BigQuery SQL：平均值、几何平均值、删除异常值、中位数

Question

我正在计算在 Stack Overflow 上得到回复的平均时间，结果毫无意义。

#standardSQL

WITH question_answers AS (
  SELECT * 
    , timestamp_diff(answers.first, creation_date, minute) minutes
  FROM (
    SELECT creation_date
      , (SELECT AS STRUCT MIN(creation_date) first, COUNT(*) c
         FROM `bigquery-public-data.stackoverflow.posts_answers` b
         WHERE a.id=b.parent_id
        ) answers
      , SPLIT(tags, '|') tags
    FROM `bigquery-public-data.stackoverflow.posts_questions` a
    WHERE EXTRACT(year FROM creation_date) > 2015
  ), UNNEST(tags) tag
  WHERE tag IN ('java', 'javascript', 'google-bigquery', 'firebase', 'php')
  AND answers.c > 0
)

SELECT tag
  , COUNT(*) questions
  , ROUND(AVG(minutes), 2) first_reply_avg_minutes
FROM question_answers
GROUP BY tag

我应该如何计算平均时间？

Answer 1

2019 年更新：分享一些持久化的公共 UDF怎么样？

第一个，中位数：

SELECT fhoffa.x.median([1,1,1,2,3,4,5,100,1000]) 

3.0

确实 - 在 Stack Overflow 上获得答案的平均时间超过 100 小时（>6000 分钟）似乎是错误的 - 并且主要是由异常值驱动的。

而不是做一个简单的AVG()你可以得到：

• 几何平均值： EXP(AVG(LOG(GREATEST(minutes,1))))
• 去除异常值后的平均值：AVG(q) FROM (SELECT q FROM QUANTILES(q, 100) LIMIT 80 OFFSET 2)).
• 中位数： all_minutes[OFFSET(CAST(ARRAY_LENGTH(all_minutes)/2 AS INT64))]

如果您使用任何这些替代方案，结果会更有意义：

正如你在这里看到的，在这种情况下，去除异常值给我们的结果类似于几何平均值——而中值报告的数字甚至更低。使用哪一种？你的选择。

WITH question_answers AS (
  SELECT * 
    , timestamp_diff(answers.first, creation_date, minute) minutes
  FROM (
    SELECT creation_date
      , (SELECT AS STRUCT MIN(creation_date) first, COUNT(*) c
         FROM `bigquery-public-data.stackoverflow.posts_answers` b
         WHERE a.id=b.parent_id
        ) answers
      , SPLIT(tags, '|') tags
    FROM `bigquery-public-data.stackoverflow.posts_questions` a
    WHERE EXTRACT(year FROM creation_date) > 2015
  ), UNNEST(tags) tag
  WHERE tag IN ('java', 'javascript', 'google-bigquery', 'firebase', 'php', 'sql', 'elasticsearch', 'apache-kafka', 'tensorflow')
  AND answers.c > 0
)

SELECT *  EXCEPT(qs, all_minutes)
  , (SELECT ROUND(AVG(q),2) FROM (SELECT q FROM UNNEST(qs) q ORDER BY q LIMIT 80 OFFSET 2)) avg_no_outliers 
  , all_minutes[OFFSET(CAST(ARRAY_LENGTH(all_minutes)/2 AS INT64)  )] median_minutes
FROM (
  SELECT tag
    , COUNT(*) questions
    , ROUND(AVG(minutes), 2) avg_minutes
    , ROUND(EXP(AVG(LOG(GREATEST(minutes,1)))),2) first_reply_avg_minutes_geom
    , APPROX_QUANTILES(minutes, 100) qs
    , ARRAY_AGG(minutes IGNORE NULLS ORDER BY minutes) all_minutes
  FROM question_answers
  GROUP BY tag
)

ORDER BY 2 DESC

MEDIAN() 来自 Elliott 的额外UDF 函数。

CREATE TEMP FUNCTION MEDIAN(arr ANY TYPE) AS ((
  SELECT
    IF(
      MOD(ARRAY_LENGTH(arr), 2) = 0,
      (arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2) - 1)] + arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))]) / 2,
      arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))]
    )
  FROM (SELECT ARRAY_AGG(x ORDER BY x) AS arr FROM UNNEST(arr) AS x)
));