根据一列对连续的行进行分组

Question

假设我从 select * from journeys:

的结果中得到这个 table

timestamp     | inJourney (1 = true and 0 = false)
--------------------------------------------------
time1         | 1
time2         | 1
time3         | 1
time4         | 0
time5         | 0
time6         | 1
time7         | 1
time8         | 1

预计：

timestamp     | inJourney (1 = true and 0 = false)
--------------------------------------------------
time1         | 1
time4         | 0
time8         | 1

注意：时间戳不重要，因为我只想统计行程数

知道我必须做什么吗？

Answer 1

这是一个缺口和孤岛问题。使用row_number():

之差

select injourney, min(timestamp), max(timestamp)
from (select t.*,
             row_number() over (order by timestamp) as seqnum,
             row_number() over (partition by injourney, order by timestamp) as seqnum_i
      from t
     ) t
group by injourney, (seqnum - seqnum_i)
order by min(timestamp);

Answer 2

这是一个间隙和孤岛问题，你可以尝试使用ROW_NUMBER window函数从结果集中获取间隙然后使用MIN

你可以试试这个。

查询#1

SELECT MIN(timestamp),inJourney 
FROM (
SELECT *,
    ROW_NUMBER() OVER(ORDER BY timestamp)  - ROW_NUMBER() OVER(PARTITION BY inJourney ORDER BY timestamp) grp
  FROM journeys
) t1
GROUP BY grp,inJourney 
ORDER BY MIN(timestamp);

| min   | injourney |
| ----- | --------- |
| time1 | 1         |
| time4 | 0         |
| time6 | 1         |

View on DB Fiddle

根据一列对连续的行进行分组

Group consecutive rows based on one column

sql

postgresql

gaps-and-islands