SQL: count all records with consecutive occurrence of same value for each device set and return the highest count

Question

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 Examp

Answer 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;