Partition Function COUNT() OVER possible using DISTINCT - sql

Question

I'm trying to write the following in order to get a running total of distinct NumUsers, like so:
NumUsers = COUNT(DISTINCT [UserAccountKey]) OVER (PARTITION BY [Mth])
Management studio doesn't seem too happy about this. The error disappears when I remove the DISTINCT keyword, but then it won't be a distinct count.
DISTINCT does not appear to be possible within the partition functions.
How do I go about finding the distinct count? Do I use a more traditional method such as a correlated subquery?
Looking into this a bit further, maybe these OVER functions work differently to Oracle in the way that they cannot be used in SQL-Server to calculate running totals.
I've added a live example here on SQLfiddle where I attempt to use a partition function to calculate a running total.

Answer 1

There is a very simple solution using dense_rank()
dense_rank() over (partition by [Mth] order by [UserAccountKey])
+ dense_rank() over (partition by [Mth] order by [UserAccountKey] desc)
- 1
This will give you exactly what you were asking for: The number of distinct UserAccountKeys within each month.

Answer 2

Necromancing:
It's relativiely simple to emulate a COUNT DISTINCT over PARTITION BY with MAX via DENSE_RANK:
;WITH baseTable AS
(
SELECT 'RM1' AS RM, 'ADR1' AS ADR
UNION ALL SELECT 'RM1' AS RM, 'ADR1' AS ADR
UNION ALL SELECT 'RM2' AS RM, 'ADR1' AS ADR
UNION ALL SELECT 'RM2' AS RM, 'ADR2' AS ADR
UNION ALL SELECT 'RM2' AS RM, 'ADR2' AS ADR
UNION ALL SELECT 'RM2' AS RM, 'ADR3' AS ADR
UNION ALL SELECT 'RM3' AS RM, 'ADR1' AS ADR
UNION ALL SELECT 'RM2' AS RM, 'ADR1' AS ADR
UNION ALL SELECT 'RM3' AS RM, 'ADR1' AS ADR
UNION ALL SELECT 'RM3' AS RM, 'ADR2' AS ADR
)
,CTE AS
(
SELECT RM, ADR, DENSE_RANK() OVER(PARTITION BY RM ORDER BY ADR) AS dr
FROM baseTable
)
SELECT
RM
,ADR
,COUNT(CTE.ADR) OVER (PARTITION BY CTE.RM ORDER BY ADR) AS cnt1
,COUNT(CTE.ADR) OVER (PARTITION BY CTE.RM) AS cnt2
-- Not supported
--,COUNT(DISTINCT CTE.ADR) OVER (PARTITION BY CTE.RM ORDER BY CTE.ADR) AS cntDist
,MAX(CTE.dr) OVER (PARTITION BY CTE.RM ORDER BY CTE.RM) AS cntDistEmu
FROM CTE
Note:
This assumes the fields in question are NON-nullable fields.
If there is one or more NULL-entries in the fields, you need to subtract 1.

Answer 3

I use a solution that is similar to that of David above, but with an additional twist if some rows should be excluded from the count. This assumes that [UserAccountKey] is never null.
-- subtract an extra 1 if null was ranked within the partition,
-- which only happens if there were rows where [Include] <> 'Y'
dense_rank() over (
partition by [Mth]
order by case when [Include] = 'Y' then [UserAccountKey] else null end asc
)
+ dense_rank() over (
partition by [Mth]
order by case when [Include] = 'Y' then [UserAccountKey] else null end desc
)
- max(case when [Include] = 'Y' then 0 else 1 end) over (partition by [Mth])
- 1
An SQL Fiddle with an extended example can be found here.

Answer 4

I think the only way of doing this in SQL-Server 2008R2 is to use a correlated subquery, or an outer apply:
SELECT datekey,
COALESCE(RunningTotal, 0) AS RunningTotal,
COALESCE(RunningCount, 0) AS RunningCount,
COALESCE(RunningDistinctCount, 0) AS RunningDistinctCount
FROM document
OUTER APPLY
( SELECT SUM(Amount) AS RunningTotal,
COUNT(1) AS RunningCount,
COUNT(DISTINCT d2.dateKey) AS RunningDistinctCount
FROM Document d2
WHERE d2.DateKey <= document.DateKey
) rt;
This can be done in SQL-Server 2012 using the syntax you have suggested:
SELECT datekey,
SUM(Amount) OVER(ORDER BY DateKey) AS RunningTotal
FROM document
However, use of DISTINCT is still not allowed, so if DISTINCT is required and/or if upgrading isn't an option then I think OUTER APPLY is your best option

Answer 5

There is a solution in simple SQL:
SELECT time, COUNT(DISTINCT user) OVER(ORDER BY time) AS users
FROM users
=>
SELECT time, COUNT(*) OVER(ORDER BY time) AS users
FROM (
SELECT user, MIN(time) AS time
FROM users
GROUP BY user
) t

Answer 6

I wandered in here with essentially the same question as whytheq and found David’s solution, but then had to review my old self-tutorial notes regarding DENSE_RANK because I use it so rarely: why DENSE_RANK instead of RANK or ROW_NUMBER, and how does it actually work? In the process, I updated that tutorial to include my version of David’s solution for this particular problem, and then thought it might be helpful for SQL newbies (or others like me who forget stuff).
The whole tutorial text can be copy/pasted into a query editor and then each example query can be (separately) uncommented and run, to see their respective results. (By default, the solution to this problem is uncommented at the bottom.) Or, each example can be copied separately into their own query-edit instance but the TBLx CTE must be included with each.
--WITH /* DB2 version */
--TBLx (Col_A, Col_B) AS (VALUES
-- ( 7, 7 ),
-- ( 7, 7 ),
-- ( 7, 7 ),
-- ( 7, 8 ))
WITH /* SQL-Server version */
TBLx (Col_A, Col_B) AS
(SELECT 7, 7 UNION ALL
SELECT 7, 7 UNION ALL
SELECT 7, 7 UNION ALL
SELECT 7, 8)
/*** Example-A: demonstrates the difference between ROW_NUMBER, RANK and DENSE_RANK ***/
--SELECT Col_A, Col_B,
-- ROW_NUMBER() OVER(PARTITION BY Col_A ORDER BY Col_B) AS ROW_NUMBER_,
-- RANK() OVER(PARTITION BY Col_A ORDER BY Col_B) AS RANK_,
-- DENSE_RANK() OVER(PARTITION BY Col_A ORDER BY Col_B) AS DENSE_RANK_
--FROM TBLx
/* RESULTS:
Col_A Col_B ROW_NUMBER_ RANK_ DENSE_RANK_
7 7 1 1 1
7 7 2 1 1
7 7 3 1 1
7 8 4 4 2
ROW_NUMBER: Just increments for the three identical rows and increments again for the final unique row.
That is, it’s an order-value (based on "sort" order) but makes no other distinction.
RANK: Assigns the same rank value to the three identical rows, then jumps to 4 for the fourth row,
which is *unique* with regard to the others.
That is, each identical row is ranked by the rank-order of the first row-instance of that
(identical) value-set.
DENSE_RANK: Also assigns the same rank value to the three identical rows but the fourth *unique* row is
assigned a value of 2.
That is, DENSE_RANK identifies that there are (only) two *unique* row-types in the row set.
*/
/*** Example-B: to get only the distinct resulting "count-of-each-row-type" rows ***/
-- SELECT DISTINCT -- For unique returned "count-of-each-row-type" rows, the DISTINCT operator is necessary because
-- -- the calculated DENSE_RANK value is appended to *all* rows in the data set. Without DISTINCT,
-- -- its value for each original-data row-type would just be replicated for each of those rows.
--
-- Col_A, Col_B,
-- DENSE_RANK() OVER(PARTITION BY Col_A ORDER BY Col_B) AS DISTINCT_ROWTYPE_COUNT_
-- FROM TBLx
/* RESULTS:
Col_A Col_B DISTINCT_ROWTYPE_COUNT_
7 7 1
7 8 2
*/
/*** Example-C.1: demonstrates the derivation of the "count-of-all-row-types" (finalized in Example-C.2, below) ***/
-- SELECT
-- Col_A, Col_B,
--
-- DENSE_RANK() OVER ( PARTITION BY Col_A ORDER BY Col_B DESC) AS ROW_TYPES_COUNT_DESC_,
-- DENSE_RANK() OVER ( PARTITION BY Col_A ORDER BY Col_B ASC) AS ROW_TYPES_COUNT_ASC_,
--
-- -- Adding the above cases together and subtracting one gives the same total count for on each resulting row:
--
-- DENSE_RANK() OVER ( PARTITION BY Col_A ORDER BY Col_B DESC)
-- +
-- DENSE_RANK() OVER ( PARTITION BY Col_A ORDER BY Col_B ASC)
-- - 1 /* (Because DENSE_RANK values are one-based) */
-- AS ROW_TYPES_COUNT_
-- FROM TBLx
/* RESULTS:
COL_A COL_B ROW_TYPES_COUNT_DESC_ ROW_TYPES_COUNT_ASC_ ROW_TYPES_COUNT_
7 7 2 1 2
7 7 2 1 2
7 7 2 1 2
7 8 1 2 2
*/
/*** Example-C.2: uses the above technique to get a *single* resulting "count-of-all-row-types" row ***/
SELECT DISTINCT -- For a single returned "count-of-all-row-types" row, the DISTINCT operator is necessary because the
-- calculated DENSE_RANK value is appended to *all* rows in the data set. Without DISTINCT, that
-- value would just be replicated for each original-data row.
-- Col_A, Col_B, -- In order to get a *single* returned "count-of-all-row-types" row (and field), all other fields
-- must be excluded because their respective differing row-values will defeat the purpose of the
-- DISTINCT operator, above.
DENSE_RANK() OVER ( PARTITION BY Col_A ORDER BY Col_B DESC)
+
DENSE_RANK() OVER ( PARTITION BY Col_A ORDER BY Col_B ASC)
- 1 /* (Because DENSE_RANK values are one-based) */
AS ROW_TYPES_COUNT_
FROM TBLx
/* RESULTS:
ROW_TYPES_COUNT_
2
*/