MySQL / BigQuery - Weighted Average & Group By - mysql

Question

I am trying to calculate a weighted average of a dataset and return the maximum value, monthly over a period of 12 months along with its' corrosponding ticket description.
I'm aware that there are tons of questions out there addressing similar problems, but I have yet to find a solution that combines the syntaxes I believe are required.
Here's some sample table data:
Month_Begin_Date
Priority
ticket_about_tag
Phone_Time
Occurances
2019-02-01
Urgent
Power Bill
22.42
36
2019-02-01
Normal
Power Bill
3.41
89
2019-05-01
Normal
Wifi Issue
45.32
12
Here's my current query for determining the weighted average:
SELECT (Month_Begin_Date,
(sum(phone_time * occurances))/sum(occurances)) AS Weighted_Average_Phone_Time
FROM database
GROUP BY month_begin_date
This returns the weighted average total for all ticket_about_tags, monthly.
But I still need to get this so that it displays the maximum weighted average grouped by ticket description. I.e. something that looks like this:
Month_Begin_Date
ticket_about_tag
Weighted_average_phone_time
2019-01-01
Power Bill
22.42
2019-02-01
Power Bill
3.41
2019-03-01
Wifi Issue
45.32
I've tried adding this as a subquery into another query in order to return the data I'm after, like so:
SELECT Month_Begin_date, Ticket_About_Tag, Phone_Average_Handle_Time
FROM database WHERE CONCAT(month_begin_date,phone_time) IN
(SELECT CONCAT (Month_Begin_Date,
(sum(phone_time * occurances))/sum(occurances)) AS Weighted_Average_Phone_Time
FROM database
GROUP BY month_begin_date
)
ORDER BY month_begin_date ASC
Thanks very much for any assistance

Answer 1

Not sure I got your question right, but using the following data:
Month_Begin_Date
Priority
Ticket_About_Tag
Phone_Time
Occurences
2019-02-01
Urgent
Power Bill
22.42
36
2019-02-01
Normal
Power Bill
3.41
89
2019-05-01
Normal
Wifi Issue
45.32
12
2019-02-01
Urgent
Wifi Issue
14.2
7
2019-02-01
Normal
Wifi Issue
30.7
5
Is this the query you're after?
SELECT
Month_Begin_Date, Ticket_About_Tag,
SUM(Phone_Time * Occurences) / SUM(Occurences) AS Weighted_Average_Phone_Time
FROM `database`
GROUP BY Month_Begin_Date, Ticket_About_Tag
ORDER BY Month_Begin_Date ASC, Ticket_About_Tag ASC;
That gives you a result like the one you posted:
Month_Begin_Date
Ticket_About_Tag
Weighted_Average_Phone_Time
2019-02-01
Power Bill
8.884880083084106
2019-02-01
Wifi Issue
21.075000206629436
2019-05-01
Wifi Issue
45.31999969482422
Response to your comment
To answer your comment you could:
SELECT
a.Month_Begin_Date,
a.Ticket_About_Tag,
b.Max_Weighted_Average_Phone_Time
FROM (
SELECT
Month_Begin_Date,
Ticket_About_Tag,
SUM(Phone_Time * Occurences) / SUM(Occurences) AS Weighted_Average_Phone_Time
FROM `database`
GROUP BY Month_Begin_Date, Ticket_About_Tag
) a
LEFT JOIN (
SELECT
b1.Month_Begin_Date,
MAX(b1.Weighted_Average_Phone_Time) AS Max_Weighted_Average_Phone_Time
FROM (
SELECT
Month_Begin_Date,
Ticket_About_Tag,
SUM(Phone_Time * Occurences) / SUM(Occurences) AS Weighted_Average_Phone_Time
FROM `database`
GROUP BY Month_Begin_Date, Ticket_About_Tag
) b1
GROUP BY b1.Month_Begin_Date
) b ON a.Month_Begin_Date = b.Month_Begin_Date
WHERE a.Weighted_Average_Phone_Time = b.Max_Weighted_Average_Phone_Time
That gives you the following output:
Month_Begin_Date
Ticket_About_Tag
Max_Weighted_Average_Phone_Time
2019-02-01
Wifi Issue
21.075000206629436
2019-05-01
Wifi Issue
45.31999969482422
There are other ways of doing this, but I think this is by far the easiest way to understand without using other SQL constructs. It reflects your need of going through the same data twice, first to aggregate by month and ticket tag, then to find the maximum of the aggregate data by month.