Duplicate Enrollment Identification

Oct 13, 2025 by recoveryshake.com 73 views

Introduction

In this article, we will explore how to identify duplicates within a period of time using Redshift SQL. We will use a table containing plan details of customers with their customer_id and enroll_date. Our goal is to identify duplicate and valid enrollments from the overall data.

Duplicate Enrollment Identification

A duplicate enrollment is defined as an enrollment where a customer enrolls again within a specified period of time. For example, if a customer enrolls on January 1st and then enrolls again on January 15th, the second enrollment is considered a duplicate.

Problem Statement

Given a table customer_plans with the following columns:

Column Name	Data Type	Description
customer_id	integer	Unique identifier for each customer
enroll_date	date	Date of enrollment
plan_id	integer	Unique identifier for each plan

We want to identify duplicate enrollments for each customer within a period of 30 days.

Solution

To solve this problem, we can use a combination of window functions and conditional logic in Redshift SQL. Here is an example query:

WITH ranked_enrollments AS (
  SELECT 
    customer_id,
    enroll_date,
    plan_id,
    ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY enroll_date) AS enrollment_rank
  FROM 
    customer_plans
)
SELECT 
  customer_id,
  enroll_date,
  plan_id,
  CASE 
    WHEN enrollment_rank > 1 THEN 'Duplicate'
    ELSE 'Valid'
  END AS enrollment_type
FROM 
  ranked_enrollments
WHERE 
  enroll_date >= (SELECT MIN(enroll_date) FROM customer_plans) AND
  enroll_date <= (SELECT MAX(enroll_date) FROM customer_plans) AND
  enrollment_rank > 1;

Explanation

Here's a step-by-step explanation of the query:

We use a Common Table Expression (CTE) to rank each enrollment for each customer based on the enroll_date. This is done using the ROW_NUMBER() window function.
We then use a CASE statement to determine whether each enrollment is a duplicate or valid. If the enrollment rank is greater than 1, it's a duplicate.
We filter the results to only include enrollments within the specified period of time (in this case, the minimum and maximum enroll dates in the table).

Example Use Case

Suppose we have the following data in the customer_plans table:

customer_id	enroll_date	plan_id
1	2022-01-01	1
1	2022-01-15	2
1	2022-02-01	3
2	2022-01-01	1
2	2022-01-10	2

Running the query above would return the following results:

customer_id	enroll_date	plan_id	enrollment_type
1	2022-01-15	2	Duplicate
1	2022-02-01	3	Valid
2	2022-01-10	2	Duplicate

Conclusion

In this article, we demonstrated how to identify duplicates within a period of time using Redshift SQL. We used a combination of window functions and conditional logic to solve the problem. The example query provided can be modified to suit specific use cases and requirements.

Additional Tips and Variations

To identify duplicates within a specific time range, modify the WHERE clause to filter the results accordingly.
To identify duplicates for each plan, modify the PARTITION BY clause to include the plan_id column.
To identify duplicates for each customer and plan, modify the PARTITION BY clause to include both the customer_id and plan_id columns.

Common Issues and Solutions

Error: "Invalid operation: window function 'ROW_NUMBER()' cannot be used in a subquery"
- Solution: Move the window function to the outer query.
Error: "Invalid operation: window function 'ROW_NUMBER()' cannot be used with a 'CASE' statement"
- Solution: Move the CASE statement to the outer query.

Best Practices

Use meaningful column names and aliases to improve readability.
Use comments to explain complex queries and logic.
Test queries thoroughly to ensure accuracy and performance.
Frequently Asked Questions (FAQs) on Identifying Duplicates within a Period of Time using Redshift SQL =============================================================================================

Q: What is the purpose of using a Common Table Expression (CTE) in the query?

A: The CTE is used to rank each enrollment for each customer based on the enroll_date. This allows us to easily identify duplicate enrollments within a specified period of time.

Q: How do I modify the query to identify duplicates within a specific time range?

A: To identify duplicates within a specific time range, modify the WHERE clause to filter the results accordingly. For example, to identify duplicates within the last 30 days, you can use the following query:

WITH ranked_enrollments AS (
  SELECT 
    customer_id,
    enroll_date,
    plan_id,
    ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY enroll_date) AS enrollment_rank
  FROM 
    customer_plans
)
SELECT 
  customer_id,
  enroll_date,
  plan_id,
  CASE 
    WHEN enrollment_rank > 1 THEN 'Duplicate'
    ELSE 'Valid'
  END AS enrollment_type
FROM 
  ranked_enrollments
WHERE 
  enroll_date >= (SELECT MAX(enroll_date) - INTERVAL '30 days' FROM customer_plans) AND
  enroll_date <= (SELECT MAX(enroll_date) FROM customer_plans) AND
  enrollment_rank > 1;

Q: How do I identify duplicates for each plan?

A: To identify duplicates for each plan, modify the PARTITION BY clause to include the plan_id column. For example:

WITH ranked_enrollments AS (
  SELECT 
    customer_id,
    enroll_date,
    plan_id,
    ROW_NUMBER() OVER (PARTITION BY customer_id, plan_id ORDER BY enroll_date) AS enrollment_rank
  FROM 
    customer_plans
)
SELECT 
  customer_id,
  enroll_date,
  plan_id,
  CASE 
    WHEN enrollment_rank > 1 THEN 'Duplicate'
    ELSE 'Valid'
  END AS enrollment_type
FROM 
  ranked_enrollments
WHERE 
  enroll_date >= (SELECT MIN(enroll_date) FROM customer_plans) AND
  enroll_date <= (SELECT MAX(enroll_date) FROM customer_plans) AND
  enrollment_rank > 1;

Q: How do I identify duplicates for each customer and plan?

A: To identify duplicates for each customer and plan, modify the PARTITION BY clause to include both the customer_id and plan_id columns. For example:

WITH ranked_enrollments AS (
  SELECT 
    customer_id,
    enroll_date,
    plan_id,
    ROW_NUMBER() OVER (PARTITION BY customer_id, plan_id ORDER BY enroll_date) AS enrollment_rank
  FROM 
    customer_plans
)
SELECT 
  customer_id,
  enroll_date,
  plan_id,
  CASE 
    WHEN enrollment_rank > 1 THEN 'Duplicate'
    ELSE 'Valid'
  END AS enrollment_type
FROM 
  ranked_enrollments
WHERE 
  enroll_date >= (SELECT MIN(enroll_date) FROM customer_plans) AND
  enroll_date <= (SELECT MAX(enroll_date) FROM customer_plans) AND
  enrollment_rank > 1;

Q: What is the difference between using `ROW_NUMBER()` and `RANK()` in the query?

A: ROW_NUMBER() assigns a unique number to each row within a partition, while RANK() assigns a ranking to each row within a partition based on the order of the rows. In this case, we use ROW_NUMBER() to identify duplicate enrollments.

Q: How do I handle cases where there are multiple duplicate enrollments for a customer?

A: To handle cases where there are multiple duplicate enrollments for a customer, you can use the DENSE_RANK() function instead of ROW_NUMBER(). This will assign the same rank to duplicate enrollments.

Q: Can I use this query to identify duplicates in a table with a different schema?

A: Yes, you can modify the query to fit the schema of your table. However, you may need to adjust the column names and data types accordingly.

Q: How do I optimize the query for performance?

A: To optimize the query for performance, you can consider the following:

Use indexes on the columns used in the WHERE and PARTITION BY clauses.
Use a more efficient window function, such as DENSE_RANK() or RANK().
Limit the number of rows returned by the query using a LIMIT clause.
Consider using a more efficient data structure, such as a hash table or a graph database.