Home Database Mysql Tutorial How to Calculate Average Values for Different Passes Using SQL in MySQL?

How to Calculate Average Values for Different Passes Using SQL in MySQL?

Oct 30, 2024 pm 06:50 PM

How to Calculate Average Values for Different Passes Using SQL in MySQL?

Calculating Average Values for Different Passes Using SQL

In MySQL, retrieving average values by grouping data can be achieved with the AVG and GROUP BY functions. However, when dealing with multiple criteria, such as calculating averages for different values of a specific column, the task becomes slightly more complex.

In this case, the goal is to extract the average of the 'val' column for each unique 'id' for varying values of 'pass', which may range from 1 to 7.

Subquery Solution

One approach is to use a subquery for each value of 'pass' within the AVG function. This will return a set of values for each 'id' and 'pass' combination.

<code class="sql">SELECT d1.id,
    (SELECT IFNULL(ROUND(AVG(d2.val), 4) ,0) FROM data_r1 d2
     WHERE d2.id = d1.id AND pass = 1) as val_1,
    (SELECT IFNULL(ROUND(AVG(d2.val), 4) ,0) FROM data_r1 d2
     WHERE d2.id = d1.id AND pass = 2) as val_2,
    [...]
    (SELECT IFNULL(ROUND(AVG(d2.val), 4) ,0) FROM data_r1 d2
     WHERE d2.id = d1.id AND pass = 7) as val_7
from data_r1 d1
GROUP BY d1.id</code>
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This query first fetches the unique 'id' values from the 'data_r1' table. Then, for each 'id', it executes seven subqueries to calculate the average value for each possible value of 'pass' and stores them as separate 'val_n' columns. The IFNULL function is used to handle any cases where the subquery returns NULL values, ensuring that the result is a number or zero.

The advantage of this approach is that it returns a single row for each 'id', making it easier to work with the results. However, it can be somewhat verbose, especially if there are many possible values for 'pass'.

Alternate Approach

If you prefer a more concise method, you can use a combination of GROUP BY and CASE statements to achieve a similar result:

<code class="sql">SELECT id,
       SUM(CASE WHEN pass = 1 THEN val END) / COUNT(DISTINCT CASE WHEN pass = 1 THEN id END) AS val_1,
       SUM(CASE WHEN pass = 2 THEN val END) / COUNT(DISTINCT CASE WHEN pass = 2 THEN id END) AS val_2,
       [...]
       SUM(CASE WHEN pass = 7 THEN val END) / COUNT(DISTINCT CASE WHEN pass = 7 THEN id END) AS val_7
FROM data_r1
GROUP BY id</code>
Copy after login

This query uses multiple CASE statements to evaluate the 'pass' value and only include the corresponding 'val' values in the calculation. The COUNT(DISTINCT ...) expressions in the denominator ensure that the average is calculated correctly, even if there are multiple 'val' values for a specific 'id' and 'pass' combination.

Once again, the IFNULL function can be used to handle any cases where the division results in NULL to ensure that the output is a number or zero.

Whichever approach you choose, these queries will effectively extract the average 'val' values for each unique 'id' for the varying values of 'pass' from the 'data_r1' table.

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