When I have discovered MySQL User-defined variables at the first time, I didn’t pay much attention to them and didn’t realize their strength. At first, one can think that they are just one of the many unnoticed MySQL features. Well, I was wrong. Here, I will try to summarize their strength, though I think there is […]
I just read SQL: Ranking without self join, in which
Shlomi
Noach shares a nice MySQL-specific trick based on user-defined variables to compute rankings.
Shlomi's trick reminds me somewhat of the trick I came across
little over a year ago to caclulate percentiles. At that time, several
people pointed out to me too that using user-defined variables in
this way can be unreliable.The problem with user-defined
variablesSo what is the problem exaclty? Well, whenever a query
assigns to a variable, and that same variable is read in another
part of the query, you're on thin ice. That's because the …
After stepping off of the GROUP_CONCAT()
solution
for calculating quantiles I figured it would be
nice to find a better way to calculate the median too.
Solution
I previously wrote on how to calculate the median using
GROUP_CONCAT()
, but I think that this is a better
way:
[Read more]
SELECT AVG(length) AS median -- take the average of left and right median
, MIN(length) AS left_median --
, MAX(length) AS right_median --
, @l AS left_median_position --
, @r AS right_median_position --
FROM (
SELECT @n, length -- @n is just here to …
My previous post on calculating percentiles with
MySQL generated some comments and good discussion. In particular,
I got some very interesting comments from Vladimir.
Basically, Vladimir was doubtful whether the
GROUP_CONCAT()
solution would be optimal in
comparison to a JOIN
. His proposal is to solve it
like this:
SELECT SUM(g1.r) sr
, g2.length l
, SUM(g1.r)/(SELECT COUNT(*) FROM film) p
FROM (SELECT COUNT(*) r, length FROM film GROUP BY length) g1
JOIN (SELECT COUNT(*) r, length FROM film GROUP BY length) g2
ON g1.length < g2.length
GROUP BY g2.length
HAVING p > 0.9
ORDER BY p
LIMIT 1
First, this query sets up two identical subqueries in …