Calculating Multiple Aggregated Values and Their Final Sum in a Single Column

As data analysis becomes increasingly important in various industries, the need for efficient ways to process and visualize data has grown significantly. In this article, we will explore how to calculate multiple aggregated values and their final sum all in one column using Postgres SQL.

Introduction to String Aggregation

String aggregation is a powerful feature in PostgreSQL that allows us to combine multiple string values into a single value. The string_agg function is used to concatenate strings with a specified delimiter. In this example, we will use the comma (,) as our delimiter.

SELECT buildingid, string_agg(distinct cast(obligatioNr as varchar(2)), ', ') as SPJ

Understanding the Problem

We have a table where we perform a string aggregation on a column named obligatioNr. The values of this field range from 0 to 12. Our goal is to calculate the total value of each number and store it in a single column.

Solution Approach

To solve this problem, we will break down the solution into several steps:

1. Grouping: We will group the data by buildingid and then perform string aggregation on the obligatioNr column.
2. Calculating Individual Totals: For each value in the aggregated string, we will calculate its individual total by counting the occurrences of that value.
3. Calculating Final Sum: We will then use these individual totals to calculate the final sum for each building.

Step 1: Grouping and String Aggregation

First, let’s group the data by buildingid and perform string aggregation on the obligatioNr column:

WITH t(v) AS (
    VALUES ('8,9'), 
           ('9,10,11')
),
m AS (SELECT unnest(string_to_array(v, ',')) u FROM t)

Step 2: Calculating Individual Totals

Next, we will calculate the individual total for each value in the aggregated string:

SELECT u || ':' || count(u) from m GROUP BY u ORDER BY u :: int;

This query counts the occurrences of each value u and stores it in a column named a.

Step 3: Calculating Final Sum

Now, we will calculate the final sum for each building by aggregating the individual totals:

SELECT string_agg(a, ',') from (
    SELECT u || ':' || count(u) as a FROM m GROUP BY u ORDER BY u :: int
) as f;

This query uses string_agg to concatenate all values in the column with a comma (,) delimiter.

Step 4: Combining Results

Finally, we will combine the original aggregated string with the final sum:

WITH t(v) AS (
    VALUES ('8,9'), 
           ('9,10,11')
),
m AS (SELECT unnest(string_to_array(v, ',')) u FROM t),
f AS (SELECT u || ':' || count(u) as a FROM m GROUP BY u ORDER BY u :: int)
SELECT v from t UNION ALL SELECT string_agg(a, ',') from f;

This query uses UNION ALL to combine the original aggregated string with the final sum.

Postgres 9.3 Limitation

In this example, we used the string_to_array and unnest functions to split the comma-separated values into individual rows. However, in Postgres 9.3, these functions are not supported.

To work around this limitation, we can use the following approach:

WITH t(v) AS (
    VALUES ('8,9'), 
           ('9,10,11')
),
m AS (SELECT v from t),
f AS (SELECT u || ':' || count(u) as a FROM m, lateral unnest(string_to_array(v, ',')) u GROUP BY u ORDER BY u :: int)
SELECT v from t UNION ALL SELECT string_agg(a, ',') from f;

In this revised query, we use the lateral keyword to access the values in the array returned by unnest. We also removed the GROUP BY clause since it is not necessary when using string_agg.

Conclusion

In conclusion, calculating multiple aggregated values and their final sum all in one column can be achieved using Postgres SQL. By breaking down the solution into several steps, we can efficiently process large datasets and provide meaningful insights.

This article demonstrated how to use string aggregation, individual totals, and final sums to solve this problem. We also explored limitations in earlier versions of Postgres and provided a revised query that works around these limitations.

Last modified on 2024-02-27