Trying to make a slope calculation work with SQL syntaxes - netezza

Question

I'm a relatively new SQL programmer, I'm trying to make the below code work in SQL. The code is for calculating slopes for a given set of data, following the exact same logic as EXCEL SLOPE function. The problem right now is that count are not allowed since aggregations are being nested. But if I create a subquery for counts and sums, i would have to group on x and y, otherwise I wouldn't have x and y in my outer query to calculate on.
CREATE TABLE TEST (X FLOAT, Y FLOAT);
INSERT INTO TEST (X, Y) VALUES (1,4.10242258729964);
INSERT INTO TEST (X, Y) VALUES (2,4.57708865242591);
INSERT INTO TEST (X, Y) VALUES (3,5.16785670619896);
INSERT INTO TEST (X, Y) VALUES (4,6.88149559336059);
select sum((x-sum(x)/count(x))^2)/sum(((x-sum(x)/count(x))*(y-sum(y)/count(y))))
from TEST

Answer 1

You can calculate the slope from sum(x * x) and sum(x * y) and avg(x) and avg(y) and n:
SELECT avg(x) AS mx,sum(x*x) AS sx2,sum(x*y) as sxy,avg(y) as my, count(x) AS n
FROM test
Then you can use:
SELECT (sxy-n*mx*my)/(sx2 - n* mx*mx)
FROM
( SELECT avg(x) AS mx,sum(x*x) AS sx2,sum(x*y) as sxy,avg(y) as my, count(x) AS n
FROM test
)

Answer 2

Here a working version of the SQL you created:
SELECT sum((x-avgx)*(x-avgx)) / sum((x-avgx)*(y-avgy))
FROM TEST, (SELECT sum(X)/count(X) as avgx, sum(Y)/count(Y) as avgy FROM TEST) average;
I looked up the excel slope function and it is defined a bit differently:
SELECT sum((x-avgx)*(y-avgy)) / sum((x-avgx)*(x-avgx))
FROM TEST,
(
SELECT
sum(X)/count(X) as avgx,
sum(Y)/count(Y) as avgy
FROM TEST
) average;
Hope this is what you needed :)

Answer 3

I usually do something along these lines (I kept the syntax as simple as possible to avoid anything that might be problematic, like CTEs):
CREATE TABLE #test (x FLOAT, y FLOAT);
INSERT INTO #test SELECT 1, 4.10242258729964;
INSERT INTO #test SELECT 2, 4.57708865242591;
INSERT INTO #test SELECT 3, 5.16785670619896;
INSERT INTO #test SELECT 4, 6.88149559336059;
SELECT
(N * SUM_XY - SUM_X * SUM_Y) / (N * SUM_X2 - SUM_X * SUM_X) AS slope
FROM
(
SELECT
COUNT(*) AS N,
SUM(x) AS SUM_X,
SUM(x * x) AS SUM_X2,
SUM(y) AS SUM_Y,
SUM(y * y) AS SUM_Y2,
SUM(x * y) AS SUM_XY
FROM
#test) a;
Just ran this, then noticed you had another answer from "SQL Hacks". I ran both versions and they get the exact same answer, but the other version is shorter :D