Impala: change the column type prior to perform the aggregation function for group by - sql

I have a table, my_table:
transaction_id | money | team
--------------------------------------------
1 | 10 | A
2 | 20 | B
3 | null | A
4 | 30 | A
5 | 16 | B
6 | 12 | B
When I group by team, I can compute max, min through query:
select team, max(money), min(money) from my_table group by team
However, I can't do avg and sum because there is null. i.e:
select team, avg(money), sum(money) from my_table group by team
would fail.
Is there a way to change the column type prior to computing the avg and sum? i.e. I want the output to be:
team | avg(money) | sum(money)
--------------------------------------
A | 20 | 40
B | 16 | 48
Thanks!

Per documentation provided by Cloudera your query should be working as-is. Both AVG Function and
SUM Function ignore null.
SELECT team, AVG(money), SUM(money)
FROM my_table
GROUP BY team
UPDATE: Per your comment, again I'm not familiar with Impala. Presumably standard SQL will work. Your error appears to be a datatype issue.
SELECT team, AVG(CAST(money AS INT)), SUM(CAST(money AS INT))
FROM my_table
GROUP BY team

Just divide the sum by the count:
SELECT team, SUM(money)/COUNT(money) AS AVG, SUM(money)
FROM team
GROUP BY team
Tested here: http://sqlfiddle.com/#!9/ba381/4

Related

Stop SQL Select After Sum Reached

My database is Db2 for IBM i.
I have read-only access, so my query must use only basic SQL select commands.
==============================================================
Goal:
I want to select every record in the table until the sum of the amount column exceeds the predetermined limit.
Example:
I want to match every item down the table until the sum of matched values in the "price" column >= $9.00.
The desired result:
Is this possible?
You may use sum analytic function to calculate running total of price and then filter by its value:
with a as (
select
t.*,
sum(price) over(order by salesid asc) as price_rsum
from t
)
select *
from a
where price_rsum <= 9
SALESID | PRICE | PRICE_RSUM
------: | ----: | ---------:
1001 | 5 | 5
1002 | 3 | 8
1003 | 1 | 9
db<>fiddle here

SQL Finding sum of rows and returning count of keys

For a database table looking something like this:
id | year | stint | sv
----+------+-------+---
mk1 | 2001 | 1 | 30
mk1 | 2001 | 2 | 20
ml0 | 1999 | 1 | 43
ml0 | 2000 | 1 | 44
hj2 | 1993 | 1 | 70
I want to get the following output:
count
-------
3
with the conditions being count the number of ids that have a sv > 40 for a single year greater than 1994. If there is more than one stint for the same year, add the sv points and see if > 40.
This is what I have written so far but it is obviously not right:
SELECT COUNT(DISTINCT id),
SUM(sv) as SV
FROM public.pitching
WHERE (year > 1994 AND sv >40);
I know the syntax is completely wrong and some of the conditions' information is missing but I'm not familiar enough with SQL and don't know how to properly do the summing of two rows in the same table with a condition (maybe with a subquery?). Any help would be appreciated! (using postgres)
You could use a nested query to get the aggregations, and wrap that for getting the count. Note that the condition on the sum must be in a having clause:
SELECT COUNT(id)
FROM (
SELECT id,
year,
SUM(sv) as SV
FROM public.pitching
WHERE year > 1994
GROUP BY id,
year
HAVING SUM(sv) > 40 ) years
If an id should only count once even it fulfils the condition in more than one year, then do COUNT(distinct id) instead of COUNT(id)
You can try like following using sum and partition by year.
select count( distinct year) from
(
select year, sum(sv) over (partition by year) s
from public.pitching
where year > 1994
) t where s>40
Online Demo

Calculate time span over a number of records

I have a table that has the following schema:
ID | FirstName | Surname | TransmissionID | CaptureDateTime
1 | Billy | Goat | ABCDEF | 2018-09-20 13:45:01.098
2 | Jonny | Cash | ABCDEF | 2018-09-20 13:45.01.108
3 | Sally | Sue | ABCDEF | 2018-09-20 13:45:01.298
4 | Jermaine | Cole | PQRSTU | 2018-09-20 13:45:01.398
5 | Mike | Smith | PQRSTU | 2018-09-20 13:45:01.498
There are well over 70,000 records and they store logs of transmissions to a web-service. What I'd like to know is how would I go about writing a script that would select the distinct TransmissionID values and also show the timespan between the earliest CaptureDateTime record and the latest record? Essentially I'd like to see what the rate of records the web-service is reading & writing.
Is it even possible to do so in a single SELECT statement or should I just create a stored procedure or report in code? I don't know where to start aside from SELECT DISTINCT TransmissionID for this sort of query.
Here's what I have so far (I'm stuck on the time calculation)
SELECT DISTINCT [TransmissionID],
COUNT(*) as 'Number of records'
FROM [log_table]
GROUP BY [TransmissionID]
HAVING COUNT(*) > 1
Not sure how to get the difference between the first and last record with the same TransmissionID I would like to get a result set like:
TransmissionID | TimeToCompletion | Number of records |
ABCDEF | 2.001 | 5000 |
Simply GROUP BY and use MIN / MAX function to find min/max date in each group and subtract them:
SELECT
TransmissionID,
COUNT(*),
DATEDIFF(second, MIN(CaptureDateTime), MAX(CaptureDateTime))
FROM yourdata
GROUP BY TransmissionID
HAVING COUNT(*) > 1
Use min and max to calculate timespan
SELECT [TransmissionID],
COUNT(*) as 'Number of records',datediff(s,min(CaptureDateTime),max(CaptureDateTime)) as timespan
FROM [log_table]
GROUP BY [TransmissionID]
HAVING COUNT(*) > 1
A method that returns the average time for all transmissionids, even those with only 1 record:
SELECT TransmissionID,
COUNT(*),
DATEDIFF(second, MIN(CaptureDateTime), MAX(CaptureDateTime)) * 1.0 / NULLIF(COUNT(*) - 1, 0)
FROM yourdata
GROUP BY TransmissionID;
Note that you may not actually want the maximum of the capture date for a given transmissionId. You might want the overall maximum in the table -- so you can consider the final period after the most recent record.
If so, this looks like:
SELECT TransmissionID,
COUNT(*),
DATEDIFF(second,
MIN(CaptureDateTime),
MAX(MAX(CaptureDateTime)) OVER ()
) * 1.0 / COUNT(*)
FROM yourdata
GROUP BY TransmissionID;

PostgreSQL return multiple rows with DISTINCT though only latest date per second column

Lets says I have the following database table (date truncated for example only, two 'id_' preix columns join with other tables)...
+-----------+---------+------+--------------------+-------+
| id_table1 | id_tab2 | date | description | price |
+-----------+---------+------+--------------------+-------+
| 1 | 11 | 2014 | man-eating-waffles | 1.46 |
+-----------+---------+------+--------------------+-------+
| 2 | 22 | 2014 | Flying Shoes | 8.99 |
+-----------+---------+------+--------------------+-------+
| 3 | 44 | 2015 | Flying Shoes | 12.99 |
+-----------+---------+------+--------------------+-------+
...and I have a query like the following...
SELECT id, date, description FROM inventory ORDER BY date ASC;
How do I SELECT all the descriptions, but only once each while simultaneously only the latest year for that description? So I need the database query to return the first and last row from the sample data above; the second it not returned because the last row has a later date.
Postgres has something called distinct on. This is usually more efficient than using window functions. So, an alternative method would be:
SELECT distinct on (description) id, date, description
FROM inventory
ORDER BY description, date desc;
The row_number window function should do the trick:
SELECT id, date, description
FROM (SELECT id, date, description,
ROW_NUMBER() OVER (PARTITION BY description
ORDER BY date DESC) AS rn
FROM inventory) t
WHERE rn = 1
ORDER BY date ASC;

MIN() Function in SQL

Need help with Min Function in SQL
I have a table as shown below.
+------------+-------+-------+
| Date_ | Name | Score |
+------------+-------+-------+
| 2012/07/05 | Jack | 1 |
| 2012/07/05 | Jones | 1 |
| 2012/07/06 | Jill | 2 |
| 2012/07/06 | James | 3 |
| 2012/07/07 | Hugo | 1 |
| 2012/07/07 | Jack | 1 |
| 2012/07/07 | Jim | 2 |
+------------+-------+-------+
I would like to get the output like below
+------------+------+-------+
| Date_ | Name | Score |
+------------+------+-------+
| 2012/07/05 | Jack | 1 |
| 2012/07/06 | Jill | 2 |
| 2012/07/07 | Hugo | 1 |
+------------+------+-------+
When I use the MIN() function with just the date and Score column I get the lowest score for each date, which is what I want. I don't care which row is returned if there is a tie in the score for the same date. Trouble starts when I also want name column in the output. I tried a few variation of SQL (i.e min with correlated sub query) but I have no luck getting the output as shown above. Can anyone help please:)
Query is as follows
SELECT DISTINCT
A.USername, A.Date_, A.Score
FROM TestTable AS A
INNER JOIN (SELECT Date_,MIN(Score) AS MinScore
FROM TestTable
GROUP BY Date_) AS B
ON (A.Score = B.MinScore) AND (A.Date_ = B.Date_);
Use this solution:
SELECT a.date_, MIN(name) AS name, a.score
FROM tbl a
INNER JOIN
(
SELECT date_, MIN(score) AS minscore
FROM tbl
GROUP BY date_
) b ON a.date_ = b.date_ AND a.score = b.minscore
GROUP BY a.date_, a.score
SQL-Fiddle Demo
This will get the minimum score per date in the INNER JOIN subselect, which we use to join to the main table. Once we join the subselect, we will only have dates with names having the minimum score (with ties being displayed).
Since we only want one name per date, we then group by date and score, selecting whichever name: MIN(name).
If we want to display the name column, we must use an aggregate function on name to facilitate the GROUP BY on date and score columns, or else it will not work (We could also use MAX() on that column as well).
Please learn about the GROUP BY functionality of RDBMS.
SELECT Date_,Name,MIN(Score)
FROM T
GROUP BY Name
This makes the assumption that EACH NAME and EACH date appears only once, and this will only work for MySQL.
To make it work on other RDBMSs, you need to apply another group function on the Date column, like MAX. MIN. etc
SELECT T.Name, T.Date_, MIN(T.Score) as Score FROM T
GROUP BY T.Date_
Edit: This answer is not corrected as pointed out by JNK in comments
SELECT Date_,MAX(Name),MIN(Score)
FROM T
GROUP BY Date_
Here I am using MAX(NAME), it will pick one name if two names were found with the same goal numbers.
This will find Min score for each day (no duplicates), scored by any player. The name that starts with Z will be picked first than the name that starts with A.
Edit: Fixed by removing group by name