Question

I’m trying to write a SQL or ActiveRecord query to answer this question: “Of people who took at least one ride, what is the average number of metro lines that they’ve departed from?”

The schema is as follows:

• Ride: A trip from one location to another. user_id for the User, location_id for the Location.
• Location: A location is a stop along a line. The location has a line_id indicating what line it’s on. A location belongs to one line.
• Line: A line is a series of related metro stops, its locations.
• User: The person who took the trip.

It looks like I need to do two things:

• Given the Rides joined to their locations, count the number of distinct [rides.user_id, locations.line_id] combinations.
• Divide that by the number of users who’ve taken at least one ride.

The result will be the average, and thus the answer to the question.

1. Does that sound right?
2. If so, what’s the best way to do that?

I’m using Rails, so if I can express this in ARel or AR syntax without having to drop into SQL, that would be great. But I’ll take what I can get.

---

A commenter asked for an example. Let’s imagine that the data looks something like this:

rides                     locations
======================    =====================
user_id    location_id    location_id   line_id
1          1              1             1
1          1              2             1
1          1              3             2
1          1              4             3
2          1              5             4
2          2              6             5
2          3
3          3
3          4
3          5
3          6

We can see that user 1 took 4 rides, user 2 took 3 rides, and user 3 took 4 rides. These rides were to [1, 3, 4] distinct locations, but only [1, 2, 4] distinct lines. Thus, the average number of lines that a given user rode was (1 + 2 + 4) / 3, or 2.33....

Answer 1

SQL:

select avg(ln) as avg_lines
from (
    select user_id, count(distinct line_id) as ln
    from rides r join locations l on (r.location_id = l.location_id)
    group by user_id
)