问题
I want to find out how many times a particular value occured consecutively for a particular partition and then display the higher count for that partition.
For Example if below is the table:
Device ID speed DateTime
--------------------------------------------------
07777778999 34 18-12-2016 17:15
07777778123 15 18-12-2016 18:10
07777778999 34 19-12-2016 19:30
07777778999 34 19-12-2016 12:15
07777778999 20 19-12-2016 13:15
07777778999 20 20-12-2016 11:15
07777778123 15 20-12-2016 9:15
07777778128 44 20-12-2016 17:15
07777778123 15 20-12-2016 17:25
07777778123 12 20-12-2016 17:35
07777778999 34 20-12-2016 17:45
07777778999 34 20-12-2016 17:55
07777778999 34 20-12-2016 18:50
07777778999 34 20-12-2016 18:55
I want to know for each device what is highest number of times the same speed appeared consecutively.
So if i partition them by device id, i would get the belo table
Device ID speed DateTime
--------------------------------------------------
07777778999 34 18-12-2016 17:15
07777778999 34 19-12-2016 19:30
07777778999 34 19-12-2016 12:15
07777778999 20 19-12-2016 13:15
07777778999 20 20-12-2016 11:15
07777778999 34 20-12-2016 17:45
07777778999 34 20-12-2016 17:55
07777778999 34 20-12-2016 18:50
07777778999 34 20-12-2016 18:55
07777778123 15 18-12-2016 18:10
07777778123 15 20-12-2016 9:15
07777778123 15 20-12-2016 17:25
07777778123 12 20-12-2016 17:35
07777778128 44 20-12-2016 17:15
-----------------------------------------------------------------
So my required output would be like
Device ID speed highcount
--------------------------------------------------
07777778999 34 4
07777778123 15 3
note that 07777778128 did not appear as there were no values which repeated consecutively```
What would be the possible way to achieve this. i was able to get the the count of all consecutive values for each device but then it doesn't give the highest rather gives count of all such consecutive groups
回答1:
This is a form of gaps-and-islands. You can use a difference of row numbers to get the islands:
select device_id, speed, count(*) as num_times
from (select t.*,
row_number() over (partition by device_id order by datetime) as seqnum,
row_number() over (partition by device_id, speed order by datetime) as seqnum_s
from t
) t
group by device_id, speed, (seqnum - seqnum_s);
Then, to get the max, use another layer of window functions:
select device_id, speed, num_times
from (select device_id, speed, count(*) as num_times,
row_number() over (partition by device_id order by count(*) desc) as seqnum
from (select t.*,
row_number() over (partition by device_id order by datetime) as seqnum,
row_number() over (partition by device_id, speed order by datetime) as seqnum_s
from t
) t
group by device_id, speed, (seqnum - seqnum_s)
) ds
where seqnum = 1;
来源:https://stackoverflow.com/questions/59037783/sql-count-all-records-with-consecutive-occurrence-of-same-value-for-each-device