Data Mastery: SQL — Grouping 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 , 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: Data Mastery: SQL Software Engineers who want to analyze their creation’s data Product Managers, Marketers, and others who want to be data-driven Beginning Data Scientists, Data Engineers, Data Analysts, or Business Intelligence Developers preparing for interviews 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. Grouping 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: Into the SELECT statement Into a GROUP BY statement 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. Try it yourself Write a SQL query that finds each user’s average score. Answer 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. Connect with me on Instagram @lauren__glass Connect with me on LinkedIn Check out my essentials list on Amazon Search for me using my nametag on Instagram!
