BigQuery SQL:平均值，几何平均值，移除离群值，中位数 [英] BigQuery SQL: Average, geometric mean, remove outliers, median

问题描述

我正在计算平均时间以得到关于Stack Overflow的答复，结果毫无意义.

I'm calculating the average time to get a reply on Stack Overflow, and the results make no sense.

#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?

How should I calculate the average time?

推荐答案

更新2019:如何共享一些第一个，中位数:

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

3.0

---

实际上-在平均超过100小时(> 6000分钟)的堆栈溢出问题上获得答案的方法似乎是错误的-很大程度上是由异常值驱动的.

---

Indeed - getting an average time to get answers on Stack Overflow of above 100 hours (>6000 minutes) seems wrong - and is largely driven by outliers.

代替简单的AVG()，您可以获得:

Instead of doing a simple AVG() you could get:

• 几何平均值:EXP(AVG(LOG(GREATEST(minutes,1))))
• 除去异常值后的平均值: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))]

• The geometric mean: EXP(AVG(LOG(GREATEST(minutes,1))))
• The mean after removing outliers: AVG(q) FROM (SELECT q FROM QUANTILES(q, 100) LIMIT 80 OFFSET 2)).
• The median: all_minutes[OFFSET(CAST(ARRAY_LENGTH(all_minutes)/2 AS INT64))]

如果使用以下任何一种选择，结果将更有意义:

The results make way more sense if you use any of those alternatives:

正如您在此处看到的，在这种情况下，除去异常值可以得出类似于几何均值的结果-而中位数报告的数字甚至更低.使用哪一个?您的选择.

As you can see here, in this case removing outliers give us results similar to the geometric mean - while the median reports even lower numbers. Which one to use? Your choice.

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

---

奖金MEDIAN() 来自Elliott的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)
));

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