bigquery-with-hyperloglog-5d3764493a5a

对于精确计数，这可以工作（但随着窗口变大而变慢）：

#standardSQL
SELECT DATE_SUB(date, INTERVAL i DAY) date_grp
 , COUNT(DISTINCT owner_user_id) unique_90_day_users
 , COUNT(DISTINCT IF(i<31,owner_user_id,null)) unique_30_day_users
 , COUNT(DISTINCT IF(i<8,owner_user_id,null)) unique_7_day_users
FROM (
  SELECT DATE(creation_date) date, owner_user_id
  FROM `bigquery-public-data.stackoverflow.posts_questions` 
  WHERE EXTRACT(YEAR FROM creation_date)=2017
  GROUP BY 1, 2
), UNNEST(GENERATE_ARRAY(1, 90)) i
GROUP BY 1
ORDER BY date_grp

近似解决方案产生的结果更快（14s vs 366s，但结果是近似的）：

#standardSQL
SELECT DATE_SUB(date, INTERVAL i DAY) date_grp
 , HLL_COUNT.MERGE(sketch) unique_90_day_users
 , HLL_COUNT.MERGE(DISTINCT IF(i<31,sketch,null)) unique_30_day_users
 , HLL_COUNT.MERGE(DISTINCT IF(i<8,sketch,null)) unique_7_day_users
FROM (
  SELECT DATE(creation_date) date, HLL_COUNT.INIT(owner_user_id) sketch
  FROM `bigquery-public-data.stackoverflow.posts_questions` 
  WHERE EXTRACT(YEAR FROM creation_date)=2017
  GROUP BY 1
), UNNEST(GENERATE_ARRAY(1, 90)) i
GROUP BY 1
ORDER BY date_grp

更新了可提供正确结果的查询 - 删除少于90天的行（在没有日期丢失时有效）：

#standardSQL
SELECT DATE_SUB(date, INTERVAL i DAY) date_grp
 , HLL_COUNT.MERGE(sketch) unique_90_day_users
 , HLL_COUNT.MERGE(DISTINCT IF(i<31,sketch,null)) unique_30_day_users
 , HLL_COUNT.MERGE(DISTINCT IF(i<8,sketch,null)) unique_7_day_users
 , COUNT(*) window_days
FROM (
  SELECT DATE(creation_date) date, HLL_COUNT.INIT(owner_user_id) sketch
  FROM `bigquery-public-data.stackoverflow.posts_questions` 
  WHERE EXTRACT(YEAR FROM creation_date)=2017
  GROUP BY 1
), UNNEST(GENERATE_ARRAY(1, 90)) i
GROUP BY 1
HAVING window_days=90
ORDER BY date_grp

Answer 2

您可以汇总日期并进行总和。什么是聚合？采取最近的日期：

select count(*) as num_users,
       sum(case when date > datediff(current_date, interval -30 day) then 1 else 0 end) as num_users_30days,
       sum(case when date > datediff(current_date, interval -60 day) then 1 else 0 end) as num_users_60days,
       sum(case when date > datediff(current_date, interval -90 day) then 1 else 0 end) as num_users_90days
from (select user_id, max(date) as max(date)
      from `consumer.events` e
      group by user_id
     ) e;

如果用户的最近日期在该期间内，则应计算该用户。

你可以得到这个＆＃34; as-of＆＃34;通过在子查询中使用where子句来确定特定日期。

在BigQuery中滚动90天活跃用户，提高性能（DAU / MAU / WAU）

2 个答案: