SQL：统计每个设备集合中所有连续出现相同值的记录，return最高计数：百分比

Question

这是下面问题的延续：

我现在可以显示所有连续出现的记录的计数，但我想以百分比显示它们。所以而不是这个输出：

Device ID        speed             highcount
--------------------------------------------------
07777778999       34               4
07777778123       15               3

我想要百分比，即最高计数/（该设备的条目数）

Device ID        speed             percentcount           
--------------------------------------------------
07777778999       34               0.44(4/9-for explaination reason 4 is higest count and 9 is num of entries of tht device so 4/9 is 0.44)
07777778123       15               0.75(3//4-same as above)

计数最高的查询是

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;

我如何修改相同的内容以获得如上所述的百分比？

让我也展示上一个问题的架构和 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

Answer 1

这是一种方法：

select ds.device_id, ds.speed, ds.num_times, (ds.num_times/ds2.num_dev) percentcount
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
left join (select device_id, count(device_id) num_dev
from t
group by device_id) ds2 
on ds.device_id = ds2.device_id
where ds.seqnum = 1

这是 DEMO，所以您可以用它来回答下一个问题:)

Answer 2

您可以简单地使用 window 函数：

select device_id, speed, num_times,
       (num_times / device_num_times) as ratio
from (select device_id, speed, count(*) as num_times,
             sum(count(*)) over (partition by device_id) as device_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;

不需要额外的子查询。

SQL：统计每个设备集合中所有连续出现相同值的记录，return最高计数：百分比

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

mysql

sql

data-analysis

aggregation