Grouping and Aggregating

Learning Goals

• Understand what an aggregate function does
• Understand what the group SQL statement does
• Understand how the group statement relates to aggregate functions
• Write aggregate/group queries in SQL
• Write aggregate/group queries in ActiveRecord
• Write aggregate/group/joins queries in ActiveRecord

Vocabulary

• SQL Group
• SQL Aggregate Function
• ActiveRecord Calculation

Recording

This recording will cover the conceptual parts of the lesson, plus some bloopers at the beginning 😁

Aggregate Functions

PostgresSQL comes with some handy built-in aggregate functions. From the Postgresql Docs: Aggregate functions compute a single result from a set of input values. Basically, aggregate functions will do some math for us.

Take a minute to look through the docs and see what the DB can do for you.

While we are talking about functions built in to Postgresql, it is important to note that nearly all relational database systems include some form of aggregating, for example MySQL.

SQL Aggregates

We’ll be working with Set List on the many-to-many-complete branch. Let’s run rails dbconsole in SetList to open up a connection to our Postgres Database. Let’s find the average length of all songs:

select avg(length) from songs;

As with all our SQl queries, this is returning a new table. It has one row and one column, the average length of the songs.

We could also get a count of all our songs:

select count(*) from songs;

get the sum of all play_counts:

select sum(play_count) from songs;

and find the longest length:

select max(length) from songs;

ActiveRecord Calculations

ActiveRecord gives us corresponding “calculations” for those aggregate functions. Take a minute to look through the ActiveRecord Calculations Docs to see what calculations are available.

Let’s open up Rails console with $ rails console and look at the ActiveRecord syntax for the above examples:

Song.average(:length)
#=> 0.14955e4

Song.count
#=> 4

Song.sum(:play_count)
#=> 8172627

Song.maximum(:length)
#=> 4378

Unlike some of the ActiveRecord methods we’ve seen so far, the order that you put the calculations does matter. For instance, if you only want to count songs with at least 10 plays:

Song.count.where("play_count > 10")

You will get this error:

NoMethodError: undefined method `where' for 4:Integer

Once ActiveRecord sees a calculation method, it performs the calculation and returns it. In this case, ActiveRecord does Song.count, returning an integer. It then throws an error when we try to call .where on that integer.

This is the same behavior we have seen with .pluck. In fact, if you look at the Calculation Docs, you’ll see that pluck is a calculation.

Grouping

A sql group by statement will take the rows that you’ve selected using select, from, and where and group them together based on a common attribute. It will then condense these rows together, usually through a calculation. Let’s look at an example. Starting with this abridged form of our songs table which uses only the first 3 artists for simplicity:

Songs Table

In our pqsl session, run:

select artist_id from songs group by artist_id;

This will return this table:

This is not very interesting information, but we can use this example to understand what the group by clause is doing. First, SQl is going to perform the select/from, which will give us this table:

Then, group by will group these rows by the artist_id:

group with artist_id = 1

group with artist_id = 2

group with artist_id = 3

Then, for each of these groups, sql will condense the rows.

group with artist_id = 1

group with artist_id = 2

group with artist_id = 3

.. and so on.

And finally put all these groups back together:

result table

This query is functionally the same as select distinct artist_id from songs;.

Let’s now look at an example that doesn’t work. Instead of selecting the artist_id, we’ll try to select the song title:

select title from songs group by artist_id;

This query will give us this error:

ERROR:  column "songs.title" must appear in the GROUP BY clause or be used in an aggregate function

Let’s break this query down in the same way to see why this isn’t working:

First, we do our select/from:

Next, sql is going to group the rows, but since the artist_id is not part of of this table, we can’t group on it, and so we get our error. We can only group on attributes that are part of the select statement.

Let’s instead try to include the artist_id in the select statement:

select title, artist_id from songs group by artist_id;

And we’ll get the same error. Let’s break this one down. First, our select/from:

Next, sql tries to group the rows. We have the artist_id so we can do this:

groups with artist_id = 1

groups with artist_id = 2

groups with artist_id = 3

Then, we try to condense the rows, but how do we condense them? If we look at the first group, there are two different titles, and sql doesn’t know what to do with them. The information is conflicting. In most cases, we are going to use aggregate functions to condense rows.

Grouping and Aggregating

Whenever we have a group by statement, we are going to need an aggregate function in our select statement to tell sql how we want to condense the rows. For example, we can get an average length of songs for each artist id:

select artist_id, avg(length) from songs group by artist_id;

For our abridged songs table (different from your seeds file), this will return:

Let’s break this query down. First, we do our select statement, which includes the artist_id and the length from songs (we are ignoring the avg for the moment):

Then, sql will group on the artist_id

artist_id = 1

artist_id = 2

artist_id = 3

And so on…

Now sql will try to condense the rows. Like the last example, there is conflicting information (the lengths are different), but this time we have used the avg aggregate function to tell sql how we want to condense that information into a single value:

artist_id = 1

artist_id = 2

artist_id = 3

Finally, sql will put the groups back together:

result table

What if we wanted to count the number of songs for each artist?

select artist_id, count(*) from songs group by artist_id;

Let’s follow our process again. First, the select statement. In this case, our select includes a *, which means everything:

Notice how we selected the artist_id twice. Then, we group on the artist_id:

artist_id = 1

artist_id = 2

artist_id = 3

In this case, there is a lot of conflicting information in our groups, but count knows how to condense them. Just count the rows:

artist_id = 1

artist_id = 2

artist_id = 3

Finally, put the groups back together:

If we don’t want our column to be labeled count, we can use an alias:

select artist_id, count(*) as "count of songs" from songs group by artist_id;

Grouping and Aggregating in ActiveRecord

Once you have a mental model for how you want to interact with your tables, you can translate the sql into the corresponding AR syntax. For example, getting a count of songs for each artist:

Song.group(:artist_id).count
#=> {1=>2, 2=>1, 3=>2}

This returns a hash where the keys are the artist_id and the values are the counts of songs.

If we wanted to get the average length of songs for each artist:

Song.group(:artist_id).average(:length)
#=> {3=>0.59e3, 2=>0.2301e4, 1=>0.4345e3}

As we mentioned before, as soon as ActiveRecord sees a calculation method, it returns the calculation immediately, so if you do something like:

Song.average(:length).group(:artist_id)

You will get:

NoMethodError: undefined method `group' for 0.33175e3:BigDecimal

Joining, Grouping, and Aggregating

If you use group with a calculation method, it will always return a hash with the grouped data and the calculation, but what if we want more than just the raw data? For instance, what if we wanted Artist objects sorted by their average length of song?

Any time we need information from two or more tables, we are going to need to join those tables. In this case, the Artist info is stored in the artists table and the Song lengths are stored in the songs table:

Artist.joins(:songs)

That joined table will look like:

In order to average the song lengths for each artist, we will need to group by the artist’s id:

Artist.joins(:songs).group(:id).average(:length)

Notice how there are two columns called id in this joined table. ActiveRecord will pick the first one, which in this case is the Artist id.

If you run this query, you’ll notice that it’s still returning us a hash. If we want to get our Artist objects, we need to include a select statement with the aggregate avg rather than using the ActiveRecord average method. This is an example of why it’s important to know SQL when writing ActiveRecord queries:

Artist.select("artists.*, avg(length)").joins(:songs).group(:id)

Let’s visualize what this is doing. First, we take the joined table, select the all the Artist columns (id and name) and the length column from Songs:

Group them by the id:

id = 1

id = 2

id = 3

And condense the rows by averaging the length:

id = 1

id = 2

id = 3

And finally, put all the groups back together:

If you run this query in the Rails console, You’ll get this return value:

=> 
[#<Artist:0x000000010685e790
  id: 1,
  name: "Prince",
  created_at: Tue, 29 Oct 2024 18:40:29.667238000 UTC +00:00,
  updated_at: Tue, 29 Oct 2024 18:40:29.667238000 UTC +00:00>,
 #<Artist:0x0000000103850b48
  id: 2,
  name: "Run The Jewels",
  created_at: Tue, 29 Oct 2024 18:40:29.670766000 UTC +00:00,
  updated_at: Tue, 29 Oct 2024 18:40:29.670766000 UTC +00:00>,
 #<Artist:0x0000000103850648
  id: 3,
  name: "Caamp",
  created_at: Tue, 29 Oct 2024 18:40:29.672881000 UTC +00:00,
  updated_at: Tue, 29 Oct 2024 18:40:29.672881000 UTC +00:00>]

We can see our Artists in this ActiveRecord::Relation, but where is our average song length? It’s there, we just can’t see it:

artists = Artist.joins(:songs).select("artists.*, avg(length)").group(:id)
artists.first.avg
#=> 0.4345e3

ActiveRecord is creating a new attribute for our returned Artists objects called avg. This is a default name. We can change it with an alias:

artists = Artist.joins(:songs).select("artists.*, avg(length) as avg_length").group(:id)
artists.first.avg_length
#=> 0.4345e3

It’s important to note that this is not a new attribute for the Artist records in our database. This is a temporary attribute that is created for the objects returned from our query.

Now that we have our average song length for each Artist, we can sort this list:

artists = Artist.joins(:songs).select("artists.*, avg(length) as avg_length").group(:id).order("avg_length")

One ActiveRecord quirk you may run into is if you try to use a symbol rather than a string:

artists = Artist.joins(:songs).select("artists.*, avg(length) as avg_length").group(:id).order(:avg_length)

Will produce:

ActiveRecord::StatementInvalid: PG::UndefinedColumn: ERROR:  column artists.avg_length does not exist

Whenever you use symbol notation, ActiveRecord assumes that you are referring to a colum of the table that corresponds to the Model you started the query with (in this case Artist relates to the artists table). We need to use string notation to tell ActiveRecord to insert the string "avg_length" as-is into our group statement, rather than look for an attribute on our model.

You may also think to group on the artist_id:

Artist.joins(:songs).select("artists.*, avg(length) as avg_length").group(:artist_id).order("avg_length")

This will produce:

ActiveRecord::StatementInvalid: PG::GroupingError: ERROR:  column "artists.id" must appear in the GROUP BY clause or be used in an aggregate function

Remember, we can only group on columns that are part of our select statement. Since we select artists.id, artists.name, and songs.length, we can’t group on artist_id even though the artist_id is the same as artists.id.

Practice Problems

Test your understanding by writing queries for the following in ActiveRecord:

1. What is the length of the longest song?
2. What is the length of each artist’s longest song?
3. What is the name of the artist with the longest song?
4. What is the average length of each artists’ songs?
5. What is the name of the artist with longest average length of song?
6. What are the names of the three artists with the least amount of total plays for all of their songs?

Extra Spicy Problem

1. Write a query to return each artist’s name and a comma separated list of all their songs, for example “Talking Heads” would have “This must be the Place, Heaven”

Checks for Understanding

• What are aggregate functions? Where do they appear in SQL statements?
• What do calculation methods in AR return?
• What does the group by statement do in sql?
• Why do we need to include an aggregate function when using group by?
• When do we need to join?