Question

我有一个包含3个属性的简单数据集。看起来像这样

Time            flag          count

9/10/2017 1:00  Blue          0
9/10/2017 1:05  Blue          1
9/10/2017 1:10  Blue          2
9/10/2017 2:00  Blue          1
9/14/2017 1:00  Red           0
9/14/2017 2:00  Red           1
9/14/2017 2:30  Red           0
9/15/2017 1:00  Green         1
9/15/2017 1:10  Green         0
9/15/2017 2:00  Green         1
9/15/2017 2:50  Green         2
9/15/2017 2:55  Green         1
9/15/2017 3:00  Green         0
9/15/2017 3:30  Green         1
9/15/2017 4:00  Green         0

我需要一个sql查询，它给我以下输出

total duration(minute)  flag    number of times >=1 duration when it's equal 
                                                        or higher than 1
60                      Blue          3              (10.09.2017  02:00:00 - 10.09.2017  01:05:00 )=55
90                      Red           1              (14.09.2017  02:30:00 -14.09.2017  02:00:00)=30
180                    Green          5              (15.09.2017  03:00:00 - 15.09.2017  01:00:00) + (15.09.2017  04:00:00 - 15.09.2017  03:30:00)=120+30=150

感谢您的帮助。谢谢你

Answer 1

以下显示了如何窗口化数据。时间跨度是另一个难题。

SELECT flag, SUM(durationSinceLast) AS totalRelevantDuration, SUM(Count) AS relevantCount
FROM(
 SELECT flag, currentTime - prevTime AS durationSinceLast, Count
 FROM(
  SELECT flag, 
       LAG(Time) OVER (PARTION BY flag ORDER BY Time) AS prevTime,
       Time AS currentTime, 
       Count
  FROM T
 )x
)y
WHERE durationSinceLast > 1
GROUP BY flag

每篇评论的修订

SELECT flag, 
     MIN(Time) AS Begin,
     MAX(Time) AS End,
     SUM(1) Count --count 1 per record
FROM T
WHERE Count >= 1
GROUP BY flag

查询将时间点更改为时间跨度

1 个答案: