Understanding Unique Row IDs in SQL using Partition: Choosing the Right Function for Cohort ID Generation

Understanding Unique Row IDs in SQL using Partition

When working with large datasets, it’s common to need a unique identifier for each row, known as a Cohort ID. This can be achieved using the PARTITION BY clause in combination with window functions like ROW_NUMBER(), RANK(), or DENSE_RANK(). In this article, we’ll delve into how to create unique Cohort IDs in SQL using partition and explore alternative approaches.

Understanding Partitioning

Partitioning is a technique used to divide large datasets into smaller, more manageable groups based on one or more columns. This allows for more efficient data processing, storage, and querying. When partitioning a dataset, the PARTITION BY clause determines which rows will be grouped together.

For example, let’s say we have a table called test_table_1 with the following columns: Asset_Type, Account, Issuer, Return_Bucket, Age_Bucket, Type_Bucket, and Risk_Bucket. We want to create Cohort IDs for each row based on these columns.

Select
Row_Number() Over(Partition By [Asset_Type],
[Account],
[Issuer],
[Return_Bucket],
[Age_Bucket],
[Type_Bucket],
[Risk_Bucket]
Order by [Account]) Account AS CohortID
into Test_Table_2
from Test_Table_1

In this example, the PARTITION BY clause groups rows together based on the specified columns. The ROW_NUMBER() function assigns a unique identifier to each row within each group.

Issues with ROW_NUMBER()

However, as seen in the original question, using ROW_NUMBER() can lead to unexpected results when there are duplicates in the partitioned columns.

Asset_Type  Account Issuer  Return_Bucket   Age_Bucket  Type_Bucket Risk_Bucket Cohort ID
Equity  Client 1    Bank A  10-15   3-6 months  Financial   Moderate    1
Equity  Client 1    Bank A  10-15   3-6 months  Financial   Moderate    1
Equity  Client 1    Bank A  10-15   6-12 months Financial   Moderate    2

In this case, both rows with Asset_Type = “Equity”, Account = “Client 1”, and Issuer = “Bank A” receive the same Cohort ID (1). This is because ROW_NUMBER() assigns a unique identifier to each row within a partition, but it doesn’t account for duplicate values in the partitioned columns.

Alternative Approach: DENSE_RANK()

To avoid this issue, we can use the DENSE_RANK() function instead of ROW_NUMBER(). DENSE_RANK() assigns a unique rank to each row within a partition, without gaps or duplicates.

Select
dense_rank() Over (order by [Asset_Type], [Account], [Issuer], [Return_Bucket],
                                   [Age_Bucket], [Type_Bucket], [Risk_Bucket]
                         ) as CohortID
into Test_Table_2
from Test_Table_1 t;

In this case, the first two rows with Asset_Type = “Equity”, Account = “Client 1”, and Issuer = “Bank A” receive rank values of 1, while the third row receives a rank value of 3.

Understanding RANK()

RANK() is another alternative to ROW_NUMBER() when dealing with duplicate values in partitioned columns. However, it has some differences:

If two rows have the same rank, they are tied.
If there are ties, all tied rows receive the same rank value.

For example, if we use RANK() instead of DENSE_RANK(), the third row would also receive a rank value of 2, because it’s tied with the second row:

Select
rank() Over (order by [Asset_Type], [Account], [Issuer], [Return_Bucket],
             [Age_Bucket], [Type_Bucket], [Risk_Bucket]
           ) as CohortID
into Test_Table_3
from Test_Table_1 t;

In this case, the first two rows receive rank values of 1 and 2, while the third row receives a rank value of 2.

Choosing the Right Function

When deciding between ROW_NUMBER(), DENSE_RANK(), or RANK(), consider the following factors:

Gaps in ranks: If you want to avoid gaps in ranks, use DENSE_RANK() instead.
Ties: If you need to handle ties, use RANK() instead.
Unique identifiers: If you need unique identifiers for each row, use ROW_NUMBER() or DENSE_RANK().

Last modified on 2024-06-22