在 BigQuery 中，计算分组中两行之间的差异答案

【问题标题】：In BigQuery, compute difference between two rows in group by在 BigQuery 中，计算分组中两行之间的差异
【发布时间】：2020-08-28 08:14:00
【问题描述】：

with
  my_stats as (
    select 24996 as competitionId, 17 as playerId, 'on' as onOff, 8 as fga, 4 as fgm, 0.50 as fgPct union all
    select 24996 as competitionId, 17 as playerId, 'off' as onOff, 5 as fga, 3 as fgm, 0.60 as fgPct union all
    select 24996 as competitionId, 24 as playerId, 'on' as onOff, 9 as fga, 6 as fgm, 0.67 as fgPct union all
    select 24996 as competitionId, 24 as playerId, 'off' as onOff, 3 as fga, 1 as fgm, 0.33 as fgPct union all
    select 24996 as competitionId, 27 as playerId, 'on' as onOff, 5 as fga, 4 as fgm, 0.8 as fgPct
  ),
  
  my_output as (
    select 24996 as competitionId, 17 as playerId, 'diff' as onOff, 3 as fga, 1 as fgm, -0.1 as fgPct union all
    select 24996 as competitionId, 24 as playerId, 'diff' as onOff, 6 as fga, 5 as fgm, 0.34 as fgPct
  )
  

select * from my_stats
select * from my_output

这里有一个简单的例子来展示我们正在努力解决的问题。我们有表my_stats，其中主键是competitionId, playerId, onOff 的组合，而onOff 列只能是“on”或“off”。对于单个competitionId, playerId then（其中有两行，一行用于“on”，一行用于“off”），我们希望从所有其他列中减去值（on - off）。

希望my_output 表清楚地说明了我们需要什么输出。在playerId = 27 的情况下，由于该播放器没有“关闭”行，因此可以简单地从输出中删除它们，因为无需进行计算。

【问题讨论】：

标签： sql count google-bigquery sum pivot

【解决方案1】：

你可以做条件聚合：

select
    competitionId,
    playerId,
    'diff' as onOff,
    sum(case when onOff = 'on' then fga   else - fga   end) fga,
    sum(case when onOff = 'on' then fgm   else - fgm   end) fga,
    sum(case when onOff = 'on' then fgpct else - fgpct end) fgpct
from my_stats
where onOff in ('on', 'off')
group by competitionId, playerId
having count(*) = 2

这会按比赛和玩家对数据进行分组，然后条件 sum()s 计算每列的“开”和“关”值之间的差异。 having 子句过滤掉没有两个记录可用的组。

【讨论】：

【解决方案2】：

基于自连接的另一种解决方案：

select
    t1.competitionId,
    t1.playerId,
    'diff' as onOff,
    t1.fga - t2.fga as fga,
    t1.fgm - t2.fgm as fgm,
    t1.fgpct - t2.fgpct as fgpct
from my_stats as t1
join my_stats as t2
  on t1.competitionId = t2.competitionId
 and t1.playerId = t2.playerId
where t1.onOff = 'on'
  and t2.onOff = 'off'

你应该检查哪种方法更有效

【讨论】：

【解决方案3】：

以下是 BigQuery 标准 SQL

#standardSQL
SELECT competitionId, playerId, 'diff' AS onOff,
  SUM(onOffSign * fga) AS fga,
  SUM(onOffSign * fgm) AS fgm,
  SUM(onOffSign * fgPct) AS fgPct  
FROM my_stats, 
  UNNEST([IF(onOff = 'on', 1, -1)]) onOffSign
GROUP BY competitionId, playerId
HAVING COUNT(1) = 2

【讨论】：