MySQL and average calculation giving different results - mysql

Question

I've created a test table called test with some sample data that looks like the following:
if I add-up each user's value and divide it by the total number of entries there is for that user_id and multiply it by 100, I get a percentage. So for example
for user_id 1 there are a total of 6 records and the values add up to 4. So 4/6*100 = 66.67%
for user_id 2 there are a total of 8 records and the values add up to 6. So 6/8*100 = 75%
I am able to get these values using the following SQL:
SELECT a.user_id, a.total_present / (SELECT count(*) as total_sessions FROM test WHERE session_date BETWEEN '2017-10-01' AND '2017-10-15' AND user_id = a.user_id) * 100 AS attendance_percentage, a.total_present
FROM (
SELECT user_id, count(*) as total_present
FROM test t
WHERE t.session_date BETWEEN '2017-10-01' AND '2017-10-15' AND t.value=1
GROUP BY user_id
) a ORDER BY a.user_id ASC
If I add the percentages up and divide by how many users then I get an average of 70.83%
I was thinking to simplify this simply just by summing up all the records and divide by total number of records and multiplying by 100 like this:
SELECT ((SELECT count(*) as total_present FROM test t WHERE t.session_date BETWEEN '2017-10-01' AND '2017-10-15' AND t.value=1) / (SELECT count(*) as total_sessions FROM test WHERE session_date BETWEEN '2017-10-01' AND '2017-10-15')) * 100 AS average_percentage
This however does not give me the same average calculation because each user has a different number of total sessions. If both users had the same amount of sessions then it matches.
So the question is, is there a way to calculate this using my simplified method but get the result more accurate to the one where I am getting each user percentage and then adding it up and dividing by total users?

Answer 1

You are complicating your query. To get the attendance_percentage by user, you can simply take the avg of value(since its already in binary) and multiply it by 100. Here is the query for that.
SELECT user_id, sum(value) as total_present, avg(value)*100 as attendance_percentage
FROM test t
WHERE t.session_date BETWEEN '2017-10-01' AND '2017-10-15'
GROUP BY user_id
And finally to calculate the average attendance_percantage percentage, just take the avg of result from above query. Like this.
select avg(attendance_percentage) from
(SELECT user_id, sum(value) as total_present, avg(value)*100 as attendance_percentage
FROM test t
WHERE t.session_date BETWEEN '2017-10-01' AND '2017-10-15'
GROUP BY user_id)q;
I have also created this sqlfiddle for you so that you can test the queries.
Edit: In case your value is not in binary, you can use your original query to find user wise attendance_percentage and simply take the avg of it using the Avg() function. So your query changes to this.
select avg(attendance_percentage) from
(SELECT a.user_id, a.total_present / (SELECT count(*) as total_sessions FROM test WHERE session_date BETWEEN '2017-10-01' AND '2017-10-15' AND user_id = a.user_id) * 100 AS attendance_percentage, a.total_present
FROM (
SELECT user_id, count(*) as total_present
FROM test t
WHERE t.session_date BETWEEN '2017-10-01' AND '2017-10-15' AND t.value=1
GROUP BY user_id
) a ORDER BY a.user_id ASC)q.
Here is the updated fiddle.

Answer 2

I would suggest you to prevent using sub-queries that easily when is not a good reason for that. To be honest, most of the times you can do that without them as well as sub-queries are usually bad on performance and not scalable at all.
However, I think your query would look something like this:
SELECT user_id, sum(IF(t.value=1,1,0)) as total_present, (total_present/count(*))*100 AS attendance_percentage
FROM test t
WHERE t.session_date BETWEEN '2017-10-01' AND '2017-10-15'
GROUP BY user_id
I hope that helps