Efficient way to calculate average time between row dates grouped by ID - sql

Question

Suppose I have a table like this:
thedate ID
2014-10-20 14:13:42.063 1
2014-10-20 14:13:43.063 1
2014-10-20 14:13:47.063 1
2014-10-20 14:12:50.063 2
2014-10-20 14:13:49.063 2
2014-10-20 14:13:54.063 2
2014-10-20 14:20:24.063 2
2014-10-20 14:13:02.063 3
To replicate a similar toybox table as in this example you can use the following code:
declare #tmp as table(thedate datetime,ID int)
insert into #tmp (thedate, ID) values
(dateadd(s,0,getdate()),1), (dateadd(s,1,getdate()),1), (dateadd(s,5,getdate()),1),
(dateadd(s,-52,getdate()),2), (dateadd(s,7,getdate()),2), (dateadd(s,12,getdate()),2),(dateadd(s,402,getdate()),2),
(dateadd(s,-40,getdate()),3)
For each ID I want the average time between the dates. Now the database is huge (lots of ID's and dates for each ID), so it has to be very efficient. I want a result like this:
ID AvgTime (seconds)
1 2,5
2 151,333333333333
3 NULL
The following code does what I want, but it is way too slow:
select
a.ID,
(select top 1 avg(cast(datediff(s,(select max(thedate)
from #tmp c where ID = b.ID
and thedate < b.thedate)
,thedate) as float)) over (partition by b.ID)
from #tmp b where ID = a.ID)
from #tmp a group by ID
Does anyone know how to do this efficiently?

Answer 1

The average is the maximum minus the minimum divided by one less than the count. You can use this to write a relatively simple query:
select id,
cast(datediff(second, min(thedate), max(thedate)) as float) / (count(*) - 1)
from #tmp
group by id;
If some of the ids only have one row, then you'll want to check for potential divide by 0:
select id,
(case when count(*) > 1
then cast(datediff(second, min(thedate), max(thedate)) as float) / (count(*) - 1)
end) as AvgDiff
from #tmp
group by id;