Create the datasets

customers <- tibble(
  customer_id = c(1,2,3,4,5),
  name = c("Alice","Bob","Charlie","David","Eve"),
  city = c("New York","Los Angeles","Chicago","Houston","Phoenix")
)

orders <- tibble(
  order_id = c(101,102,103,104,105,106),
  customer_id = c(1,2,3,2,6,7),
  product = c("Laptop","Phone","Tablet","Desktop","Camera","Printer"),
  amount = c(1200,800,300,1500,600,150)
)

1. Inner Join

q1 <- inner_join(customers, orders, by = "customer_id")
q1

## # 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

a. How many rows are in the result?

nrow(q1)

## [1] 4

Answer: 4 rows.

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

The inner join only keeps rows where customer_id exists in both tables.
Customers without orders and orders without matching customers are excluded.

c. Display the result

q1

## # 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

2. Left Join

q2 <- left_join(customers, orders, by = "customer_id")
q2

## # 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

a. How many rows are in the result?

nrow(q2)

## [1] 6

Answer: 6 rows.

b. Explain why this number differs from the inner join result.

A left join keeps all rows from the customers table, even if they do not have matching orders.

c. Display the result

q2

## # 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

3. Right Join

q3 <- right_join(customers, orders, by = "customer_id")
q3

## # 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

a. How many rows are in the result?

nrow(q3)

## [1] 6

Answer: 6 rows.

b. Which customer_ids have NULL (NA) name and city? Explain why.

q3

## # 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

Customer IDs 6 and 7 have NA values because they exist in the orders table but not in the customers table.

c. Display the result

q3

## # 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

4. Full Join

q4 <- full_join(customers, orders, by = "customer_id")
q4

## # A tibble: 8 × 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
## 7           6 <NA>    <NA>             105 Camera     600
## 8           7 <NA>    <NA>             106 Printer    150

a. How many rows are in the result?

nrow(q4)

## [1] 8

Answer: 8 rows.

b. Identify any rows where there is information from only one table.

Customers with no orders have NA for order information, and orders with customer IDs not in the customers table have NA for customer information.

c. Display the result

q4

## # A tibble: 8 × 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
## 7           6 <NA>    <NA>             105 Camera     600
## 8           7 <NA>    <NA>             106 Printer    150

5. Semi Join

q5 <- semi_join(customers, orders, by = "customer_id")
q5

## # 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. How many rows are in the result?

nrow(q5)

## [1] 3

Answer: 3 rows.

b. How does this result differ from the inner join result?

A semi join returns only rows from the customers table that have matching orders, but it does not include columns from the orders table.

c. Display the result

q5

## # 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

6. Anti Join

q6 <- anti_join(customers, orders, by = "customer_id")
q6

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

a. Which customers are in the result?

q6

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

Answer: David and Eve.

b. Explain what this result tells you about these customers.

These customers appear in the customers table but do not appear in the orders table, meaning they have not placed any orders.

c. Display the result

q6

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

7. Practical Application

a. Which join would you use to find all customers, including those who have not placed orders?

Left join.

b. Which join would you use to find only customers who have placed orders?

Inner join.

c. Write the R code for both scenarios

all_customers <- left_join(customers, orders, by = "customer_id")
customers_with_orders <- inner_join(customers, orders, by = "customer_id")

d. Display the result

all_customers

## # 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

customers_with_orders

## # 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

8. Challenge Question

Create a summary showing each customer’s name, city, total number of orders, and total amount spent.

q8 <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarise(
    total_orders = sum(!is.na(order_id)),
    total_spent = sum(amount, na.rm = TRUE),
    .groups = "drop"
  )

q8

## # 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

This table shows each customer, their number of orders, and total amount spent. Customers without orders are still included.

Assignment 4

Aidan Magner

2026-02-27

Create the datasets

1. Inner Join

a. How many rows are in the result?

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

c. Display the result

2. Left Join

a. How many rows are in the result?

b. Explain why this number differs from the inner join result.

c. Display the result

3. Right Join

a. How many rows are in the result?

b. Which customer_ids have NULL (NA) name and city? Explain why.

c. Display the result

4. Full Join

a. How many rows are in the result?

b. Identify any rows where there is information from only one table.

c. Display the result

5. Semi Join

a. How many rows are in the result?

b. How does this result differ from the inner join result?

c. Display the result

6. Anti Join

a. Which customers are in the result?

b. Explain what this result tells you about these customers.

c. Display the result

7. Practical Application

a. Which join would you use to find all customers, including those who have not placed orders?

b. Which join would you use to find only customers who have placed orders?

c. Write the R code for both scenarios

d. Display the result

8. Challenge Question