SQL Group Byチュートリアル:カウント、合計、平均、および句の説明

このGROUP BY条項は強力ですが、考えるのが難しい場合もあります。

8年後でも、使用するたびGROUP BYに立ち止まって、実際に何をしているのかを考えなければなりません。

この記事では、GROUP BY句を作成する方法、句がクエリに対してどのように機能するか、および句を使用して集計を実行し、データに関する洞察を収集する方法について説明します。

これが私たちがカバーするものです:

  • データベースの設定
  • サンプルデータの設定(販売の作成)
  • どのように機能しGROUP BYますか?
  • GROUP BY節を書く
  • 集計(COUNTSUMAVG
  • 複数のグループでの作業
  • で関数を使用する GROUP BY
  • でグループをフィルタリングする HAVING
  • 暗黙的なグループ化を使用した集計

データベースの設定

クエリを作成する前に、データベースを設定する必要があります。

これらの例ではPostgreSQLを使用しますが、ここに示すクエリと概念は、他の最新のデータベースシステム(MySQL、SQL Serverなど)に簡単に変換できます。

PostgreSQLデータベースを操作するには、インタラクティブなPostgreSQLコマンドラインプログラムであるpsqlを使用できます。一緒に仕事を楽しんでいる別のデータベースクライアントがある場合も問題ありません。

まず、データベースを作成しましょう。PostgreSQLがすでにインストールされているので、createdb ターミナルでコマンドを実行して新しいデータベースを作成できます。私は私のと呼んだfcc

$ createdb fcc 

次に、コマンドを使用してインタラクティブコンソールを起動し、次psqlを使用して作成したデータベースに接続します\c

$ psql psql (11.5) Type "help" for help. john=# \c fcc You are now connected to database "fcc" as user "john". fcc=# 
注:psqlこれらの例の出力は読みやすくするためにクリーンアップしたので、ここに表示されている出力が端末で表示されているものと正確に一致していなくても心配しないでください。

これらの例に従って、これらのクエリを自分で実行することをお勧めします。これらの例を読むだけでなく、それらを通して作業することで、はるかに多くのことを学び、覚えることができます。

データの設定(販売の作成)

この例では、さまざまな店舗の場所でのさまざまな製品の販売記録を格納するテーブルを使用します。

このテーブルsalesをと呼びます。これは、店舗の売上を簡単に表したものです。場所の名前、製品名、価格、および販売された時刻です。

実際のアプリケーションでこのテーブルを作成する場合は、他のテーブル(locationsまたはなどproducts)への外部キーを設定します。ただし、GROUP BY概念を説明するために、単純なTEXT列を使用します。

テーブルを作成して、いくつかの販売データを挿入しましょう。

CREATE TABLE sales( location TEXT, product TEXT, price DECIMAL, sold_at TIMESTAMP ); INSERT INTO sales(location, product, price, sold_at) VALUES ('HQ', 'Coffee', 2, NOW()), ('HQ', 'Coffee', 2, NOW() - INTERVAL '1 hour'), ('Downtown', 'Bagel', 3, NOW() - INTERVAL '2 hour'), ('Downtown', 'Coffee', 2, NOW() - INTERVAL '1 day'), ('HQ', 'Bagel', 2, NOW() - INTERVAL '2 day'), ('1st Street', 'Bagel', 3, NOW() - INTERVAL '2 day' - INTERVAL '1 hour'), ('1st Street', 'Coffee', 2, NOW() - INTERVAL '3 day'), ('HQ', 'Bagel', 3, NOW() - INTERVAL '3 day' - INTERVAL '1 hour'); 

HQDowntown1stStreetの3つの場所があります

CoffeeBagelの2つの製品があり、これらの売上を異なるsold_at値で挿入して、異なる日時に販売されているアイテムを表します。

今日、昨日、一昨日からの販売があります。

どのように機能しGROUP BYますか?

GROUP BY句がどのように機能するかを説明するために、最初に例を見てみましょう。

さまざまな国で生まれた人々でいっぱいの部屋があったと想像してみてください。

国ごとの部屋の人々の平均身長を知りたい場合は最初にこれらの人々に出生国に基づいてグループに分けるように依頼します。

それらがグループに分けられると、そのグループ内の平均身長を計算できます。

これがGROUP BY条項の仕組みです。最初に、行をグループ化する方法を定義します。次に、グループに対して計算または集計を実行できます。

複数のグループ

データは、必要な数のグループまたはサブグループにグループ化できます。

たとえば、出生国に基づいてグループに分けるように人々に求めた後、それらの国の各グループに、目の色に基づいてさらにグループに分けるように指示することができます

これにより、生まれた国目の色の組み合わせに基づいた人々のグループができます。

これで、これらの小さなグループのそれぞれの平均身長を見つけることができ、より具体的な結果が得られます。国ごと、目の色ごとの平均身長です。

GROUP BY句は、多くの場合、あなたはフレーズを使用することができるような状況のために使用されているあたりに何かまたはそれぞれに何か

  • 出生国あたりの平均身長
  • 目と髪の色の組み合わせごとの合計人数
  • 製品あたりの総売上高

GROUP BY節を書く

GROUP BY句は非常に簡単です、ライト、我々はちょうどキーワードを使用しGROUP BY、我々はによってグループ化するフィールド(複数可)を指定した後と:

SELECT ... FROM sales GROUP BY location;

この単純なクエリは、salesデータをlocation列ごとにグループ化します。

グループ化は完了しましたが、何を入力しますSELECTか?

選択するのは明らかです。locationグループ化するので、少なくとも作成したグループの名前を確認したいと思います。

SELECT location FROM sales GROUP BY location; 

結果は私たちの3つの場所です:

 location ------------ 1st Street HQ Downtown (3 rows) 

If we look at our raw table data (SELECT * FROM sales;), we'll see that we have four rows with a location of HQ, two rows with a location of Downtown, and two rows with a location of 1st Street:

 product | location | price | sold_at ---------+------------+-------+---------------------------- Coffee | HQ | 2 | 2020-09-01 09:42:33.085995 Coffee | HQ | 2 | 2020-09-01 08:42:33.085995 Bagel | Downtown | 3 | 2020-09-01 07:42:33.085995 Coffee | Downtown | 2 | 2020-08-31 09:42:33.085995 Bagel | HQ | 2 | 2020-08-30 09:42:33.085995 Bagel | 1st Street | 3 | 2020-08-30 08:42:33.085995 Coffee | 1st Street | 2 | 2020-08-29 09:42:33.085995 Bagel | HQ | 3 | 2020-08-29 08:42:33.085995 (8 rows) 

By grouping on the location column, our database takes these inputs rows and identifies the unique locations among them—these unique locations serve as our "groups."

But what about the other columns in our table?

If we try to select a column like product that we didn't group by...

SELECT location, product FROM sales GROUP BY location; 

...we run into this error:

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

The problem here is we've taken eight rows and squished or distilled them down to three.

We can't just return the rest of the columns like normal—we had eight rows, and now we have three.

What do we do with the remaining five rows of data? Which of the eight rows' data should be displayed on these three distinct location rows?

There's not a clear and definitive answer here.

To use the rest of our table data, we also have to distill the data from these remaining columns down into our three location groups.

This means that we have to aggregate or perform a calculation to produce some kind of summary information about our remaining data.

Aggregations (COUNT, SUM, AVG)

Once we've decided how to group our data, we can then perform aggregations on the remaining columns.

These are things like counting the number of rows per group, summing a particular value across the group, or averaging information within the group.

To start, let's find the number of sales per location.

Since each record in our sales table is one sale, the number of sales per location would be the number of rows within each location group.

To do this we'll use the aggregate function COUNT() to count the number of rows within each group:

SELECT location, COUNT(*) AS number_of_sales FROM sales GROUP BY location; 

We use COUNT(*) which counts all of the input rows for a group.

(COUNT() also works with expressions, but it has slightly different behavior.)

Here's how the database executes this query:

  • FROM sales — First, retrieve all of the records from the sales table
  • GROUP BY location — Next, determine the unique location groups
  • SELECT ... — Finally, select the location name and the count of the number of rows in that group

We also give this count of rows an alias using AS number_of_sales to make the output more readable. It looks like this:

 location | number_of_sales ------------+----------------- 1st Street | 2 HQ | 4 Downtown | 2 (3 rows) 

The 1st Street location has two sales, HQ has four, and Downtown has two.

Here we can see how we've taken the remaining column data from our eight independent rows and distilled them into useful summary information for each location: the number of sales.

SUM

In a similar way, instead of counting the number of rows in a group, we could sum information within the group—like the total amount of money earned from those locations.

To do this we'll use the SUM() function:

SELECT location, SUM(price) AS total_revenue FROM sales GROUP BY location; 

Instead of counting the number of rows in each group we sum the dollar amount of each sale, and this shows us the total revenue per location:

 location | total_revenue ------------+--------------- 1st Street | 5 HQ | 9 Downtown | 5 (3 rows) 

Average (AVG)

Finding the average sale price per location just means swapping out the SUM() function for the AVG() function:

SELECT location, AVG(price) AS average_revenue_per_sale FROM sales GROUP BY location; 

Working with multiple groups

So far we've been working with just one group: location.

What if we wanted to sub-divide that group even further?

Similar to the "birth countries and eye color" scenario we started with, what if we wanted to find the number of sales per product per location?

To do this all we need to do is add the second grouping condition to our GROUP BY statement:

SELECT ... FROM sales GROUP BY location, product;

By adding a second column in our GROUP BY we further sub-divide our location groups into location groups per product.

これでproduct列ごとにグループ化されたので、SELECT!で返すことができます。

ORDER BY出力を読みやすくするために、これらのクエリにいくつかの句をスローします。)

SELECT location, product FROM sales GROUP BY location, product ORDER BY location, product; 

新しいグループ化の結果を見ると、独自の場所と製品の組み合わせがわかります。

 location | product ------------+--------- 1st Street | Bagel 1st Street | Coffee Downtown | Bagel Downtown | Coffee HQ | Bagel HQ | Coffee (6 rows) 

グループができたので、残りの列データをどのように処理しますか?

さて、以前と同じ集計関数を使用して、場所ごとの製品ごとの販売数を見つけることができます。

SELECT location, product, COUNT(*) AS number_of_sales FROM sales GROUP BY location, product ORDER BY location, product; 
 location | product | number_of_sales ------------+---------+----------------- 1st Street | Bagel | 1 1st Street | Coffee | 1 Downtown | Bagel | 1 Downtown | Coffee | 1 HQ | Bagel | 2 HQ | Coffee | 2 (6 rows) 
TheReader™の演習として場所ごとの各製品の総収益(合計)を見つけます。

で関数を使用する GROUP BY

次に、1日あたりの総売上高を調べてみましょう。

If we follow a similar pattern as we did with our locations and group by our sold_at column...

SELECT sold_at, COUNT(*) AS sales_per_day FROM sales GROUP BY sold_at ORDER BY sold_at; 

...we might expect to have each group be each unique day—but instead we see this:

 sold_at | sales_per_day ----------------------------+--------------- 2020-08-29 08:42:33.085995 | 1 2020-08-29 09:42:33.085995 | 1 2020-08-30 08:42:33.085995 | 1 2020-08-30 09:42:33.085995 | 1 2020-08-31 09:42:33.085995 | 1 2020-09-01 07:42:33.085995 | 1 2020-09-01 08:42:33.085995 | 1 2020-09-01 09:42:33.085995 | 1 (8 rows) 

It looks like our data isn't grouped at all—we get each row back individually.

But, our data is actually grouped! The problem is each row's sold_at is a unique value—so every row gets its own group!

The GROUP BY is working correctly, but this is not the output we want.

The culprit is the unique hour/minute/second information of the timestamp.

Each of these timestamps differ by hours, minutes, or seconds—so they are each placed in their own group.

We need to convert each of these date and time values into just a date:

  • 2020-09-01 08:42:33.085995 =>2020-09-01
  • 2020-09-01 09:42:33.085995 =>2020-09-01

Converted to a date, all of the timestamps on the same day will return the same date value—and will therefore be placed into the same group.

To do this, we'll cast the sold_at timestamp value to a date:

SELECT sold_at::DATE AS date, COUNT(*) AS sales_per_day FROM sales GROUP BY sold_at::DATE ORDER BY sold_at::DATE; 

In our GROUP BY clause we use ::DATE to truncate the timestamp portion down to the "day." This effectively chops off the hours/minutes/seconds of the timestamp and just returns the day.

In our SELECT, we also return this same expression and give it an alias to pretty up the output.

For the same reason we couldn't return product without grouping by it or performing some kind of aggregation on it, the database won't let us return just sold_at—everything in the SELECT must either be in the GROUP BY or some kind of aggregate on the resulting groups.

The result is the sales per day that we originally wanted to see:

 date | sales_per_day ------------+--------------- 2020-08-29 | 2 2020-08-30 | 2 2020-08-31 | 1 2020-09-01 | 3 (4 rows) 

Filtering groups with HAVING

Next let's look at how to filter our grouped rows.

To do this, let's try to find days where we had more than one sale.

Without grouping, we would normally filter our rows by using a WHERE clause. For example:

SELECT * FROM sales WHERE product = 'Coffee'; 

With our groups, we may want to do something like this to filter our groups based on the count of rows...

SELECT sold_at::DATE AS date, COUNT(*) AS sales_per_day FROM sales WHERE COUNT(*) > 1 -- filter the groups? GROUP BY sold_at::DATE; 

Unfortunately, this doesn't work and we receive this error:

ERROR:  aggregate functions are not allowed in WHERE

Aggregate functions are not allowed in the WHERE clause because the WHERE clause is evaluated before the GROUP BY clause—there aren't any groups yet to perform calculations on.

But, there is a type of clause that allows us to filter, perform aggregations, and it is evaluated after the GROUP BY clause: the HAVING clause.

The HAVING clause is like a WHERE clause for your groups.

To find days where we had more than one sale, we can add a HAVING clause that checks the count of rows in the group:

SELECT sold_at::DATE AS date, COUNT(*) AS sales_per_day FROM sales GROUP BY sold_at::DATE HAVING COUNT(*) > 1; 

This HAVING clause filters out any rows where the count of rows in that group is not greater than one, and we see that in our result set:

 date | sales_per_day ------------+--------------- 2020-09-01 | 3 2020-08-29 | 2 2020-08-30 | 2 (3 rows) 

Just for the sake of completeness, here's the order of execution for all parts of a SQL statement:

  • FROM — Retrieve all of the rows from the FROM table
  • JOIN — Perform any joins
  • WHERE — Filter rows
  • GROUP BY - Form groups
  • HAVING - Filter groups
  • SELECT - Select the data to return
  • ORDER BY - Order the output rows
  • LIMIT - Return a certain number of rows

Aggregates with implicit grouping

The last topic we'll look at is aggregations that can be performed without a GROUP BY—or maybe better said they have an implicitgrouping.

These aggregations are useful in scenarios where you want to find one particular aggregate from a table—like the total amount of revenue or the greatest or least value of a column.

For example, we could find the total revenue across all locations by just selecting the sum from the entire table:

SELECT SUM(price) FROM sales; 
 sum ----- 19 (1 row) 

So far we've done $19 of sales across all locations (hooray!).

Another useful thing we could query is the first or last of something.

For example, what is the date of our first sale?

To find this we just use the MIN() function:

SELECT MIN(sold_at)::DATE AS first_sale FROM sales; 
 first_sale ------------ 2020-08-29 (1 row) 

(To find the date of the last sale just substitute MAX()for MIN().)

Using MIN / MAX

While these simple queries can be useful as a standalone query, they're often parts of filters for larger queries.

For example, let's try to find the total sales for the last day that we had sales.

One way we could write that query would be like this:

SELECT SUM(price) FROM sales WHERE sold_at::DATE = '2020-09-01'; 

This query works, but we've obviously hardcoded the date of 2020-09-01.

09/01/2020 may be the last date we had a sale, but it's not always going to be that date. We need a dynamic solution.

This can be achieved by combining this query with the MAX() function in a subquery:

SELECT SUM(price) FROM sales WHERE sold_at::DATE = ( SELECT MAX(sold_at::DATE) FROM sales ); 

In our WHERE clause we find the largest date in our table using a subquery: SELECT MAX(sold_at::DATE) FROM sales.

Then, we use this max date as the value we filter the table on, and sum the price of each sale.

Implicit grouping

I say that these are implicit groupings because if we try to select an aggregate value with a non-aggregated column like this...

SELECT SUM(price), location FROM sales; 

...we get our familiar error:

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

GROUP BY is a tool

As with many other topics in software development, GROUP BY is a tool.

There are many ways to write and re-write these queries using combinations of GROUP BY, aggregate functions, or other tools like DISTINCT, ORDER BY, and LIMIT.

Understanding and working with GROUP BY's will take a little bit of practice, but once you have it down you'll find an entirely new batch of problems are now solvable to you!

If you liked this post, you can follow me on twitter where I talk about database things and how to succeed in a career as a developer.

Thanks for reading!

John