SQL RANK() Function

To limit the number of records that match a query, use the "COUNT()" Function. Only the number of records in the table that fit the given filter will be returned by the COUNT() function. When using COUNT() with the FILTER clause in PostgreSQL, the syntax is as follows.

We can verify that the column containing the results count has a name assigned to it using the AS keyword.

Example 1: Let's test a statement utilizing the COUNT keyword in psql using the student table. Sub-select can be used to filter or group rows if necessary following window calculations. Let's examine the following query.

select contact_id, count(*),
rank() over (PARTITION BY contact_id order by count(*) desc) “”rank””
from hh_xref
group by contact_id
order by “”rank”” desc;

Example 2: But this gives me the ranking from least to greatest:

SELECT BloggerName,Topic,[Year],Total,
Rank() OVER (Partition by BloggerName Order by Total DESC) as 'Ranking',
COUNT(*) OVER(PARTITION BY BloggerName) AS BloggerNameCount
FROM BlogCount;

Gives each row a unique number starting at 1, with the exception of rows having duplicate values, in which case the same ranking is given to all rows and a gap is put in the order for each duplicate ranking.

Duplicate removal from a table can be a difficult procedure. There are numerous ways to locate and delete records. It takes a great deal of careful consideration to create a query to keep certain entries and delete others. I recently used SQL Server's RANK function to find duplicate data and designate them for deletion, which is a quick and easy approach.

Within a result set's partitioned grouping, the SQL Server RANK function delivers the rank, or row number, of each row. In this article, I won't go into great detail on the RANK function. More details about it are available here.

Example 1: I make a query with RANK that generates sets of duplicate records and gives a number to each row in the set of duplicate records. In the table, the row number will be reset for every duplicate set. The number is then used to determine which rows should be eliminated.

SELECT Id, CustomerId, OrderAmount, OrderDate,
RANK() OVER (PARTITION BY CustomerId, 
OrderAmount ORDER BY Id DESC) AS RowNumberWithinDuplicateSet
FROM duplicates

Example 2: This example uses a value that is duplicated in the fees_paid column:

SELECT
RANK(150) WITHIN GROUP (ORDER BY fees_paid) AS rank_val
FROM student;

Result:

It shows 4, because the value of 150 is the 4th lowest in the fees_paid column in the student table.

In SQL, RANK allocates a rank to each group. If the value stays the same, both numbers will receive the same rank and the subsequent rank will be ignored.

Example 1: The hypothetical new row (num=4, odd=1) is ranked inside each group of rows from test4 in the example that follows, where groups are identified by the values in the odd column:

SELECT * FROM test4;

Output:

SELECT odd, RANK(4,1) WITHIN GROUP (ORDER BY num, odd)
FROM test4 GROUP BY odd;

In both cases, the rank of the hypothetical new row is 4. In the group odd=0, the new row.

Example 2:According to the total sales revenue generated by each order date, this query ranks customers. It should be noted that the ORDER BY clause of the RANK() OVER is a computed column created by aggregating data using the SUM function and GROUP BY.

select b.CustomerID, 
	b.OrderDate,
	sum(a.UnitPrice * (1 - a.Discount) * a.Quantity) as Sales,
    RANK() OVER (PARTITION BY b.OrderDate ORDER BY sum(a.UnitPrice * (1 - a.Discount) * a.Quantity)) as sales_rank
from order_details as a
join orders as b on a.OrderID=b.OrderID
group by b.CustomerID, 
	b.OrderDate
order by b.OrderDate, sales_rank;

Example 3: Imagine you want to know which department has the greatest average income as well as the quantile of each department's average compensation. The following query divides the information into departments, calculates the average salary for each department, ranks the averages, and displays the average salary quantile.

SELECT WORKDEPT, INT(AVG(SALARY)) AS AVERAGE, 
 RANK() OVER(ORDER BY AVG(SALARY) DESC) AS AVG_SALARY,  
 FROM EMPLOYEE 
 GROUP BY  WORKDEPT 

Result:

WORK 	DEPTAVERAGE	 AVG_SALARY QUANTILE
B01	41,250		 1	     1
A00	40,850		 2	     1
E01	40,175	 	 3	     1
C01	29,722	 	 4	     2
D21	25,668	 	 5	     2
D11	25,147	 	 6	     2
E21	24,086	  	 7	     3
E11	21,020	         8	     3

In this instance, the rank function takes an input, indicating that the results should be divided into three sets of equal size. The two lowest number quantiles each contain an extra row because the result set cannot be divided evenly by the number of quantiles.

Example 4:

select *, RANK () OVER (
PARTITION BY item_group
ORDER BY
price
) from Basket;

SQL Server's RANK() function, a component of the window function, can be used to determine the rank of each row within a result set partition.

A partition's rows with identical values will have the same rank. The first row in a partition has rank one. The ranks may not be consecutive since the RANK() method multiplies the number of tied rows by the tied rank to determine the rank of the following row.

The RANK function gives identical values within the same partition the same rank number when it finds them. The following number in the ranking will also include duplicate numbers in addition to the preceding rank. As a result, this function does not always rank rows in sequential order.

The following points should be remembered while using this function:

• With the OVER() clause, it always functions.
• Each row is given a rank depending on the ORDER BY clause.
• Each row is given a rank in descending order.
• Rows are always given a rank, beginning with one with each new division.

NOTE: Rank assigns temporary values for rows within the partition when the query is executed.

Syntax:

The following shows the syntax of the RANK() function:

RANK() OVER (
    [PARTITION BY partition_expression, ... ]
    ORDER BY sort_expression [ASC | DESC], ...
)

Explanation of syntax:

• The rows of the result set partitions to which the function is applied are first divided by the PARTITION BY clause.
• Second, the logical sort order of the rows in each partition to which the function is applied is specified by the ORDER BY clause.
• The top-N and bottom-N reports benefit from the use of the RANK() function.

Example 1: Let us use RANK() to assign ranks to the rows in the result set of geek_demo table:

SELECT Name, 
RANK () OVER (
ORDER BY Name
) AS Rank_no 
FROM geek_demo;

Output:

Name	Rank_no
A	1
B	2
B	2
C	4
C	4
D	6
E	7

Example 2: The following example uses the RANK() function to assign ranks to the products by their list prices:

SELECT product_id,
	product_name,
	list_price,
	RANK () OVER ( 
		ORDER BY list_price DESC
	) price_rank 
FROM
	production.products;

Because the PARTITION BY clause was omitted in this case, the entire result set was regarded as one partition by the RANK() method.

Each row in the result set is given a rank by the RANK() function after being sorted by list price from high to low.

Example 3: Let's examine SQL Server's RANK() function in more detail. The sentence that follows will number each row using the rank function:

SELECT first_name, last_name, city,   
RANK () OVER (ORDER BY city) AS Rank_No   
FROM rank_demo; 

Rank over() and other window functions can be used to create serial numbers among a collection of rows. To accomplish this objective, we can establish start and last variables in SAS.

In a BY group, FIRST.VARIABLE assigns a value of 1 to the first observation and a value of 0 to every subsequent observation.

In a BY group, LAST.VARIABLE provides a value of 1 to the final observation and a value of 0 to every other observation.

Before utilizing FIRST and LAST Variables, the data collection must be sorted BY group.

Syntax:

FIRST_VALUE | LAST_VALUE
( expression [ IGNORE NULLS | RESPECT NULLS ] ) OVER
(
[ PARTITION BY expr_list ]
[ ORDER BY order_list frame_clause ]
)

FIRST_VALUE returns the value of the given expression in relation to the window frame's first row. The expression's value in relation to the last row in the frame is returned by the LAST_VALUE function.

Example 1: The seating capacity for each venue in the VENUE table is returned in the example below, with the results organized by capacity (high to low). The FIRST VALUE function is used to choose the name of the location that corresponds to the frame's first row, or the row with the most seats in this instance. Since the results are divided up by states, a new initial value is chosen whenever the VENUESTATE value changes. The window frame is unbounded, thus each row in the window is chosen with the same first value.

For California, Qualcomm Stadium has the highest number of seats (70561), so this name is the first value for all of the rows in the CA partition.

select venuestate, venueseats, venuename,
first_value(venuename)
over(partition by venuestate
order by venueseats desc
rows between unbounded preceding and unbounded following)
from (select * from venue where venueseats >0)
order by venuestate;

Example:

When I use the rank function, it ranks all the NULL values as one (1) and the values start after that. There are multiple rows with NULL values. Usually, there are 30 Nulls in the report, thus the rank starts at 31.

Is there a way to tell the rank function to disregard NULL values? Or is there another way to make it function?

if([Measure1] is null)
then (null)
else(
rank([Measure1] asc within set [Store])
)

RANK() OVER(PARTITION BY)—Multiple Columns; Single Column Although we'll mostly utilize the SQL RANK() function, it's important to keep in mind that the RANK function in Introduction to SQL Server is a window function that ranks each row within a partition of a result set. A partition's rows with identical values will have the same rank. The first row in a partition has rank one.

Select
rank() over(order by col1 desc) as Col1Rank,
rank() over(order by col2 desc) as Col2Rank, etc, etc

If the SORT BY was ascending rather than descending, the CASE statement that was earlier provided would include the NULL entries in the rank. This would cause the ranking to begin at 5 instead of 1, which is probably not what is wanted.

Example 1: By putting an initial sort criteria on whether the value IS NULL or not, you can force the nulls to the bottom to ensure that they aren't included in the rank:

SELECT
    CASE WHEN Value IS NULL THEN NULL
         ELSE RANK() OVER 
               (ORDER BY CASE WHEN Value IS NULL THEN 1 ELSE 0 END, VALUE DESC) 
    END AS RANK,
    USER_ID,
    VALUE
FROM yourtable

Example 2: Table of values in ranking.

When I perform a RANK() OVER (ORDER BY VALUE DESC) as RANK, I get the following results (in a hypothetical table):

RANK | USER_ID   | VALUE
1    | 33        | 30000
2    | 10        | 20000
3    | 45        | 10000
4    | 12        | 5000
5    | 43        | 2000
6    | 32        | NULL
6    | 13        | NULL
6    | 19        | NULL
6    | 28        | NULL

The issue is that I don't want the rows with NULL for a VALUE to have a rank; instead, I need a solution to make their rank NULL. I have not yet found any information on how I might be able to do this by searching the internet.

You can try a CASE statement:

SELECT
    CASE WHEN Value IS NULL THEN NULL
    ELSE RANK() OVER (ORDER BY VALUE DESC)
    END AS RANK,
    USER_ID,
    VALUE
FROM yourtable

FIRST_VALUE Below is a simple explanation of the FIRST_VALUE analytical function. Here, the analytic phrase is explained in greater depth.

Syntax:

FIRST_VALUE 
  { (expr) [ {RESPECT | IGNORE} NULLS ]
  | (expr [ {RESPECT | IGNORE} NULLS ])
  }
  OVER (analytic_clause)

Example 1: The FIRST_VALUE analytic function is similar to the FIRST analytic function, allowing you to return the first result from an ordered set.

SELECT empno,
       deptno,
       sal,
       FIRST_VALUE(sal) IGNORE NULLS 
         OVER (PARTITION BY deptno ORDER BY sal) AS lowest_in_dept
FROM   emp;

LAST_VALUE Below is a simple explanation of the LAST_VALUE analytical function. Here, the analytic phrase is explained in greater depth.

Syntax:

LAST_VALUE
  { (expr) [ { RESPECT | IGNORE } NULLS ]
  | (expr [ { RESPECT | IGNORE } NULLS ])
  OVER (analytic_clause)

Example 2: The LAST_VALUE analytic function is similar to the LAST analytic function, allowing you to return the last result from an ordered set. Using the default windowing clause the result can be a little unexpected.

SELECT empno,
       deptno,
       sal,
       LAST_VALUE(sal) IGNORE NULLS
         OVER (PARTITION BY deptno ORDER BY sal) AS highest_in_dept
FROM   emp;

Example 3: Cumulative sum sorted by date functions precisely as rank excluding nulls; assign 1 to non-null entries and 0 where the column is null.

Another option is to use RANK in combination with ROW NUMBER, which will take into account ties in the Date column and function just like RANK with NULLs:

select dt,
       col,
       case when col is not null then 
           rank() over (order by dt)
       else 
           rank() over (order by dt) - row_number() over (partition by rnDiff order by dt)
       end rnk
from (
    select dt, col,
           row_number() over (order by dt) -
               row_number() over (partition by coalesce(col, 0) order by dt) rnDiff
    from @tbl
) a
order by dt

Example: example for sql rank over condition :

SELECT v1, 
    v2, 
    CASE 
        WHEN v1 >= 95 
        THEN RANK() OVER (
            PARTITION BY CASE WHEN v1 >= 95 THEN 1 ELSE 2 END
            ORDER BY v2 DESC) 
      END AS Ranking
FROM (VALUES (100, 500), (90, 300), (89,800), (95,400)) AS Value(v1, v2)

This is due to the fact that the RANK() function examines all rows returned by the query, not only those that the CASE expression has examined. The rows will effectively be ranked into two groups—those that satisfy your criterion (>= 95) by adding the same expression to the PARTITION BY.

Use the RANK() method along with the PARTITION BY clause to partition rows and rank them according to where they are within the partition.

The rank function is applied to specified partitions of rows that have been divided up by the partition by clause. Always after the RANK clause is the OVER() clause. The ORDER BY clause of OVER() must be present. You should include a PARTITION BY clause inside the OVER() clause if you're returning ranks within a partition.

Note: If you're not utilizing partitions, you can just place the ORDER BY phrase in OVER() instead of PARTITION BY.

The empty parenthesis must be given, even if the RANK function does not accept any arguments. The rankings are determined within the subset of rows that each partition defines if the OVER clause includes the optional window PARTITION clause.

Syntax:

The OVER clause has two capabilities, each with its own behavior! Here are the corresponding syntaxes:

OVER (PARTITION BY ...)

OVER (PARTITION BY ... ORDER BY ...)

With the exception of not altering the number of rows, window functions are quite similar to aggregations.

The OVER keyword can be used to organize rows inside a partition or to create groups of rows (aggregate functions) (ranking functions).

Example 1: The keyword OVER can be used to calculate cumulative sums or determine rankings .

SELECT id_intermediary, jurisdiction,     
cnt_entities, sum(cnt_entities) OVER (PARTITION BY id_intermediary) 
AS entities_by_intermediary FROM nb_entities ;

Example 2: You can divide rows in a database that contains information on employees, for instance, according to the departments they are employed in.

A good example is provided below. The developers in the same department are given ranks using the rank function.

SELECT *, rank() OVER 
(partition BY department ORDER BY salary DESC) AS rank_number FROM developers;

The rows in the aforementioned query are first divided into departments. The records in each division are then sorted by salary in descending order by the order by clause.

Example 3: The OVER() clause, which specifies a user-specified set of rows within a query result set, is used by four ranking window procedures. By specifying a column or comma-separated columns to define the partition, you can also include the PARTITION BY clause in the definition of the OVER() clause, which sets the set of rows that the window function will analyze. The ORDER BY clause, which specifies the sorting standards within the partitions via which the function will process the data, may also be added.

The above query can be changed to have more than one partition by adding the PARTITION BY clause, as seen in the T-SQL query below:

SELECT *, 
RANK() OVER(PARTITION BY Student_Score  ORDER BY Student_Score) AS RowNumberRank
FROM StudentScore

The ranking result will have no meaning, as the rank will be done according to Student_Score values per each partition, and the data will be partitioned according to the Student_Score values. And due to the fact that each partition will have rows with the same Student_Score values, the rows with the same Student_Score values in the same partition will be ranked with a value equal to 1. Thus, when moving to the second partition, the rank will be reset, starting again with the number 1, having all ranking values equal to 1.

Let's examine the PARTITION BY clause with many columns as a last example. It doesn't differ all that much from using PARTITION BY with a single column. Look at this:

SELECT
  RANK() OVER(PARTITION BY city, first_name
    ORDER BY exam_date ASC) AS ranking,
  city,
  first_name,
  last_name,
  exam_date
FROM exam_result;

We're using PARTITION BY with two columns in the aforementioned query: city and first name. This implies that we will have separate ranks for each unique combination of city and first name.

Example 1: Using the random() method, divide the retrieved data into two groups (group A and group B in distinct columns); however, random values change each time the query is executed, modifying the rank window function. Although I am aware that you must use "set seed to value;," I am having trouble understanding how. Is it possible to maintain the consistency of the random values even after executing the queries again:

select name,
   random(),
   rank() over (order by random())
from user

Example 2:

SELECT * 
FROM (
SELECT *, rank() OVER (PARTITION BY category ORDER BY random()) as rn
FROM table ) sub
WHERE rn = 1;

Items with lesser values will be ranked lower since the .rank() method ranks data in ascending order (i.e., starting at 1). We can specify the ascending=False argument to reverse this behavior and have the values rank in descending order.

Example 1: Let’s see what this looks like when we rank the the same column in different orders:

# Reversing ranking order of a Pandas Dataframe
df['Score_Ranked_Asc'] = df['Score'].rank()
df['Score_Ranked_Desc'] = df['Score'].rank(ascending=False)
print(df)

# Returns:
#    Name  Count  Score  Score_Ranked_Asc  Score_Ranked_Desc
# 0   Nik    100   22.0               2.0                4.0
# 1  Kate    100   33.0               3.0                3.0
# 2  Evan    105   11.0               1.0                5.0
# 3  Kyra     75    NaN               NaN                NaN
# 4  Piet     75   77.0               4.0                2.0
# 5  Maya    150   99.0               5.0                1.0

Here, we can see that while the missing NaN values are still considered the same, the ranked values start at different ends.

You'll discover how to use the method= argument to rank identical items using various ways in the following section.

Example 2: You just need to reverse your sort order.

SELECT InvoiceNumber, (InvoiceTotal - PaymentTotal - CreditTotal) AS Balance, 
RANK () OVER (ORDER BY (InvoiceTotal - PaymentTotal - CreditTotal) DESC) AS BalanceRank
FROM Invoices
WHERE InvoiceDueDate <= '2012-04-30' AND (InvoiceTotal - PaymentTotal - CreditTotal) > 0;

Example 1: Showing all rows tied for position N may be appropriate. This indicates that all four of the customers who placed eight orders at store 1 will be included in the top three consumers per location. You need the fourth row to have the same row number as the third row in order to do this. This is done by the rank function, so all you have to do is update row_number to be as follows:

with rws as (
select store_id, customer_id, count (*) num_orders,
  rank () over (
  partition by store_id
  order by count(*) desc
  ) rn
  from   co.orders
  group  by store_id, customer_id
)
  select * from rws
  where  rn <= 3
order  by store_id, rn, customer_id;

Example 2: Even if OrderYear can naturally work as a rank value, there is just one difference: RANK() function is introduced to generate a rank value.

The basic set is this. Here, the RANK() method is used to generate a rank value for each product category.

select c.CategoryName,
	year(b.OrderDate) as OrderYear,
	round(sum(a.UnitPrice*a.Quantity*(1-a.Discount)),2) as YearlySales,
	RANK() OVER (PARTITION BY c.CategoryName ORDER BY b.OrderDate) as RankValue
from Order_Details as a
inner join Orders b on a.OrderID = b.OrderID
inner join Products as p on a.ProductID=p.ProductID
inner join Categories as c on c.CategoryID=p.CategoryID
group by c.CategoryName,year(b.OrderDate)
order by c.CategoryName, OrderYear;

Example 1: In a derived table, use rank(). Use the order by feature and partition by CellID according to your specifications. To get the top 10 rows for each category, you filter on rn in the main query.

select T.CellID,
       T.PathID,
       T.Duration
from (
     select T.CellID,
            T.PathID,
            T.Duration,
            rank() over(partition by T.CellID order by T.Duration desc) as rn
     from dbo.YourTable as T
     ) as T
where T.rn <= 10;

Example 2: Two statements with identical semantics perform better when TOP n is specified than when QUALIFY RANK is specified.

SELECT TOP 10 WITH TIES * 
FROM sales ORDER BY county;

SELECT * 
FROM sales 
QUALIFY RANK() OVER (ORDER BY county) <= 10;

Example 3: list of the top 10 salaries along with their ranking. The following query generates the ranking number for you:

SELECT EMPNO, SALARY, 
   RANK() OVER(ORDER BY SALARY DESC),
   DENSE_RANK() OVER(ORDER BY SALARY DESC),
   ROW_NUMBER() OVER(ORDER BY SALARY DESC)  
FROM EMPLOYEE
FETCH FIRST 10 ROWS ONLY;

This query returns the following information.

Table 1. Results of the previous query

EMPNO    SALARY        RANK DENSE_RANK ROW_NUMBER
000010	52,750.00	1	1	1
000110	46,500.00	2	2	2
200010	46,500.00	2	2	3
000020	41,250.00	4	3	4
000050	40,175.00	5	4	5
000030	38,250.00	6	5	6
000070	36,170.00	7	6	7
000060	32,250.00	8	7	8
000220	29,840.00	9	8	9
200220	29,840.00	9	8	10

In this instance, the top 10 salaries were returned in descending order. Each salary's relative ranking is displayed in the RANK column. At position 2, you'll see that there are two rows with the same salary. The same rank value is assigned to each of those rows. The value four is allocated to the subsequent row. For each row, RANK returns a value that is one greater than the sum of the rows that came before it. When there are duplicates, there are gaps in the numerical order.

Example 4: The RANK() OVER window function acts like ROW_NUMBER, but may return more or less than n rows in case of tie conditions, e.g. to return the top-10 youngest persons:

SELECT * FROM (
  SELECT
    RANK() OVER (ORDER BY age ASC) AS ranking,
    person_id,
    person_name,
    age
  FROM person
) AS foo
WHERE ranking <= 10

The code above may yield more than ten rows, for instance, eleven rows if there are two people of the same age.

Main Article :- Sql difference between RANK() and DENSE_RANK() Functions

Create new columns in a table and compare the ranks generated by RANK and DENSE_RANK.

The methods RANK() and DENSE_RANK. With other columns, these functions are employed in SELECT queries. We utilize the OVER() function after RANK or DENSE_RANK, which requires an ORDER BY clause with the name of the column to sort before determining a ranking.

With gaps in the ranking when there are ties, the RANK() function in SQL Server returns the position of a value within the partition of a result set.

With no gaps in the ranking when there are ties, the DENSE_RANK() function in SQL Server returns the position of a value within the partition of a result set.

Main Differences Between RANK and DENSE_RANK

• The SQL language includes a function called rank that aids programmers in classifying various sets of data, while another function called dense rank performs a similar task without omitting any numbers.
• While the goal of dense ranking is to just evaluate the rankings of a single column, the goal of ranking is to analyze the given rank of every room.
• Rank is written as RANK () while dense rank is written as DENSE_RANK ().
• Similar numbers are given the same rank in the ranking, but typically the number after that rank is skipped. However, in dense ranking, no number is left out, and the order of the rankings is strictly numerical.
• While the dense ranking is only appropriate for generating data from a certain collection of columns or rows, the ranking makes it simple to obtain results from large amounts of data.

Example 1: Both RANK and RANK_DENSE work on partitions of data:

SELECT RANK() OVER(PARTITION BY month ORDER BY sold products DESC) AS r,
  DENSE_RANK() OVER(PARTITION BY month ORDER BY sold products DESC) AS dr,
  first_name,
  last_name,
  month,
  sold products
FROM sales_assistant;

Example 2: Let us understand this difference with an example and then observe the results while using these two functions:

To compare the results of two searches, one using RANK() and the other using DENSE_RANK(), we shall run them both. To illustrate the distinction, we will use the NORTHWIND database's ORDERS table. The request will return a list of Customers in ascending order of the most orders each has placed.

Using the RANK() function

SELECT RANK() OVER (ORDER BY TotCnt DESC) AS TopCustomers, 
CustomerID, TotCnt
FROM (SELECT CustomerID, COUNT(*) AS TotCnt
FROM Orders Group BY CustomerID) AS Cust

Using the DENSE_RANK() function

SELECT DENSE_RANK() OVER (ORDER BY TotCnt DESC) AS TopCustomers, 
CustomerID, TotCnt
FROM (SELECT CustomerID, COUNT(*) AS TotCnt
FROM Orders Group BY CustomerID) AS Cust

Example 3:

It all boils down to how these two functions handle identical values, which is where their differences lie. Consider two children who are in the same grade and who both get 90s on their math test.

Depending on where they rank in relation to the other values, RANK and DENSE_RANK will assign the grades the same rank. The next available ranking value will then be skipped by RANK, however DENSE_RANK will continue to use the following chronological ranking value.

With RANK, the next lowest value would be given a rank of 4, passing over 3, if the two 90s are given a ranking of 2. With DENSE_RANK, no values would be skipped and the next-lowest value would be given a rank of 3.

Let’s compare the outcomes of both of these functions.

SELECT student_name, 
RANK() OVER(ORDER BY grades DESC) AS rank_w_rank, DENSE_RANK() 
OVER(ORDER BY grades DESC) AS rank_w_dense_rank

Explanation:

• As you can see once more, rank 2 is not present in the RANK column, whereas rank 2 and rank 4 are present in the DENSE_RANK column, despite there being 5 rows in the table.
• Now that you know how to use RANK and DENSE_RANK, you hopefully know when to utilize each. Normally, I use SQL's DENSE_RANK function as my default rank function. I believe there are more issues if you proceed without skipping a number in the order of chronological ranking.

Ranking or placement of a certain row in relation to other rows in accordance with a specific ordering scheme. There are four ranking functions in SQL Server that can be used to assign ranks. These are NTILE, ROW_NUMBER, RANK, and DENSE_RANK.

To get an increasing integer value, utilize the functions RANK, DENSE_RANK, and ROW_NUMBER. They begin with a value determined by the ORDER BY clause's condition. The ORDER BY clause is necessary for the proper operation of each of these features. If the data is partitioned, each partition's integer counter is reset to 1.

RANK

• The RANK function delivers the same result for each group, but it skips the next group's sequence number.
• The RANK function assigns increment numbers to the results, just like ROW_NUMBER does. However, it gives the identical numbers to the identical tied values. The given numbers are therefore not unique.

DENSE_RANK

• Similar to the RANK function, the DENSE_RANK function continues the sequence value for the subsequent group.
This function is very similar to RANK. Dense Rank, however, does not reveal gaps in the rankings. For instance, DENSE_RANK assigns 1,2,2,2,3 when RANK assigns a ranking of 1,4,4,4,5. There are no jump numbers in the dense ranking results.

ROW_NUMBER

For each row in a result set, an increasing number (1, 2, 3,...) is returned. The final result won't have any redundant ranking values.

The RANK, DENSE_RANK and ROW_NUMBER Functions have the following similarities:

1. An order by clause is necessary for each of them.
2. They all give back a rising integer with a base value of 1.
3. As we've seen, all of these functions reset the returned integer value to 1 when used with a PARTITION BY clause.
4. These functions produce the same results if the column used in the ORDER BY clause does not include any duplicate data.
5. Difference between dense rank, rank(), and row number().
6. Only when there were duplicates will number 6 be revealed.
7. Row number provides consistent ranking even in cases of duplication
8. Both dense rank and rank produce the same ranking, but dense rank lacks a jump whereas rank does.
9. Inside the OVER clause, there should be an ORDER BY clause.
10. Can have PARTITION BY clause inside the OVER clause.
11. The only difference between RANK, DENSE_RANK and ROW_NUMBER function is when there are duplicate values in the column being used in ORDER BY Clause.

Syntax:

Ranking functions are windowed functions. The syntax is the same for ROW_NUMBER, RANK, and DENSE_RANK. However, for NTILE, it has a slightly different syntax, which includes the number_of_groups, as shown in the table below.

ROW_NUMBER, RANK, or DENSE_RANK	function_name()
OVER ([partition_by_clause] order_by_clause)

The syntax uses the following arguments:

• function_name. It can be ROW_NUMBER, RANK, DENSE_RANK, or NTILE.
• partition_by_clause. It separates the results into partitions. This argument is optional (not mandatory).
• order_by_clause. It determines the sorting in which the ranking numbers are assigned to the result set.

Example 1: ROW_NUMBER, RANK, DENSE_RANK are functions in SQL server and returns numeric output by different sequence order.

SELECT 
ROW_NUMBER() OVER(ORDER by Employee_City ASC) AS ROWNUM_CITYWISE,
ROW_NUMBER() OVER(PARTITION BY Employee_City ORDER by Employee_City ASC) AS ROWNUM_PART_CITYWISE,
RANK() OVER(ORDER BY Employee_City ASC) AS RANK_CITYWISE,
DENSE_RANK() OVER(ORDER BY Employee_City ASC) AS DENSRANK_CITYWISE,
Employee_Id,
Employee_Name,
Employee_Address,
Employee_City,
Employee_State,
Employee_Salary
FROM Employee

If the PARTITION clause was applied to the Employee_Address column, then ROW_NUMBER provides a series and unique number for each group, as shown in the result above. If not, ROW_NUMBER returns a unique and sequential value for the entire set of records.

Example 2: When there are duplicate records, the distinction between rank, row_number, and dense_rank is obvious. Since we are rating records in all of our examples based on their salaries, you will be able to tell the difference between these three ranking functions if two records have the same salary.

select e.*,
row_number() over (order by salary desc) row_number, 
rank() over (order by salary desc) rank,
dense_rank() over (order by salary desc) as dense_rank 
from #Employee e

And this output demonstrates the distinction between the rankings produced by the rank() and dense_rank() functions. Your confusion concerning the rank, desnse_rank, and row_number functions will be removed by this.

Example 3: If you go back to the Cars table in the ShowRoom database, you can see it contains lots of duplicate values. Let’s try to find the RANK, DENSE_RANK, and ROW_NUMBER of the Cars1 table ordered by power. Execute the following script:

SELECT name,company, power,
RANK() OVER(ORDER BY power DESC) AS [Rank],
DENSE_RANK() OVER(ORDER BY power DESC) AS [Dense Rank],
ROW_NUMBER() OVER(ORDER BY power DESC) AS [Row Number]
FROM Cars;

Example 1: A ranked list of records prior to a certain date and all records after it. I questioned whether I could use a CASE statement with the RANK function.

select * from (
   select ml.*,
   case
    when [time] < @startTime
    then rank() over(partition by [itemID] order by [timeColumn] desc)
   else 1
   end rk
  from someHistoricTable ml
  where [timeColumn] < @stopTime ) T1
  where rk=1 order by [time] desc

The case/rank bit divides the time, ranking records that were earlier than the @startTime and returning '1' for records that were later. The outer pick then just includes all rows with a "rk" of 1, which are either records during the time period or the row with the greatest ranking prior.

Example 2: You need to add a PARTITION BY to your RANK() function, as in the following:

SELECT v1, 
    v2, 
    CASE 
    WHEN v1 >= 95 
    THEN RANK() OVER (
    PARTITION BY CASE WHEN v1 >= 95 THEN 1 ELSE 2 END
    ORDER BY v2 DESC) 
    END AS Ranking
FROM (VALUES (100, 500), (90, 300), (89,800), (95,400)) AS Value(v1, v2)

This is due to the fact that the RANK() function examines all rows returned by the query, not only those that the CASE expression has evaluated. The rows will effectively be ranked into two groups—those that satisfy your criterion (>= 95)—by adding the same expression to the PARTITION BY.

Example 1: Utilize aggregate functions while ranking. Even while it sounds terrifying, it's actually a rather rational concept when understood properly. Your database initially calculates the aggregate functions before generating a ranking based on the results. Check out this illustration using AVG():

SELECT
  RANK() OVER(ORDER BY AVG(points) DESC) AS ranking,
  city,
  AVG(points) AS average_points
FROM exam_result
GROUP BY city;

As you can see, there isn't much to distinguish this query from the others we've seen so far. In ranking functions, you may easily employ aggregate functions. Use of the GROUP BY clause is crucial to keep in mind. The aggregate functions are computed first, as was already indicated. As a result, when using GROUP BY, you can only use the expressions you're grouping by or aggregate functions.

For instance, you would need to add this other column in the GROUP BY clause if you wanted to use another column for ordering so that the rows are ordered by this other column if the average number of points is the same.

The aforementioned query provides the typical number of points earned by participants in each city.

ranking	city		average_points
1	San Francisco	80
2	San Diego	76
3	Los Angeles	55

Example 2: To group table entries with matching data, use the SELECT query with the GROUP BY clause. This is what we've done to prevent output redundancy. The RANK function's GROUP BY clause will now be used.

After creating the table employee, let's verify the rank function query with the group by clause.

select product_id,
   sum(sold),
   rank () over (order by sum(sold) desc) as rank,
   dense_rank () over (order by sum(sold) desc) as dense_rank
   from sales
   where to_char(time_id, 'yyyy-mm') = '2001-06'
   group by product_id;

To regulate the order in which sources (tables or files) are connected in a dataflow, utilize join rank. To build the join, the highest rated source is accessed first.

Example: In the Query editor, provide the larger table a higher join rank value for a join between two tables and, if practical, cache the smaller table.

SELECT p.id, p.name, COUNT(s.sale) AS sale_count, 
RANK() OVER (ORDER BY COUNT(s.sale)) AS sale_rank FROM people p
JOIN sales s ON s.people_id = p.id
GROUP BY p.id;

Without employing the PARTITION BY Clause, the RANK function in SQL Server is used. The RANK function will treat the entire result set as a single partition and offer consecutive numbering beginning from 1 when we did not mention the PARTITION BY Clause. This is true except when there is a tie.

PARTITION BY clause is optional.

The OVER clause may include a PARTITION BY clause. All of these functions return a returned integer value that is reset to 1 when used with a PARTITION BY clause.

Example 1: Here is an illustration of how to utilize the RANK function without the PARTITION BY clause. Here, the Salary column's Order By Clause is used. As a result, the rank will be determined by the Salary column.

SELECT Name, Department, Salary,
RANK() OVER (ORDER BY Salary DESC) AS [Rank]
FROM Employees

You will receive the following output after running the aforementioned query. As you can see in the output below, there won't be a partition, thus all the rows are assigned with sequential numbers beginning with 1, with the exception of ties, such as when the salary is 8000 and 65000, in which case both rows receive the same rank.

SELECT student_id,
first_name,
last_name,
marks,
RANK() OVER (ORDER BY marks DESC) merit_list
FROM students;

In this instance, we can see that ranking without a partition function does not produce a useful result because all of the rows in the result set are treated as a single group.

Example 2: Without PARTITION BY clause

SELECT Id , Name, Salary, Gender,
RANK() OVER (ORDER BY Salary DESC) AS [Rank],
DENSE_RANK() OVER (ORDER BY Salary DESC) AS DenseRank
FROM Employees

Output:

Id	Name	Gender	Salary	Rank	DenseRank
1	Aark	Male	8000	1	1
2	Micky	Male	8000	1	1
9	Tommy	Male	7000	3	2
10	Ronald	Male	6800	4	3
7	Donald	Male	6500	5	4
6	Mary	Female	6000	6	5
3	Mini	Female	5000	7	6
8	Jodi	Female	4500	8	7
4	Sara	Female	4000	9	8
5	Goffy	Male	3500	10	9

Example 3: RANK without partition

The following sample SQL uses RANK function without PARTITION BY clause:

SELECT TXN.*, RANK() OVER (ORDER BY TXN_DT) AS ROW_RANK FROM VALUES 
(101,10.01, DATE'2021-01-01'),
(101,102.01, DATE'2021-01-01'),
(102,93., DATE'2021-01-01'),
(103,913.1, DATE'2021-01-02'),
(101,900.56, DATE'2021-01-03')
AS TXN(ACCT,AMT, TXN_DT);

Result:

ACCT    AMT     TXN_DT 	        ROW_RANK
101     10.01   2021-01-01      1
101     102.01  2021-01-01      1
102     93.00   2021-01-01      1
103     913.10  2021-01-02      4
101     900.56  2021-01-03      5

Without missing any numbers, rank rows in a table.

Employed RANK to rank a certain group of athletes. However, you'll frequently find multiple groupings in real-world data. Without splitting your data, the ranks of one group will be impacted by the values of another.

Example 1: Please see below example.

Step 1: Table Creation

CREATE TABLE #test(
apples int NOT NULL,
) ON [PRIMARY]
GO

Step 2: Insert Data

insert into #test( apples ) values ( 10 )
insert into #test( apples ) values ( 10 )
insert into #test( apples ) values ( 20 )
insert into #test( apples ) values ( 30 )

Step 3:

select *, RANK() over (order by apples) as theRank from #test
 
drop table #test
go

Output:

Example 2: While RANK skips numerals when two values are same, it is not a natural approach to allocate rankings. Most people rank the nation following them as third if two nations tie for second place.

WITH Athlete_Medals AS (
  SELECT
    Country, Athlete, COUNT(*) AS Medals
  FROM Summer_Medals
  WHERE
    Country IN ('JPN', 'KOR')
    AND Year >= 2000
  GROUP BY Country, Athlete
  HAVING COUNT(*) > 1)

SELECT
  Country,
  -- Rank athletes in each country by the medals they've won
  ___,
  ___ OVER (PARTITION BY ___
                ORDER BY Medals DESC) AS Rank_N
FROM Athlete_Medals
ORDER BY Country ASC, RANK_N ASC;

Example 3: In this post, I want to cover off a little bit of RANK() usage in SQL, but then focus on how to use ranking to get a unique rank number for each row. I will cover both SQL Server and Teradata.

SELECT Region, 
Store, Sales
,RANK() OVER (ORDER BY sales DESC) AS SalesRank
FROM SalesTable

Example 4: Your stores, their sales, and the ranking of each store's sales will be listed here. Let's imagine we only want to display the top 4 stores in each region when we want to display the ranking by region. Your data is "Partitoned," and the outcomes are "Qualified":

SELECT Region, 
Store, Sales
,RANK() OVER (PARTITION BY Region ORDER BY sales DESC) AS SalesRank
FROM SalesTable
QUALIFY SalesRank >= 4

Example 1: A column may be assessed and compared to all other rows using the Ranking function (RANK), either based on high or low order, to produce the output set. By default, the ranking column's descending order, which corresponds to descending rank, will be used to sort the order.

NAME	SUB1	SUB2	SUB3
Ria	45	54	87
Robert	45	44	67
Joel	85	40	67
Roshan	45	94	67
Joldrine 45	44	97

Query:

SELECT NAME
,SUM(SUB1 + SUB2 + SUB3) AS TOTAL_MARKS
,RANK() OVER (ORDER BY TOTAL_MARKS DESC ) AS STUDENT_RANK
FROM STUDENT
GROUP BY 1;

Output:

NAME	TOTAL_MARKS	STUDENT_RANK
Roshan	   206			1
Joel	   192			2
Joldrine   186			3
Ria  	   186			3
Robert	   156			5

Example 2: The RANK function can be included to determine which state received the most packages. Notably, an ORDER BY phrase is no longer used in the main query itself:

SELECT ship_to_state, SUM(Actual_Total_Package_Qty ) AS Total_Packages,
RANK() OVER (ORDER BY SUM(Actual_Total_Package_Qty ) DESC) AS PackageRanking
FROM `handy-bonbon-142723.qvc_sample_data.sample_qvc_data`
GROUP BY ship_to_state

Postgresql's where clause and rank function. The line below uses a CTE and the DENSE_RANK function to return the most expensive item in each item group. In a SELECT statement, the WHERE clause often follows the FROM clause.

Example 1: The WHERE clause will utilize the condition to filter the rows returned by the SELECT clause. Two tables, items and items group, are present. Common table expressions (CTE) in PostgreSQL are used to streamline complex queries.

Since it is a mutual table expression, we can make references to a variety of PostgresqlSQL statements, including SELECT, INSERT, UPDATE, and DELETE. Let's examine the following query.

WITH cte AS(
SELECT item_id,
	item_name,
	group_id,
	price,
	DENSE_RANK () OVER ( 
PARTITION BY group_id
	ORDER BY price DESC
	) price_rank 
	FROM
	items
) 
SELECT item_id, 
item_name, price
FROM 
	cte
WHERE  price_rank = 1;

Example 2: Rank in a Where Clause.

with Cte AS(
  Select DebtorID
        ,rank() over (partition by DebtorID order by BalanceDate) as RankBalanceDate
        ,BalanceDate
        ,Balance
        ,UnallocatedBalance
        ,Overdue
    From Debtorbalances
)
select * 
from Cte
where 
RankBalanceDate = 1;

Example 3:

SELECT DailyValueChange, 
BUSINESS_DATE, RANK() OVER (order by DailyValueChange) AS RANK_Vals
FROM Table
WHERE (BUSINESS_DATE = @CurrentBusDate) AND (RANK_Vals = 100);

It tells me RANK_Vals is an invalid column name when I try to update the stored procedure, which is false because if I run it without the where clase, it executes and provides all results.

Example 4: In this example, the where clause is used to return products with a rank of less than or equal to three for each product ranked by list price in each brand.

SELECT * FROM (
SELECT
	product_id,
	product_name,
	brand_id,
	list_price,
	RANK () OVER ( 
		PARTITION BY brand_id
		ORDER BY list_price DESC
	) price_rank 
	FROM
	production.products
) t
WHERE price_rank <= 3;

Example 5:

The "RANK" function cannot be used directly in the "WHERE" clause because of the logical order in which SQL queries must be executed.

As the "RANK" function may be assigned to the source, I would use the "Construct from query" option with a SQL query to create a new base view by performing the following actions:

Click the "Create base view" button at the top of the selected data source after opening it.

Select the ‘Create from query’ option.

Enter the view name and the SQL query as follows

select < column_name>, 
rank_column from (select < column_name>,
RANK() OVER (PARTITION BY < column_name> ORDER BY < column_name>) rank_column 
from < table_name>) where rank_column =< value>

Click on ‘Save’ to create a base view.

From this base view, I would filter the “RANK” function without creating a selection view on top of this view.

Because it is an iterator, RANKX builds a row context. In The expression must be evaluated using a table. The goal of RANKX is to return the ranking for each row in the table argument by examining each row in the table and conducting its evaluation. In both computed columns and calculated measures, the RANKX function is available.

Syntax

RANKX(<table>, <expression>, <value>, <order>, <ties>)

When there is a tie, the RANKX function is unable to determine which item should be chosen. If we have two rank values of 3, for example, RANKX will return a rank of 3 for both of our values. In the default state of RANKX or if we pass SKIP into the ties argument, take note that there will be a "gap" in the rank values. Since our rank values for this example are 1, 2, 3, and 5, the following rank value will be 5. In our given example, the following rank value after our tied RANKX value of 3 will be 4, since if we pass DENSE in the ties argument, it will not allow for gaps in the value of RANKX.

Example 1: Practical RANKX examples:

Let's begin by looking at the simplest RANKX example that is utilized in a calculated column. Here, we want to order the Total Sales across all of the table's rows. Let's apply the calculation below:

RANX Total Sales =
RANKX(
‘Sales Table’,
‘Sales Table'[Total Sales]
)

Example 2:

Rank =
IF(
    NOT ISBLANK( ‘fact'[Avg Fact]),
    RANKX(CROSSJOIN(ALLSELECTED(‘dim Date'[date].[Month]),ALLSELECTED(‘dim'[Group])),
    [Avg Fact],,DESC)
)