BigQuery SQL:平均值、几何平均值、删除异常值、中位数 [英] BigQuery SQL: Average, geometric mean, remove outliers, median

问题描述

我正在计算在 Stack Overflow 上获得回复的平均时间，但结果毫无意义.

#standardSQLWITH question_answers AS (选择 *, timestamp_diff(answers.first, creation_date, minute) 分钟从 (选择创建日期, (SELECT AS STRUCT MIN(creation_date) first, COUNT(*) c从`bigquery-public-data.stackoverflow.posts_answers` bWHERE a.id=b.parent_id) 答案, SPLIT(tags, '|') 标签从`bigquery-public-data.stackoverflow.posts_questions` aWHERE EXTRACT(从创建日期开始的年份)>2015年), UNNEST(tags) 标签WHERE 标签('java'、'javascript'、'google-bigquery'、'firebase'、'php')AND 答案.c >0)选择标签, COUNT(*) 个问题, ROUND(AVG(minutes), 2) first_reply_avg_minutesFROM question_answers按标签分组

我应该如何计算平均时间?

解决方案

2019 年更新: 分享一些

正如您在此处看到的，在这种情况下，移除异常值会为我们提供类似于几何平均值的结果 - 而中位数报告的数字甚至更低.使用哪一种?您的选择.

WITH question_answers AS (选择 *, timestamp_diff(answers.first, creation_date, minute) 分钟从 (选择创建日期, (SELECT AS STRUCT MIN(creation_date) first, COUNT(*) c从`bigquery-public-data.stackoverflow.posts_answers` bWHERE a.id=b.parent_id) 答案, SPLIT(tags, '|') 标签从`bigquery-public-data.stackoverflow.posts_questions` aWHERE EXTRACT(从创建日期开始的年份)>2015年), UNNEST(tags) 标签WHERE 标签('java'、'javascript'、'google-bigquery'、'firebase'、'php'、'sql'、'elasticsearch'、'apache-kafka'、'tensorflow')AND 答案.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))] medium_minutes从 (选择标签, COUNT(*) 个问题, 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_minutesFROM question_answers按标签分组)按 2 个 DESC 订购

<小时>

奖金 MEDIAN() 来自 Elliott 的 UDF 函数.

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

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

How should I calculate the average time?

解决方案

Update 2019: How about sharing some persisted public UDFs?

First one, median:

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

3.0

---

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.

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

• 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))]

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

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

---

Bonus MEDIAN() UDF function from Elliott.

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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