There are so many exciting projects out there in the Data World. Artificial Intelligence, Machine Learning, Neural Nets, Blockchain, and more are sweeping the technology industry. In order to get to the cutting-edge stuff, first and foremost, data needs to be stored, evaluated, and tested. The best place to do that is SQL (or a library that operates with SQL-like commands, see my article on Python’s Pandas library).
This series Data Mastery: SQL, will teach you the essential subjects. These are not exhaustive tutorials. Instead they are focused preparation guides — with brevity and efficiency in mind. It is meant for:
See my resource list of the books I used to prepare for my big interview
Each article will contain a brief technical explanation of the subject, an example question, and an answer. Follow up articles will contain challenging questions so you can evaluate your learning.
This series does not come with accompanying data sets. The advantage to this is when you are on the drawing board, whether in an interview or project design, you do not have test data to play with. You have to think abstract.
The next level of using aggregation functions is when you want to see a breakdown. We will put the column name of the breakdown in two places in the query:
Let’s recall our daily_user_score table from earlier in the series:
date | userid | sessionid | score
------------------------------------------
2018–09–01 | 983489272 | 125 | 112
2018–09–01 | 234342423 | 34 | 112
2018–09–01 | 567584329 | 207 | 618
2018–09–02 | 983489272 | 126 | 410
2018–09–02 | 983489272 | 127 | 339
Say we wanted to focus on the number of users and their average score per day. The query will look like this:
SELECT date,
COUNT(DISTINCT userid) AS number_of_users,
AVG(score) AS average_score,
FROM daily_user_score
GROUP BY date;
This query returns:
date | number_of_users | average_score
-----------------------------------------------
2018–09–01 | 3 | 280.66
2018–09–01 | 1 | 374.5
It is important to know that GROUP BY includes the functionality of SELECT DISTINCT. If you put a column name in the GROUP BY statement, the results will only include unique values.
Write a SQL query that finds each user’s average score.
SELECT userid, AVG(score) AS average_score
FROM daily_user_score
GROUP BY userid;
This query returns:
userid | average_score
----------------------
983489272 | 287
234342423 | 112
567584329 | 618
Thanks for reading! If you have questions feel free to comment & I will try to get back to you.
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