Ranking Columns in SQL Based on Row Day Difference and Partition

Introduction

When working with data, it’s not uncommon to need to rank rows within a partition based on certain conditions. In this article, we’ll explore how to achieve this using the RANK() function in SQL, specifically when dealing with row day differences and partitions.

Understanding RANK()

The RANK() function is used to assign a ranking to each row within a result set that are related to the rows in the DENSE_RANK() function. The main difference between these two functions lies in how they handle ties. DENSE_RANK() assigns the same rank to tied rows without gaps, whereas RANK() skips ranks when there are ties.

To use RANK(), you need to first identify the column(s) that will serve as the ranking criteria. In our case, we’ll be using a combination of row day differences and partitions.

Using PARTITION

When working with partitioning in SQL, it’s essential to understand how it affects the way rows are ordered. Partitioning allows you to divide a table into smaller parts based on certain conditions. When ranking rows within a partition, the RANK() function will ignore any rows from other partitions.

In our example query, we’re partitioning by hotel_id. This means that each partition is separate and independent of the others.

Using CTEs (Common Table Expressions)

CTEs are temporary result sets that can be referenced within a SQL statement. They’re particularly useful when dealing with complex queries or recursive operations.

In our query, we’ll use two CTEs: hotel_rows and ranks. The first CTE generates row numbers for each visit on a per-hotel basis using the ROW_NUMBER() function. The second CTE uses recursion to generate the ranking values based on row day differences and partitions.

** hotel_rows **

This is the first CTE that generates row numbers for each visit on a per-hotel basis.

WITH RECURSIVE hotel_rows AS (
  SELECT hotel_id, customer_id, visit_date,
         ROW_NUMBER() OVER (PARTITION BY hotel_id ORDER BY visit_date) AS rn
  FROM hotel
  ORDER BY hotel_id, visit_date
)

Here’s what’s happening:

We’re using a recursive CTE (WITH RECURSIVE) to define our first CTE.
The ROW_NUMBER() function assigns a unique number to each row within the partition (in this case, by hotel_id).
We’re ordering the rows by visit_date for consistency.

** ranks **

This is the second CTE that generates ranking values based on row day differences and partitions.

ranks AS (
  SELECT hotel_id, customer_id, visit_date, rn, 1 AS `rank`
  FROM hotel_rows
  WHERE rn = 1
  UNION ALL
  SELECT h.hotel_id, h.customer_id, h.visit_date, h.rn,
         r.rank + (h.visit_date > r.visit_date + INTERVAL 2 DAY)
  FROM hotel_rows h
  JOIN ranks r ON h.hotel_id = r.hotel_id
              AND h.rn = r.rn + 1
)

Here’s what’s happening:

We’re using the UNION ALL operator to combine two result sets: the first CTE and a recursive operation.
The first part of the UNION ALL selects rows from the hotel_rows CTE where the row number is 1. This corresponds to the first visit for each hotel.
In the second part of the UNION ALL, we’re joining the hotel_rows CTE with itself (ranks) on both hotel_id and rn.
We’re adding a recursive clause that increments the ranking if the current row’s visit date is more than 2 days after the previous row’s visit date.

** Using RANK() **

Finally, we’re using the RANK() function to assign rankings to each row within the result set.

SELECT hotel_id, customer_id, visit_date, rank
FROM ranks
ORDER BY hotel_id, visit_date;

Here’s what’s happening:

We’re selecting the desired columns from the ranks CTE.
We’re ordering the rows by hotel_id and then visit_date.
The RANK() function assigns a ranking to each row based on its position within the partition.

Conclusion

In this article, we’ve explored how to rank rows within a partition in SQL using the RANK() function. We’ve used CTEs to define our recursive operation and combined it with partitions for more efficient querying.

By understanding how to use RANK(), you can efficiently assign rankings to your data based on various conditions. Whether you’re working with simple or complex queries, mastering this technique will help you extract valuable insights from your data.

Last modified on 2024-01-23