Question

Is there a faster way to select the distinct count of users from a table? Perhaps using row_number, partitioning, or cross apply?

I just can’t think of it right now.

Example:

Table UsageLog

UserId     Date     StoreNumber
Alice      200901   342
Alice      200902   333
Alice      200902   112
Bob        200901   112
Bob        200902   345
Charlie    200903   322

Here’s my current query:

select count(distinct userID), date
from
   UsageLog
where
   date between 200901 and 200902
group by date

My actual table has millions of rows and all columns are actually integers.

Is there a faster way to get the list of users?

Edit:

I already have nonclustered indexes on all individual columns. For some reason, the execution plan shows that I am still doing a table scan. I guess I should create a clustered index on Date. I’ll see how that works…

Answer 1

SELECT DISTINCT() is the way to go. The problem is that you are hitting the date index tipping point, so your plan goes for the clustered index scan instead. See the link for Kimberley L. Tripp article what a ‘tipping point’ is.

You need a covering index:

CREATE INDEX idx_UsageLog_date_user_id ON UsageLog(date) INCLUDE (userID);

Clustered index will also work, but has other side effects as well. If the clustered index on date is OK with the rest of your data access patterns, then is better than the covering index I propose.

Update:

The reverse order index you tried on (userID, date) also works, will range seek each userID. In fact is better than the (date, userID) or (date) INCLUDE (userID) because it returns the userIDs pre-sorted so the DISTINCT does not introduce the additional sort.

Still I recommend going over the link I posted to understand why ‘index on each individual columns’ was not helping.