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[英]compute time difference between rows and conditional group by using PostgreSQL
[英]In BigQuery, compute difference between two rows in group by
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
这是一个简单的例子来演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
、 competitionId, playerId, onOff
的组合,并且onOff
列只能是“on”或“off”。 对于单个competitionId, playerId
然后(其中有两行,一行表示“on”,一行表示“off”),我们想从所有其他列中减去值(on - off)。
希望my_output
表明确我们需要什么输出。 在playerId = 27
的情况下,由于该玩家没有“关闭”行,因此可以简单地将其从输出中删除,因为无需进行计算。
您可以进行条件聚合:
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
这按比赛和球员对数据进行分组,然后条件sum()
计算每一列的“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。
另一种基于自联接的解决方案:
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'
您应该检查哪种方法更有效
下面是 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
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