How Can I Count Distinct Values Within Partitions in SQL Server?

Counting Distinct Values Over Partitions with COUNT() OVER

Overcoming the challenge of counting distinct values when using the COUNT() OVER partition function is a common issue in SQL Server. This article explores an alternative approach to achieve the desired result using a combination of dense_rank() and arithmetic operations.

Original Query and Error

The original query attempts to calculate a running total of distinct values using COUNT(DISTINCT) OVER:

NumUsers = COUNT(DISTINCT [UserAccountKey]) OVER (PARTITION BY [Mth])

However, SQL Server generates an error due to the incompatibility of DISTINCT with partition functions.

Alternative Solution Using Dense_rank()

A practical solution to this issue is to utilize the dense_rank() function in conjunction with simple arithmetic operations:

dense_rank() over (partition by [Mth] order by [UserAccountKey]) 
+ dense_rank() over (partition by [Mth] order by [UserAccountKey] desc) 
- 1

This expression calculates the running total of distinct values within each month. Here's how it works:

• The first term, dense_rank() over (partition by [Mth] order by [UserAccountKey]), assigns a rank to each value in [UserAccountKey] within each month, starting from 1.
• The second term, dense_rank() over (partition by [Mth] order by [UserAccountKey] desc), ranks values in the opposite order, also starting from 1.
• Subtracting the second term from the first produces a result of 0 for duplicate values and the rank of the first occurrence for unique values.

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