How to return average of each condition in SQL

Question

I am struggling with this SQL question for several days. I am quite new to SQL. Really appreciate your time and effort.

Q: returns the average arrival delay time for each day of the week.

Expect results :

+--------------+---------------+
| weekday_name |   avg_delay   |
+--------------+---------------+
|    Friday    | 14.4520127056 |
|    Monday    | 10.5375015249 |
|   Thursday   | 8.47985564693 |
|  Wednesday   |  8.4561902339 |
|   Saturday   | 7.54455459234 |
|   Tuesday    | 4.63152453983 |
|    Sunday    | 4.21165978081 |
+--------------+---------------+

I already have two table ready: flight_delays and weekdayName My sql code in ipython book:

SELECT distinct w.weekday_name, AVG(f.arr_delay) as average_delay
FROM flight_delays as f, weekdayName as w
WHERE f.day_of_week = w.dayofweek and w.dayofweek <= 7
ORDER BY AVG(arr_delay)

it only returns:

weekday_name    average_delay
Sunday          8.295147670495197

So it actually average all seven days' results. But I want to average results of each day. Could you please explain where is my mistake. Thanks a lot.

Answer 1

First, learn to use proper join and group by syntax. Also, I don't think and w.dayofweek <= 7 is needed.

Does this do what you want?

SELECT w.weekday_name, AVG(f.arr_delay) as average_delay
FROM flight_delays f join
     weekdayName w
     on f.day_of_week = w.dayofweek 
GROUP BY w.weekday_name
ORDER BY AVG(arr_delay)

Answer 2

Made some adjustments:

SELECT w.weekday_name, AVG(f.arr_delay) as average_delay
FROM flight_delays f INNER JOIN weekdayName w
ON f.day_of_week = w.dayofweek
GROUP BY w.weekday_name
ORDER BY AVG(arr_delay);

If you are doing aggregation(here AVG ) in your SELECT , you need to provide non-aggregation fields in the GROUP BY .