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, usecount(*)
for the total. Faster, simpler. -
This assumes you don’t have
NULL
values. I.e. both columns are definedNOT 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 beforeOR
. 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()
isbigint
by default.
A divisionbigint / 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.