Teradata how can I use count(distinct)? - sql

Question

I saw some questions similar but I still couldn't figure out.
There is many columns, but to short, I want to count distinct by g1 column.
I thought I could just use COUNT(DISTINCT)...?
Please help me for this problem.
Thank you so much in advance
G1
C1
expected results
1
A
2
1
A
2
1
B
2
2
A
3
2
B
3
2
A
3
2
C
3
3
A
1
3
A
1
3
A
1
3
A
1

Answer 1

Most (all?) databases do not support using COUNT(DISTINCT ...) as an analytic function. So in this case, I would suggest just joining to a subquery which finds the distinct counts:
SELECT t1.G1, t1.C1, t2.cnt
FROM yourTable t1
INNER JOIN
(
SELECT G1, COUNT(DISTINCT C1) AS cnt
FROM yourTable
GROUP BY G1
) t2
ON t2.G1 = t1.G1
ORDER BY
t1.G1, t1.C1;

Answer 2

You can get a windowed count distinct, but you always need two functions, one possible way is:
SELECT G1, C1, max(dr) over (partition by G1) as cnt
FROM
(
SELECT G1, C1,
dense_rank()
over (partition by G1
order by C1) AS dr
FROM yourTable
) as dt
Depending on your data and actual query this might perform better than Tim's query :-)
Of course, this can be modified for NULLable columns flagging the 1st occurance:
SELECT G1, C1, sum(flag) over (partition by G1) as cnt
FROM
(
SELECT G1, C1,
case
when lag(C1)
over (partition by G1
order by C1) = C1
then null
else C1
end as flag
FROM yourTable
) as dt

Answer 3

You can use the below query:
SELECT count(C1) from TableName Group by G1

Answer 4

If your database doesn't use count(distinct) as a window function, you can use the handy trick of the sum of dense_rank():
select t.*,
(-1 +
dense_rank() over (partition by g1 order by c1 asc) +
dense_rank() over (partition by g1 order by c1 desc)
) as expected
from t;
Given that count(distinct) as a window function is easily implemented this way, I am surprised that many databases do not support this functionality directly.
One nuance: This counts NULL as a valid value. You don't have NULL values in your sample data so I don't think this affects you. But, if you wanted an exact equivalent:
select t.*,
( (case when count(*) over (partition by g1) = count(c1) over (partition by g1)
then -1 else -2
end) +
dense_rank() over (partition by g1 order by c1 asc) +
dense_rank() over (partition by g1 order by c1 desc)
) as expected
from t;