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How to Calculate the Percentage of Matching Values in a SQL SELECT Query?
How to Calculate the Percentage of Matching Values in a SQL SELECT Query?
Calculate Percentages from SUM() in the Same SELECT Query
In the my_obj table, where the integer fields value_a and value_b are present, the goal is to determine the percentage of times value_a equals value_b.
Incorrect Attempt:
select sum(case when o.value_a = o.value_b then 1 else 0 end) as nb_ok,
sum(case when o.value_a != o.value_b then 1 else 0 end) as nb_not_ok,
compute_percent(nb_ok,nb_not_ok)
from my_obj as o
group by o.property_name;This approach fails because the nb_ok column does not exist within the query.
Optimized Solution:
SELECT property_name
,(count(value_a = value_b OR NULL) * 100) / count(*) AS pct
FROM my_obj
GROUP BY 1;Explanation:
- The operator precedence of = over OR ensures that the comparison is evaluated first.
- count(value_a = value_b OR NULL) ensures that NULL values are ignored.
- The division of (count(value_a = value_b OR NULL) * 100) by count(*) calculates the percentage.
- count(*) represents the total number of rows, including NULL values.
Result:
property_name | pct --------------+---- prop_1 | 17 prop_2 | 43
Alternative for Fractional Digits:
SELECT property_name
,round((count(value_a = value_b OR NULL) * 100.0) / count(*), 2) AS pct
FROM my_obj
GROUP BY 1;This variation introduces a fractional digit to preserve decimal places.
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