Compute percents from SUM() in the same SELECT sql query - sql

Question

In the table my_obj there are two integer fields:
(value_a integer, value_b integer);
I try to compute how many time value_a = value_b, and I want to express this ratio in percents.
This is the code I have tried:
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;
compute_percent is a stored_procedure that simply does (a * 100) / (a + b)
But PostgreSQL complains that the column nb_ok doesn't exist.
How would you do that properly ?
I use PostgreSQL 9.1 with Ubuntu 12.04.

Answer 1

There is more to this question than it may seem.
Simple version
This is much faster and simpler:
SELECT property_name
,(count(value_a = value_b OR NULL) * 100) / count(*) AS pct
FROM my_obj
GROUP BY 1;
Result:
property_name | pct
--------------+----
prop_1 | 17
prop_2 | 43
How?
You don't need a function for this at all.
Instead of counting value_b (which you don't need to begin with) and calculating the total, use count(*) for the total. Faster, simpler.
This assumes you don't have NULL values. I.e. both columns are defined NOT NULL. The information is missing in your question.
If not, your original query is probably not doing what you think it does. If any of the values is NULL, your version does not count that row at all. You could even provoke a division-by-zero exception this way.
This version works with NULL, too. count(*) produces the count of all rows, regardless of values.
Here's how the count works:
TRUE OR NULL = TRUE
FALSE OR NULL = NULL
count() ignores NULL values. Voilá.
Operator precedence governs that = binds before OR. You could add parentheses to make it clearer:
count ((value_a = value_b) OR FALSE)
You can do the same with
count NULLIF(<expression>, FALSE)
The result type of count() is bigint by default.
A division bigint / bigint, truncates fractional digits.
Include fractional digits
Use 100.0 (with fractional digit) to force the calculation to be numeric and thereby preserve fractional digits.
You may want to use round() with this:
SELECT property_name
,round((count(value_a = value_b OR NULL) * 100.0) / count(*), 2) AS pct
FROM my_obj
GROUP BY 1;
Result:
property_name | pct
--------------+-------
prop_1 | 17.23
prop_2 | 43.09
As an aside:
I use value_a instead of valueA. Don't use unquoted mixed-case identifiers in PostgreSQL. I have seen too many desperate question coming from this folly. If you wonder what I am talking about, read the chapter Identifiers and Key Words in the manual.

Answer 2

Probably the easiest way to do is to just use a with clause
WITH data
AS (SELECT Sum(CASE WHEN o.valuea = o.valueb THEN 1 ELSE 0 END) AS nbOk,
Sum(CASE WHEN o.valuea != o.valueb THEN 1 ELSE 0 END) AS nbNotOk,
FROM my_obj AS o
GROUP BY o.property_name)
SELECT nbok,
nbnotok,
Compute_percent(nbok, nbnotok)
FROM data

Answer 3

You might also want to try this version:
WITH all(count) as (SELECT COUNT(*)
FROM my_obj),
matching(count) as (SELECT COUNT(*)
FROM my_obj
WHERE valueA = valueB)
SELECT nbOk, nbNotOk, Compute_percent(nbOk, nbNotOk)
FROM (SELECT matching.count as nbOk, all.count - matching.count as nbNotOk
FROM all
CROSS JOIN matching) data