Inner Join (3 points) Perform an inner join between the customers and orders datasets.

Q1 <- inner_join(customers, orders)

## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 4 rows.

Why are some customers or orders not included in the result?

Because the customers and orders that are not included did not have a match in the other table.

Display the result

## # A tibble: 4 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300

Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.

Q2 <- left_join(customers , orders)

## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 6 rows.

Explain why this number differs from the inner join result.

Because the left join keeps all customers and the customers without orders appear with NA values.

Display the result

## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA

Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

Q3 <- right_join(customers , orders)

## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 6 rows.

Which customer_ids in the result have NULL for customer name and city? Explain why.

Orders exist for 6 and 7, but no matching customers because right join keeps all rows from orders.

Display the result

## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           6 <NA>    <NA>             105 Camera     600
## 6           7 <NA>    <NA>             106 Printer    150

Full Join (3 points) Perform a full join between customers and orders.

Q4 <- full_join(customers , orders)

## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 8 rows.

Identify any rows where there’s information from only one table. Explain these results.

The full join returns 8 rows because it includes all customers and all orders, with customers 4 and 5 having no orders and orders 6 and 7 having no customers.

Display the result

## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA

Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

Q5 <- semi_join(customers , orders)

## Joining with `by = join_by(customer_id)`

How many rows are in the result?

The resulting data has 3 rows.

How does this result differ from the inner join result?

The semi join returns 3 rows because it includes only customers 1,2, and 3 who have at least one order and does not include order details like the inner join.

Display the result

## # A tibble: 3 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago

Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

Q6 <- anti_join(customers , orders)

## Joining with `by = join_by(customer_id)`

Which customers are in the result?

The anti join returns customers 4 (David) and 5(Eve), showing that these customers have not placed any orders.

Explain what this result tells you about these customers.

The result tells us that David and Eve do not have any matching records in the orders table, meaning that they have not placed any orders.

Display the result

## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix

Practical Application (4 points) Imagine you’re analyzing customer behavior.

Which join would you use to find all customers, including those who haven’t placed any orders? Why?

I would used a left join because it keeps all customers even if they do not have matching orders.

Which join would you use to find only the customers who have placed orders? Why?

I would use a semi join because it returns only customers who have at least one matching order.

Write the R code for both scenarios.

Q7 <- left_join(customers, orders)

## Joining with `by = join_by(customer_id)`

Q7_orders_only <- semi_join(customers, orders, by = "customer_id")

Display the result

## # A tibble: 3 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago

## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA

Challenge Question (3 points) Create a summary that shows each customer’s name, city, total number of orders, and total amount spent. Include all customers, even those without orders. Hint: You’ll need to use a combination of joins and group_by/summarize operations.

## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_spent
##         <dbl> <chr>   <chr>              <int>       <dbl>
## 1           1 Alice   New York               1        1200
## 2           2 Bob     Los Angeles            2        2300
## 3           3 Charlie Chicago                1         300
## 4           4 David   Houston                0           0
## 5           5 Eve     Phoenix                0           0

Assignment 4

Ryan Armstrong

2026-02-17

Inner Join (3 points) Perform an inner join between the customers and orders datasets.

How many rows are in the result?

Why are some customers or orders not included in the result?

Display the result

Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.

How many rows are in the result?

Explain why this number differs from the inner join result.

Display the result

Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

How many rows are in the result?

Which customer_ids in the result have NULL for customer name and city? Explain why.

Display the result

Full Join (3 points) Perform a full join between customers and orders.

How many rows are in the result?

Identify any rows where there’s information from only one table. Explain these results.

Display the result

Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

How many rows are in the result?

How does this result differ from the inner join result?

Display the result

Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

Which customers are in the result?

Explain what this result tells you about these customers.

Display the result

Practical Application (4 points) Imagine you’re analyzing customer behavior.

Which join would you use to find all customers, including those who haven’t placed any orders? Why?

Which join would you use to find only the customers who have placed orders? Why?

Write the R code for both scenarios.

Display the result

Challenge Question (3 points) Create a summary that shows each customer’s name, city, total number of orders, and total amount spent. Include all customers, even those without orders. Hint: You’ll need to use a combination of joins and group_by/summarize operations.